Split (1)

train · 15k rows

text stringlengths 235 3.08k
8=23 16=24 32=25 64=26 Table 10.1: Cell growth data. 134 CHAPTER 10. EXPONENTIAL FUNCTIONS that the population of cells will double every hour, then reasoning as above will lead us to conclude that the formula N(t) = (3 106)2t × gives the population of cells after t hours. Now, as long as t represents a non-negative integer, we know how to calculate N(t). For example, if t = 6, then N(t) = (3 = (3 × × 106)26 106)(2 106)64 106. = (3 × = 192 × ) The key point is that computing N(t) only involves simple arithmetic. But what happens if we want to know the population of cells after 6.37 hours? That would require that we work with the formula N(t) = (3 106)26.37 × and the rules of arithmetic do not sufﬁce to calculate N(t). We are stuck, since we must understand the meaning of an expression like 26.37. In order to proceed, we will need to review the algebra required to make sense of raising a number (such as 2) to a non-integer power. We need to understand the precise meaning of expressions like: 26.37, 2√5, 2−π, etc. 10.1 Functions of Exponential Type this is a variable x y = b this is a fixed positive integer b y = x this is a fixed number this is a variable Exponential Picture Monomial Picture Figure 10.2: Viewing the difference between exponential and monomial functions. On a symbolic level, the class of functions we are trying to motivate is easily introduced. We have already studied the monomials y = xb, where x was our input variable and b was a ﬁxed positive integer exponent. What happens if we turn this around, interchanging x and b, deﬁning a new rule: y = f(x) = bx. (10.1) 10.1. FUNCTIONS OF EXPONENTIAL TYPE 135 We refer to x as the power and b the base. An expression of this sort is called a function of exponential type. Actually, if your algebra is a bit rusty, it is easy to initially confuse functions of exponential type and monomials (see Figure 10.2). 10.1.1 Reviewing
the Rules of Exponents To be completely honest, making sense of the expression y = bx for all numbers x requires the tools of Calculus, but it is possible to establish a reasonable comfort level 0 by handling the case when x is a rational number. If b and n is a positive integer (i.e. n = 1, 2, 3, 4,... ), then we can try to solve the equation ≥ tn = b. (10.2) A solution t to this equation is called an nth root of b. This leads to complications, depending on whether n is even or odd. In the odd case, for any real number b, notice that the graph of y = b will always cross the graph of y = tn exactly once, leading to one solution of (10.2). On the other hand, if n is even and b < 0, then the graph of y = tn will miss the graph of y = b, implying there are no solutions to the equation in (10.2). (There will be complex solutions to equations such as t2 = −1, involv√−1, but we ing the imaginary complex numbers are only working with real numbers in this course.) Also, again in the case when n is even, it can happen that there are two solutions to (10.2). We do not want to constantly worry about this even/odd distinction, so we will henceforth assume b > 0. To eliminate possible ambiguity, we will single out a particular nth-root; we deﬁne the symbols1.5 -1 -0.5 0 0.5 1 1.5 -1 -2 -3 3 2 1 0 n=2 n=4 n=6 n=3 n=5 n=7 -1.5 -1 -0.5 0 0.5 1 1.5 -1 -2 -3 Figure 10.3: Even and odd monomials. n√b = b 1 n = the largest real nth root of b. (10.3) ± Thus, whereas 1 are both 4th-roots of 1, we have deﬁned 4√1 = 1. In order to manipulate y = bx for rational x, we need to recall some basic facts from algebra. Important Facts 10.1.1 (Working with rational exponents). For all positive integers p and q, and any real number base b > 0
, we have p q = b q√b p = q√bp. For any rational numbers r and s, and for all positive bases a and b: 136 CHAPTER 10. EXPONENTIAL FUNCTIONS y-axis n y=t n even = solution y-axis n y=t n odd y=b t-axis y=b* no solution or two solutions y=b t-axis exactly one solution Figure 10.4: How many solutions to tn = b?. 1. Product of power rule: brbs = br+s 2. Power of power rule: (br)s = brs 3. Power of product rule: (ab)r = arbr 4. Zero exponent rule: b0 = 1 5. Negative power rule: b−r = 1 br These rules have two important consequences, one theoretical and the other more practical. On the ﬁrst count, recall that any rational number r can be written in the form r = p q, where p and q are integers. Consequently, using these rules, we see that the expression y = bx deﬁnes a function of x, whenever x is a rational number. On the more practical side of things, using the rules we can calculate and manipulate certain expressions. For example, 27 2 3 = 3√27 2 = 32 = 9; 8− 5 3 = 3√8 −5 = 2−5 = 1 25 = 1 32. The sticky point which remains is knowing that f(x) = bx actually deﬁnes a function for all real values of x. This is not easy to verify and we are simply going to accept it as a fact. The difﬁculty is that we need the fundamentally new concept of a limit, which is the starting point of a Calculus course. Once we know the expression does deﬁne a function, we can also verify that the rules of Fact 10.1.1 carry through for all real 10.2. THE FUNCTIONS Y = A0BX 137 exponent powers. Your calculator should have a “y to the x key”, allowing you to calculate expressions such as π√2 involving non-rational powers. Here are the key modeling functions we will work with in this Chapter. Deﬁnition 10.1.2. A function of exponential type has the form A(x) = A0
bx, for some b > 0, b = 1, and A0 6 = 0. We will refer to the formula in Deﬁnition 10.1.2 as the standard exponential form. Just as with standard forms for quadratic functions, we sometimes need to do a little calculation to put an equation in standard form. The constant A0 is called the initial value of the exponential function; this is because if x represents time, then A(0) = A0b0 = A0 is the value of the function at time x = 0; i.e. the initial value of the function. Example 10.1.3. Write the equations y = 83x and y = 7 exponential form. 2x−1 in standard 1 2 Solution. In both cases, we just use the rules of exponents to maneuver the given equation into standard form: and y = 83x = (83)x = 512x 2x−1 2x 1 2 1 2 x 2 −1 2! x 1 4 = 14 10.2 The Functions y = A0bx We know f(x) = 2x deﬁnes a function of x, so we can study basic qualitative features of its graph. The data assembled in the solution of the “Doubling Effect” beginning this Chapter, plus the rules of exponents, produce a number of points on the graph. This graph exhibits four key qualitative features that deserve mention: 6 138 CHAPTER 10. EXPONENTIAL FUNCTIONS x 2x...... -2 1/4 -1 1/2 0 1 2 3... 1 2 4 8... Point on the graph of y = 2x... (-2, 1/4) (-1, 1/2) (0, 1) (1, 2) (2, 4) (3, 8)... x y = 2 (3,8) (−1,1/2) (−2,1/4) (2,4) (1,2) (0,1) −1 1 (a) Data points from y = 2x. (b) Graph of y = 2x. Figure 10.5: Visualizing y = 2x. • • • • The graph is always above the horizontal axis; i.e. values are always positive. the function The graph has y-intercept 1 and is increasing. The graph becomes closer and closer to the horizontal axis as we the x-axis
is a horizontal asymptote for the leftmove left; i.e. hand portion of the graph. The graph becomes higher and higher above the horizontal axis as we move to the right; i.e., the graph is unbounded as we move to the right. The special case of y = 2x is representative of the function y = bx, but there are a few subtle points that need to be addressed. First, recall we are always assuming that our base b > 0. We will consider three separate cases: b = 1, b > 1, and 0 < b < 1. 10.2.1 The case b = 1 In the case b = 1, we are working with the function y = 1x = 1; this is not too exciting, since the graph is just a horizontal line. We will ignore this case. 10.2.2 The case b > 1 If b > 1, the graph of the function y = bx is qualitatively similar to the situation for b = 2, which we just considered. The only difference is the exact amount of “concavity” in the graph, but the four features highlighted above are still valid. Figure 10.6(a) indicates how these graphs 10.2. THE FUNCTIONS Y = A0BX 139 compare for three different values of b. Functions of this type exhibit what is typically referred to as exponential growth; this codiﬁes the fact that the function values grow rapidly as we move to the right along the x-axis. y-axis y-axis all graphs pass through (0,1) x-axis all graphs pass through (0,1) x-axis (a) Graph of y = bx, b > 1. (b) Graph of y = bx, 0 < b < 1. Figure 10.6: Visualizing cases for b. 10.2.3 The case 0 < b < 1 We can understand the remaining case 0 < b < 1, by using the remarks above and our work in Chapter 13. First, with this condition on b, notice that 1 b > 1, so the graph of y = is of the type in Figure 10.6(a). Now, using the rules of exponents: 1 b x y = −x = 1 b 1 b x −1! = bx. −x 1 b By the reﬂection principle, the graph of y = is obtained by re
ﬂecting the graph of y = ( 1 b)x about the y-axis. Putting these remarks to- gether, if 0 < b < 1, we conclude that the graph of y = bx will look like Figure 10.6(b). Notice, the graphs in Figure 10.6(b) share qualitative features, mirroring the features outlined previously, with the “asymptote” and “unbounded” portions of the graph interchanged. Graphs of this sort are often said to exhibit exponential decay, in the sense that the function values rapidly approach zero as we move to the right along the x-axis. Important Facts 10.2.1 (Features of Exponential Type Functions). Let b be a positive real number, not equal to 1. The graph of y = bx has these four properties: 1. The graph is always above the horizontal axis. 2. The graph has y-intercept 1. 3. If b > 1 (resp. 0 < b < 1), the graph becomes closer and closer to the horizontal axis as we move to the left (resp. move to the right); this says the x-axis is a horizontal asymptote for the left-hand portion of the graph (resp. right-hand portion of the graph). 140 CHAPTER 10. EXPONENTIAL FUNCTIONS 4. If b > 1 (resp. 0 < b < 1), the graph becomes higher and higher above the horizontal axis as we move to the right (resp. move to the left); this says that the graph is unbounded as we move to the right (resp. move to the left). If A0 > 0, the graph of the function y = A0bx is a vertically expanded or compressed version of the graph of y = bx. If A0 < 0, we additionally reﬂect about the x-axis. 10.3 Piano Frequency Range A sound wave will cause your eardrum to move back and forth. In the case of a so-called pure tone, this motion is modeled by a function of the form d(t) = A sin(2πft), where f is called the frequency, in units of “periods/unit time”, called “Hertz” and abbreviated “Hz”. The coefﬁcient A is related to the actual displacement of the eardrum, which is
, in turn, related to the loudness of the sound. A person can typically perceive sounds ranging from 20 Hz to 20,000 Hz. A# C# D# F# G# A# C# D# F# G#A# C#D# F# G# A# C# D# F# G#A# C# D# F# G# A# C# D# F# G# A# C# D# F# G#A# A B DC 220 Hz middle C Figure 10.7: A piano keyboard. A piano keyboard layout is shown in Figure 10.7. The white keys are labelled A, B, C, D, E, F, and G, with the sequence running from left to right and repeating for the length of the keyboard. The black keys ﬁt into this sequence as “sharps”, so that the black key between A and B is “A sharp”, denoted A#. Thus, starting at any A key, the 12 keys to the right are A, A#, B, C, C#, D, D#, E, F, F#, G, and G#. The sequence then repeats. Notice that between some adjacent pairs of white keys there is no black key. A piano keyboard is commonly tuned according to a rule requiring that each key (white and black) has a frequency 21/12 times the frequency of the key to its immediate left. This makes the ratio of adjacent keys always the same (21/12), and it means that keys 12 keys apart have a ratio of frequencies exactly equal to 2 (since (21/12)12 = 2). Two such keys are 10.3. PIANO FREQUENCY RANGE 141 said to be an octave apart. Assuming that the key A below middle C has a frequency of 220 Hz, we can determine the frequency of every key on the keyboard. For instance, the A# to the right of this key has frequency 233.08188Hz. The B to the right of this 220 key has frequency 233.08188 1.059463094... 246.94165Hz. 21/12 = 220 ≈ 21/12 × × × ≈ 142 CHAPTER 10. EXPONENTIAL FUNCTIONS 10.4 Exercises Problem 10.1. Let’s brush up on the required calculator skills. Use a calculator to approximate: (a
) 3π (b) 42+√5 (c) ππ (d) 5−√3 2 (e) 3π (f) √11π−7 Problem 10.2. Put each equation in standard exponential form: (a) y = 3(2−x) (b) y = 4−x/2 (c) y = ππx (d) y = 1 1 3 3+ x 2 (e) y = 5 0.3452x−7 (f) y = 4(0.0003467)−0.4x+2 (d) Anja, a third member of your lab working with the same yeast cells, took these 106 cells after two measurements: 7.246 106 cells after 6 hours. 4 hours; 16.504 Should you be worried by Anja’s results? If Anja’s measurements are correct, does your model over estimate or under estimate the number of yeast cells at time t? × × Problem 10.4. middle C. (a) Find the frequency of (b) Find the frequency of A above middle C. (c) What is the frequency of the lowest note on the keyboard? Is there a way to solve this without simply computing the frequency of every key below A220? (d) The Bosendorfer piano is famous, due in part, to the fact it includes additional keys at the left hand end of the keyboard, extending to the C below the bottom A on a standard keyboard. What is the lowest frequency produced by a Bosendorfer? Problem 10.3. A colony of yeast cells is estimated to contain 106 cells at time t = 0. After collecting experimental data in the lab, you decide that the total population of cells at time t hours is given by the function y = 106e0.495105t. Problem 10.5. You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number of dimes. On the ﬁrst square you place one dime. On the second square you stack 2 dimes. Then you continue, always doubling the number from the previous square. (a) How many cells are present after one hour? (a) How many dimes will you have stacked on the 10th square? (b) (True or False) The population of yeast cells will double every 1.4 hours. (b) How many
dimes will you have stacked on the nth square? (c) Cherie, another member of your lab, looks at your notebook and says :...that formula is wrong, my calculations predict the formula for the number of yeast cells is given by the function (c) How many dimes will you have stacked on the 64th square? (d) Assuming a dime is 1 mm thick, how high will this last pile be? y = 106(2.042727)0.693147t. Should you be worried by Cherie’s remark? (e) The distance from the earth to the sun is approximately 150 million km. Relate the height of the last pile of dimes to this distance. 10.4. EXERCISES 143 63 64 calculates the fraction of hemoglobin saturated with oxygen at a given pressure p. (a) The graphs of M(p) and H(p) are given 100; which p below on the domain 0 is which? ≤ ≤ 1 2 3 10 9 8 Problem 10.6. Myoglobin and hemoglobin are oxygen carrying molecules in the human body. Hemoglobin is found inside red blood cells, which ﬂow from the lungs to the muscles through the bloodstream. Myoglobin is found in muscle cells. The function p 1 + p Y = M(p) = calculates the fraction of myoglobin saturated with oxygen at a given pressure p torrs. For example, at a pressure of 1 torr, M(1) = 0.5, which means half of the myoglobin (i.e. 50%) is oxygen saturated. (Note: More precisely, you need to use something called the “partial pressure”, but the distinction is not important for this problem.) Likewise, the function Y = H(p) = p2.8 262.8 + p2.8 fraction 1 0.8 0.6 0.4 0.2 20 40 60 80 p 100 (b) If the pressure in the lungs is 100 torrs, what is the level of oxygen saturation of the hemoglobin in the lungs? (c) The pressure in an active muscle is 20 torrs. What is the level of oxygen saturation of myoglobin in an active muscle? What is the level of hemoglobin in an active muscle? (d) Deﬁne the efﬁciency of oxygen transport at a given pressure p to be M(p) − H(p). What is
the oxygen transport efﬁciency at 20 torrs? At 40 torrs? At 60 torrs? Sketch the graph of M(p) − H(p); are there conditions under which transport efﬁciency is maximized (explain)? 144 CHAPTER 10. EXPONENTIAL FUNCTIONS Chapter 11 Exponential Modeling Example 11.0.1. A computer industry spokesperson has predicted that the number of subscribers to geton.com, an internet provider, will grow exponentially for the ﬁrst 5 years. Assume this person is correct. If geton.com has 100,000 subscribers after 6 months and 750,000 subscribers after 12 months, how many subscribers will there be after 5 years? Solution. The solution to this problem offers a template for many exponential modeling applications. Since, we are assuming that the number of subscribers N(x), where x represents years, is a function of exponential type, N(x) = N bx, ◦ for some N values of N(x): ◦ and b > 1. We are given two pieces of information about the N(0.5) = 100,000; i.e., N0b0.5 = 100,000, and N(1) = 750,000; i.e., N0b = 750,000. We can use these two equations to solve for the two unknowns N as follows: If we divide the second equation by the ﬁrst, we get ◦ and b = 7.5 b1 b0.5 b0.5 = b1/2 = √b = 7.5 ∴ b = 56.25. Plugging this value of b into either equation (say the ﬁrst one), we can = 100,000 (56.25)0.5 = 13,333. We conclude that the number of solve for N geton.com subscribers will be predicted by : N ◦ ◦ N(x) = 13,333(56.25)x. In ﬁve years, we obtain N(5) = 7,508,300,000,000 subscribers, which exceeds the population of the Earth (which is between 5 and 6 billion)! 145 146 Q CHAPTER 11. EXPONENTIAL MODELING There are two important conclusions we can draw from this problem. First, the given information provides us with two points on the graph
of the function N(x): P 0.2 0.4 0.6 0.8 1 P = (0.5, 100,000) Q = (1, 750,000). 800000 600000 400000 200000 Figure 11.1: Finding the equation for N(x) = N0bx. More importantly, this example illustrates a very important principal we can use when modeling with functions of exponential type. Important Fact 11.0.2. A function of exponential type can be determined if we are given two data points on its graph.!!! CAUTION!!! When you use the above strategy to ﬁnd the base b of the exponential model, make sure to write down a lengthy decimal approximation. As a rule of thumb, go for twice as many signiﬁcant digits as you are otherwise using in the problem. 11.1 The Method of Compound Interest You walk into a Bank with P0 dollars (usually called principal), wishing to invest the money in a savings account. You expect to be rewarded by the Bank and paid interest, so how do you compute the total value of the account after t years? The future value of the account is really a function of the number of years t elapsed, so we can write this as a function P(t). Our goal is to see that P(t) is a function of exponential type. In order to compute the future value of the account, the Bank provides any savings account investor with two important pieces of information: r = annual (decimal) interest rate n = the number of compounding periods per year The number n tells us how many times each year the Bank will compute the total value P(t) of the account. For example, if n = 1, the calculation is done at one-year intervals; if n = 12, the calculation is done each month, etc. The bank will compute the value of your account after a typical compounding period by using the periodic rate of return r n. For example, if the interest rate percentage is 12% and the compounding period is monthly (i.e., n = 12), then the annual (decimal) interest rate is 0.12 and the periodic rate is 0.12 12 = 0.01.!!! CAUTION!!! The number r always represents the decimal interest rate, which is a decimal between 0 and 1. If you are given the interest rate percentage (which is a positive number between 0 and 100), you
need to convert to a decimal by dividing by 100. 11.1. THE METHOD OF COMPOUND INTEREST 147 11.1.1 Two Examples Let’s consider an example: P0 = $1,000 invested at the annual interest percentage of 8% compounded yearly, so n = 1 and r = 0.08. To compute the value P(1) after one year, we will have P(1) = P0 + (periodic rate)P0 = P0 + rP0 = P0(1 + r) = $1,000(1 + 0.08) = $1,080. To compute the value after two years, we need to apply the periodic rate to the value of the account after one year: P(2) = P(1) + (periodic rate)P(1) = P0(1 + r) + rP0(1 + r) = P0(1 + r)2 = $1,000(1 + 0.08)2 = $1,166.40. Notice, the amount the Bank has paid after two years is $166.40, which is slightly bigger than twice the $80 paid after one year. To compute the value after three years, we need to apply the periodic rate to the value of the account after two years: P(3) = P(2) + (periodic rate)P(2) = P0(1 + r)2 + rP0(1 + r)2 = P0(1 + r)3 = $1,000(1 + 0.08)3 = $1,259.71. Again, notice the amount the Bank is paying after three years is $259.71, which is slightly larger than three times the $80 paid after one year. Continuing on in this way, to ﬁnd the value after t years, we arrive at the formula P(t) = P0(1 + r)t = $1,000(1.08)t. In particular, after 5 and 10 years, the value of the account (to the nearest dollar) will be $1,469 and $2,159, respectively. As a second example, suppose we begin with the same $1,000 and the same annual interest percentage 8%, but now compound monthly, so n = 12 and r = 0.08. The value of the account after one compounding period is
P(1/12), since a month is one-twelfth of a year. Arguing as before, paying special attention that the periodic rate is now r 12, we have n = 0.08 P(1/12) = P0 + (periodic rate)P0 = P0 1 +.08 12 = $1,000(1 + 0.006667) = $1,006.67. 148 CHAPTER 11. EXPONENTIAL MODELING After two compounding periods, the value is P(2/12), P(2/12) = P(1/12) + (periodic rate)P(1/12) = P0 1 + = P0 1 +.08 12.08 12 0.08 12 P0 1 +.08 12 + 2 = $1,000(1 + 0.006667)2 = $1,013.38. Continuing on in this way, after k compounding periods have elapsed, the value will be P, which is computed as k 12 P(k/12) = P0 1 +.08 12 k. It is possible to rewrite this formula to give us the value after t years, noting that t years will lead to 12t compounding periods; i.e., set k = 12t in the previous formula: P(t) = P0 1 +.08 12 12t For example, after 1, 5 and 10 years, the value of the account, to the nearest dollar, would be $1,083, $1,490, and $2,220. 11.1.2 Discrete Compounding The two examples above highlight a general formula for computing the future value of an account. Important Fact 11.1.1 (Discrete compounding). Suppose an account is opened with P0 principal. If the decimal interest rate is r and the number of compounding periods per year is n, then the value P(t) of the account after t years will be P(t) = P0 1 + nt. r n Notice, the future value P(t) is a function of exponential type; the base, which will be greater than one. Since P0 > 0, the is the number graph will be qualitatively similar to the ones pictured in Figure 10.6(a). 1 + r n Example 11.1.2. At birth, your Uncle Hans secretly purchased a $5,000 U.S
. Savings Bond for $2,500. The conditions of the bond state that the U.S. Government will pay a minimum annual interest rate of r = 8.75%, compounded quarterly. Your Uncle has given you the bond as a gift, subject to the condition that you cash the bond at age 35 and buy a red Porsche. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 149 On your way to the Dealer, you receive a call from your tax accountant informing you of a 28% tax on the capital gain you realize through cashing in the bond; the capital gain is the selling price of the bond minus the purchase price. Before stepping onto the showroom ﬂoor, compute how much cash will you have on hand, after the U.S. Government shares in your proﬁts. Solution. The value of your bond after 35 years is computed by the formula in Fact 11.1.1, using P0 = $2,500, r = 0.0875, n = 4, and t = 35. Plugging this all in, we ﬁnd that the selling price of the bond is P(35) = $2,500 1 + 0.0875 4 4(35) = $51,716.42. The capital gain will be $51,716.42 - $2,500 = $49,216.42 and the tax due is $(49,216.42)(0.28) = $13,780.60. You are left with $51,716.42 $13,780.60 = $37,935.82. Better make that a used Porsche! 11.2 The Number e and the Exponential Function What happens to the future value of an investment of P0 dollars as the number of compounding periods is increased? For example, return to our earlier example: P0 = $1000 and an annual interest percentage of 8%. After 1 year, the table below indicates the value of the investment for various compounding periods: yearly, quarterly, monthly, weekly, daily, and hourly. n Compounding Period Value after 1 year (to nearest dollar) 1 yearly $1,000(1 + 0.08)1 = $1,080.00 4 quarterly $1,000 1 + 0.08 4 4 = $1,082.43 12 monthly $1,000 52 weekly $1,000 365 daily $1,000
1 + 0.08 12 12 1 + 0.08 52 52 1 + 0.08 365 365 = $1,083.00 = $1,083.22 = $1,083.28 8,760 hourly $1,000 1 + 0.08 8,760 8,760 =$1,083.29 150 CHAPTER 11. EXPONENTIAL MODELING We could continue on, considering “minute” and “second” compounding and what we will ﬁnd is that the value will be at most $1,083.29. This illustrates a general principal: Important Fact 11.2.1. Initially increasing the number of compounding periods makes a signiﬁcant difference in the future value; however, eventually there appears to be a limiting value. Let’s see if we can understand mathematically why this is happening. The ﬁrst step is to recall the discrete compounding formula: P(t) = P0 1 + nt. r n If our desire is to study the effect of increasing the number of compounding periods, this means we want to see what happens to this formula as n gets BIG. To analyze this, it is best to rewrite the expression using a substitution trick: Set z = n z. Plugging in, we have r, so that n = rz and r n = 1 P(t) = P0 1 + nt rzt r n 1 z 1 z 1 + 1 + = P0 = P0 z rt. (11.1) z 1 + 1 z So, since r is a ﬁxed number and z = n r, letting n get BIG is the same as letting z become BIG in (11.1). This all means we need to answer as z becomes this new question: What happens to the expression large? On the one hand, the power in the expression is getting large; at the same time, the base is getting close to 1. This makes it very tricky to make quick predictions about the outcome. It is best to ﬁrst tabulate 1 + 1 some numerical data for the values of y = g(z) = and look at a z 100: See Figure 11.2. z plot of this function graph on the domain 0.01 ≤ ≤ z approaches the 1 + 1 You can see from this plot, the graph of y = z “dashed”
horizontal asymptote, as z becomes BIG. We will let the letter “e” represent the spot where this horizontal line crosses the vertical axis and 2.7182818. This number is only an approximation, since e is known to e be an irrational number. What sets this irrational number apart from the ones you are familiar with (e.g. √2, π, etc.) is that deﬁning the number e requires a “limiting” process. This will be studied a lot more in your Calculus course. The new number e is a positive number greater than 1, so we can study the function: ≈ z y = ex. (11.2) Since e > 1, the graph will share the properties in Figure 10.6(a). This function is usually referred to as THE exponential function. Scientiﬁc calculators will have a key of the form “ exp(x) ” or “ ex ”. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 151 z z 1 + 1 z 2 1 2.25 2 2.37037 3 2.4414 4 2.65329 20 100 2.70481 1000 2.71692 109 2.71828 (a) Data points for 1 + 1 z. z 2.5 2 1.5 1 0.5 20 40 60 80 100 (b) The graph of 1 + 1 z z. Figure 11.2: What happens when z get very large? 11.2.1 Calculator drill Plugging in x = 1, you can compute an approximation to e on your calculator; you should get e = 2.7183 to four decimal places. Make sure you can compute expressions like e3, eπ, and √e = e1/2; to four decimal places, you should get 20.0855, 23.1407, and 1.6487. 11.2.2 Back to the original problem... We can now return to our future value formula (11.1) and conclude that as the number of compounding periods increases, the future value is approaching a limiting value: P(t) = P0 1 + 1 z z rt = P0ert. The right hand limiting formula Q(t) = P0ert computes the future value using what is usually referred to as continuous compounding. From the investors viewpoint, this is the best
possible scheme for computing future value. ⇒ Important Fact 11.2.2 (Continuous compounding). The future value of P0 dollars principal invested at an annual decimal interest rate of r under continuous compounding after t years is Q(t) = P0ert; this value is alnt, for any discrete compounding 1 + r ways greater than the value of P0 n scheme. In fact, P0ert is the limiting value. 152 CHAPTER 11. EXPONENTIAL MODELING 11.3 Exercises Problem 11.1. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wage was $2.30 per hour. Assume the minimum wage grows according to an exponential model w(t), where t represents the time in years after 1960. (a) Find a formula for w(t). (b) What does the model predict for the min- imum wage in 1960? (c) If the minimum wage was $5.15 in 1996, is this above, below or equal to what the model predicts. Problem 11.2. The Pinedale, Wyoming, is experiencing a population boom. In 1990, the population was 860 and ﬁve years later it was 1210. town of (a) Find a linear model l(x) and an exponential model p(x) for the population of Pinedale in the year 1990+x. (b) What do these models estimate the population of Pinedale to be in the year 2000? Problem 11.3. In 1989, research scientists published a model for predicting the cumulative number of AIDS cases reported in the United States: a(t) = 155 t − 1980 10 3, (thousands) where t is the year. This paper was considered a “relief”, since there was a fear the correct model would be of exponential type. Pick two data points predicted by the research model a(t) to construct a new exponential model b(t) for the number of cumulative AIDS cases. Discuss how the two models differ and explain the use of the word “relief”. Problem 11.4. Deﬁne two new functions: and y = cosh(x) = y = sinh(x) = ex + e−x 2 ex − e−x 2. These are called the basic hyperbolic trigonometric functions. (a) Sketch rough
graphs of these two func- tions. (b) The graph of the equation x2 − y2 = 1 is shown below; this is called the unit hyperbola. For any value a, show that the point (x,y) = (cosh(a), sinh(a)) is on the unit hyperbola. (Hint: Verify that [cosh(x)]2 − [sinh(x)]2 = 1, for all x.) y (−1,0) (1,0) x (c) A hanging cable is modeled by a portion of the graph of the function y = a cosh( x − h a ) + C, for appropriate constants a, h and C. The constant h depends on how the coordinate system is imposed. A cable for a suspension bridge hangs from two 100 ft. high towers located 400 ft. apart. Impose a coordinate system so that the picture is symmetric about the y-axis and the roadway coincides with the x-axis. The hanging cable constant is a = 500 and h = 0. Find the minimum distance from the cable to the road. towers cable d 400 ft 100 ft roadway Chapter 12 Logarithmic Functions If we invest P0 = $1,000 at an annual rate of r = 8% compounded continuously, how long will it take for the account to have a value of $5000? The formula P(t) = 1,000e0.08t gives the value after t years, so we need to solve the equation: 5,000 = 1,000e0.08t 5 = e0.08t. Unfortunately, algebraic manipulation will not lead to a further simpliﬁcation of this equation; we are stuck! The required technique involves the theory of inverse functions. Assuming we can ﬁnd the inverse function of f(t) = et, we can apply f−1(t) to each side of the equation and solve for t: f−1(5) = f−1(e0.08t) = 0.08t (12.5)f−1(5) = t The goal in this section is to describe the function f−1, which is usually denoted by the symbol f−1(t) = ln(t) and called the natural logarithm function. On your calculator, you will ﬁnd a button dedicated to this function and we can now
compute ln(5) = 1.60944. Conclude that the solution is t = 20.12 years. 12.1 The Inverse Function of y = ex If we sketch a picture of the exponential function on the domain of all real numbers and keep in mind the properties in Fact 10.2.1, then every horizontal line above the x-axis intersects the graph of y = ex exactly once: See Figure 12.1(a). The range of the exponential function will consist of all possible y-coordinates of points on the graph. Using the graphical techniques of Chapter 6, we can see that the range of will be all POSITIVE real numbers: See Figure 12.1(b). 153 154 CHAPTER 12. LOGARITHMIC FUNCTIONS PSfrag x y = e range = all positive numbers y = ex graphs of y=c, c>1 cross exponential graph exactly once. these horizontal lines miss graph of exponential function. −1 1 −1 1 domain = all real numbers (a) Horizontal line test for y = ex. (b) The domain and range for y = ex. Figure 12.1: Properties needed to ﬁnd the inverse of f(x) = ex. By the horizontal line test, this means the exponential function is one- to-one and the inverse rule f−1(c) will deﬁne a function f−1(c) =    reflecting line y = x x y = e y=ln (x) −1 1 Figure 12.2: Visualizing the y = ln(x). soluthe unique tion of the equation c = ex!, if c > 0 (undeﬁned), if c 0. ≤ (12.1) This inverse function is called the natural logarithm function, denoted ln(c). We can sketch the graph of the the natural logarithm as follows: First, by Fact 9.2.1, the domain of the function ln(y) = x is just the range of the exponential function, which we noted is all positive numbers. Likewise, the range of the function ln(y) = x is the domain of the exponential function, which we noted is all Interchanging x and y, the graph of the
real numbers. natural logarithm function y = ln(x) can be obtained by ﬂipping the graph of y = ex across the line y = x: Important Facts 12.1.1 (Graphical features of natural log). The function y = ln(x) has these features: • • • The largest domain is the set of positive numbers; e.g. ln(−1) makes no sense. The graph has x-intercept 1 and is increasing. The graph becomes closer and closer to the vertical axis as we approach x = 0; i.e. the y-axis is a vertical asymptote for the graph. The graph is unbounded as we move to the right. • Any time we are working with an inverse function, symbolic properties are useful. Here are the important ones related to the natural logarithm. Important Facts 12.1.2 (Natural log properties). We have the following properties: 12.1. THE INVERSE FUNCTION OF Y = EX 155 (a) For any real number x, ln(ex) = x. (b) For any positive number x, eln(x) = x. (c) ln(bt) = t ln(b), for b > 0 and t any real number; (d) ln(ba) = ln(a) + ln(b), for all a, b > 0; (e) ln b a = ln(b) − ln(a), for all a, b > 0. The properties (c)-(e) are related to three of the rules of exponents in Facts 10.1.1. Here are the kinds of basic symbolic maneuvers you can pull off using these properties: Examples 12.1.3. (i) ln(83) = 3 ln(8) = 6.2383; ln(6π) = ln(6) + ln(π) = 2.9365; ln ln(3) − ln(5) = −0.5108. 3 5 = (ii) ln √x = ln = 1 2 ln(x); ln x5 = ln x1/2 x5 x2+1 ln(x + 1); ln Examples 12.1.4. Given the equation 3x+1 = 12, we can
solve for x: − ln x2 + 1 = 5 ln(x) − ln x2 + 1. x2 − 1 = ln((x − 1)(x + 1)) = ln(x − 1) + 3x+1 = 12 3x+1 ln = ln(12) (x + 1) ln(3) = ln(12) ln(12) ln(3) x = − 1 = 1.2619. Example 12.1.5. If $2,000 is invested in a continuously compounding savings account and we want the value after 12 years to be $130,000, what is the required annual interest rate? If, instead, the same $2,000 is invested in a continuously compounding savings account with r = 6.4% annual interest, when will the exact account value be be $130,000? Solution. In the ﬁrst scenario, 130,000 = 2,000e12r 65 = e12r ln(65) = ln ln(65) = 12r ln(65) 12 e12r r = = 0.3479. 156 CHAPTER 12. LOGARITHMIC FUNCTIONS This gives an annual interest rate of 34.79%. In the second scenario, we study the equation 130,000 = 2,000e(0.064)t 65 = e(0.064)t ln(65) 0.064 t = = 65.22. So, it takes over 65 years to accumulate $130,000 under the second scheme. 12.2 Alternate form for functions of exponential type The standard model for an exponential function is A(t) = A0bt, for some = 0. Using the properties of the natural logarithm b > 0, b function, = 1, and A0 6 bt = eln(b) t = et ln(b). This means that every function as in Deﬁnition 10.1.2 can be re-written using the exponential function et. Another way of saying this is that you really only need the function keys “ et ” and “ ln(t) ” on your calculator. Important Fact 12.2.1 (Observation). A function of exponential type can be written in the form A(t) = A0eat, for some constants A0
6 = 0 and a = 0. By studying the sign of the constant a, we can determine whether the function exhibits exponential growth or decay. For example, given the function A(t) = eat, if a > 0 (resp. a < 0), then the function exhibits exponential growth (resp. decay). Examples 12.2.2. (a) The function A(t) = 200 (2t) exhibits exponential growth and can be re-written as: A(t) = 200 et ln(2) = 200e0.69315t (b) The function A(t) = 4e−0.2t exhibits exponential decay and can be re- written as: A(t) = 4e−0.2t = 4 e−0.2 t = 4 0.81873t. 6 6 12.3. THE INVERSE FUNCTION OF Y = BX 157 12.3 The Inverse Function of y = bx For some topics in Chemistry and Physics (e.g. acid base equilibria and acoustics) it is useful to have on hand an inverse function for y = bx, = 1. Just as above, we would show that f(x) = bx is where b > 0 and b one-to-one, the range is all positive numbers and obtain the graph using ideas in Figure 12.2. We will refer to the inverse rule as the logarithm function base b, denoted logb(x), deﬁned by the rule: the unique solution of the equation c = bx!, if c > 0 (undeﬁned), if c 0. ≤ logb(c) =    We will need to consider two cases, depending on the magnitude of b: The important qualitative features of the logarithm function y = logb(x) mirror Fact 12.1.1: x y = b reflecting line y = x x y = b reflecting line y = x y=log (x)b −1 1 −1 1 y=log (x) b The case b > 1 The case 0 < b < 1 Figure 12.3: Cases to consider for b. Important Facts 12.3.1 (Graphical features of general logs). The
function y = logb(x) has these features: • • • The largest domain is the set of positive numbers; e.g. logb(−1) is not deﬁned. The graph has x-intercept 1 and is increasing if b > 1 (resp. decreasing if 0 < b < 1). The graph becomes closer and closer to the vertical axis as we approach x = 0; this says the y-axis is a vertical asymptote for the graph. 6 158 • CHAPTER 12. LOGARITHMIC FUNCTIONS The graph is unbounded as we move to the right. Important Facts 12.3.2 (Log properties). Fix a positive base b, b = 1. (a) For any real number x, logb(bx) = x. (b) For any positive number x, blogb(x) = x. (c) logb(rt) = t logb(r), for r > 0 and t any real number; s) = logb(r) − logb(s), for all r, s > 0. (d) logb(rs) = logb(r) + logb(s), for all r, s > 0; (e) logb( r It is common to simplify terminology and refer to the function logb(x) as the log base b function, dropping the longer phrase “logarithm”. Some scientiﬁc calculators will have a key devoted to this function. Other calculators may have a key labeled “log(x)”, which is usually understood to mean the log base 10. However, many calculators only have the key “ln(x)”. This is not cause for alarm, since it is always possible to express logb(x) in terms of the natural log function. Let’s see how to do this, since it is a great application of the Log Properties listed in Fact 12.3.2. Suppose we start with y = logb(x). We will rewrite this in terms of the natural log by carrying out a sequence of algebraic steps below; make sure you see why each step is justiﬁed. y = logb(x) by = x ln(by) = ln(x) y ln(b) = ln(x) ln(x) ln(
b) y =. We have just veriﬁed a useful conversion formula: Important Fact 12.3.3 (Log conversion formula). For x a positive number and b > 0, b = 1 a base, logb(x) = ln(x) ln(b). For example, log10(5) = log0.02(11) = log20 1 2 = ln(5) ln(10) ln(11) ln(0.02) ln( 1 2) ln(20) = 0.699 = −0.613 = −0.2314 6 6 12.4. MEASURING THE LOUDNESS OF SOUND 159 The conversion formula allows one to proceed slightly differently when solving equations involving functions of exponential type. This is illustrated in the next example. Example 12.3.4. Ten years ago, you purchased a house valued at $80,000. Your plan is to sell the house at some point in the future, when the value is at least $1,000,000. Assume that the future value of the house can be computed using quarterly compounding and an annual interest rate of 4.8%. How soon can you sell the house? Solution. We can use the future value formula to obtain the equation 1,000,000 = 80,000 1 + 12.5 = (1.012)4t 0.048 4 4t Using the log base b = 1.012, log1.012(12.5) = log1.012 log1.012(12.5) = 4t (1.012)4t t = ln(12.5) 4 ln(1.012) = 52.934. Since you have already owned the house for 10 years, you would need to wait nearly 43 years to sell at the desired price. 12.4 Measuring the Loudness of Sound As we noted earlier, the reception of a sound wave by the ear gives rise to a vibration of the eardrum with a deﬁnite frequency and a deﬁnite amplitude. This vibration may also be described in terms of the variation of air pressure at the same point, which causes the eardrum to move. The perception that rustling leaves and a jet aircraft sound different involves two concepts: (1) the fact that the frequencies involved may differ; (2) the intuitive notion of
“loudness”. This loudness is directly related to the force being exerted on the eardrum, which we refer to as the intensity of the sound. We can try to measure the intensity using some sort of scale. This becomes challenging, since the human ear is an amazing instrument, capable of hearing a large range of sound intensities. For that reason, a logarithmic scale becomes most useful. The sound pressure level β of a sound is deﬁned by the equation β = 10 log10, I I0 (12.2) where I0 is an arbitrary reference intensity which is taken to correspond with the average faintest sound which can be heard and I is the intensity 160 CHAPTER 12. LOGARITHMIC FUNCTIONS of the sound being measured. The units used for β are called decibels, (Historically, the units of loudness were called bels, abbreviated “db”. in honor of Alexander Graham Bell, referring to the quantity log10.) Notice, in the case of sound of intensity I = I0, we have a sound pressure level of I I0 β = 10 log10 I0 I0 = 10 log10(1) = 10(0) = 0. We refer to any sound of intensity I0 as having a sound pressure level at the threshold of hearing. At the other end of the scale, a sound of intensity the maximum the eardrum can tolerate has an average sound pressure level of about 120 db. The Table 12.4(a) gives a hint of the sound pressure levels associated to some common sounds. Source of Noise Threshold of pain Riveter Busy Street Trafﬁc Ordinary Conversation Quiet Auto Background Radio Whisper Rustle of Leaves Threshold of Hearing Sound Pressure Level in db 120 95 70 65 50 40 20 10 0 pain threshold Zone of Hearing 120 100 80 60 40 20 0 db 20 100 1000 10,000 20,000 hearing threshold Hz (a) Sources of noise levels. (b) Graphing noise levels. Figure 12.4: Considering noise levels. It turns out that the above comments on the threshold of hearing and pain are really only averages and depend upon the frequency of the given sound. In fact, while the threshold of pain is on average close to 120 db across all frequencies between 20 Hz and 20,000 Hz, the threshold of hearing is much more sensitive to frequency. For example, for a tone of 20 Hz
(something like the ground-shaking rumble of a passing freight train), the sound pressure level needs to be relatively high to be heard; 100 db on average. As the frequency increases, the required sound pressure level for hearing tends to drop down to 0 db around 2000 Hz. An examination by a hearing specialist can determine the precise sensitivities of your ear across the frequency range, leading to a plot of your “envelope of hearing”; a sample plot is given in Figure 12.4(b). Such a plot would differ from person to person and is helpful in isolating hearing problems. Example 12.4.1. A loudspeaker manufacturer advertises that their model no. 801 speaker produces a sound pressure level of 87 db when a reference test tone is applied. A competing speaker company advertises that 12.4. MEASURING THE LOUDNESS OF SOUND 161 their model X-1 speaker produces a sound pressure level of 93 db when fed the same test signal. What is the ratio of the two sound intensities produced by these speakers? If you wanted to ﬁnd a speaker which produces a sound of intensity twice that of the no. 801 when fed the test signal, what is its sound pressure level? Solution. If we let I1 and I2 refer to the sound intensities of the two speakers reproducing the test signal, then we have two equations: 87 = 10 log10 93 = 10 log10 I1 I0 I2 I0 Using log properties, we can solve the ﬁrst equation for I1: 87 = 10 log10 I1 I0 log10(I1) = 8.7 + log10(I0) 10log10(I1) = 108.7+log10(I0) = 10 log10(I1) − 10 log10(I0) I1 = 108.710log10(I0) = 108.7I0. Similarly, we ﬁnd that I2 = 109.3I0. This means that the ratio of the intensities will be I2 I1 = 109.3I0 108.7I0 = 100.6 = 3.98. This means that the test signal on the X − 1 speaker produces a sound pressure level nearly 4 times that of the same test signal on the no. 801 speaker. To ﬁnish the problem, imagine a third speaker which produces a sound pressure level β
, which is twice that of the ﬁrst speaker. If I3 is the corresponding intensity of the sound, then as above, I3 = 10(β/10)I0. We are assuming that I3 = 2I1, so this gives us the equation I1 = 108.7I0 = 1 2 1 2 I3 10(β/10)I0 log10 108.7 = log10 8.7 = log10 1 2 1 2 10(β/10) + log10 8.7 = −0.30103 + 90 = β β 10 10(β/10) 162 CHAPTER 12. LOGARITHMIC FUNCTIONS So, the test signal on the third speaker must produce a sound pressure level of 90 db. 12.5. EXERCISES 12.5 Exercises Problem 12.1. These problems will help you develop your skills with logarithms. (a) Compute: log5 3, loge 11, log√2 π, log2 10, log10 2. (b) Solve for x: 35 = ex, log3 x = e, log3 5 = xe3. (c) Solve each of these equations for x in terms of y: y = 10x, 3y = 10x, y = 103x. Problem 12.2. As light from the surface penetrates water, its intensity is diminished. In the clear waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of depth. Thus, the intensity will have the form of a general exponential function. (a) If the intensity of light at the water’s sur, ﬁnd a formula for I(d), the inface is I tensity of light at a depth of d meters. Your formula should depend on I and d. ◦ ◦ (b) At what depth will the light intensity be decreased to 1% of its surface intensity? Problem 12.3. Rewrite each function in the eat, for appropriate constants A form y = A and a. ◦ ◦ (a) y = 13(3t) (b) y = 2( 1 8 )t (c) y = −7(1.567)t−3 (d) y = −17(2.005)−t (e) y = 3(14.24)4t Problem 12.4. (a)
If you invest Po dollars at 7% annual interest and the future value is computed by continuous compounding, how long will it take for your money to double? 163 (c) A rule of thumb used by many people to determine the length of time to double an investment is the rule of 70. The rule says it takes about t = 70 r years to double the investment. Graphically compare this rule to the one isolated in part b. of this problem. Problem 12.5. The length of some ﬁsh are modeled by a von Bertalanffy growth function. For Paciﬁc halibut, this function has the form L(t) = 200 (1 − 0.956 e−0.18t) where L(t) is the length (in centimeters) of a ﬁsh t years old. (a) What is the length of a new-born halibut at birth? (b) Use the formula to estimate the length of a 6–year–old halibut. (c) At what age would you expect the hal- ibut to be 120 cm long? (d) What is the practical (physical) signiﬁcance of the number 200 in the formula for L(t)? Problem 12.6. A cancerous cell lacks normal biological growth regulation and can divide continuously. Suppose a single mouse skin cell is cancerous and its mitotic cell cycle (the time for the cell to divide once) is 20 hours. The number of cells at time t grows according to an exponential model. (a) Find a formula C(t) for the number of cancerous skin cells after t hours. (b) Assume a typical mouse skin cell is 10−4 cm. Find the spherical of radius 50 combined volume of all cancerous skin cells after t hours. When will the volume of cancerous cells be 1 cm3? × (b) Suppose you invest Po dollars at r% aninterest and the future value is nual computed by continuous compounding. If you want the value of the account to double in 2 years, what is the required interest rate? Problem 12.7. Your Grandfather purchased a house for $55,000 in 1952 and it has increased in value according to a function y = v(x), where x is the number of years owned. These questions probe the future value of the house under various mathematical models. 164 CHAPTER 12. LOGARITHMIC FUNCTIONS
(a) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a linear function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (b) Suppose the value of the house is $75,000 in 1962 and $120,000 in 1967. Assume v(x) is a quadratic function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (c) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a function of exponential type. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? Problem 12.8. Solve the following equations for x: (a) log3(5) = log2(x) (b) 10log2(x) = 3 x (c) 35 = 7 (d) log2(ln(x)) = 3 (e) ex = 105 (f) 23x+5 = 32 (d) There also was an exponentially-growing population of anteaters on board. At the start of the voyage there were 17 anteaters, and the population of anteaters doubled every 2.8 weeks. How long into the voyage were there 200 ants per anteater? Problem 12.10. The populations of termites and spiders in a certain house are growing exponentially. The house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders. How long (in days) does it take the popula- tion of spiders to triple? Problem 12.11. In 1987, the population of Mexico was estimated at 82 million people, with an annual growth rate of 2.5%. The 1987 population of the United States was estimated at 244 million with an annual growth rate of 0.7 %. Assume that both populations are growing exponentially. (a) When will Mexico double its 1987 popu- lation? (b) When will the United States and Mexico have the same population? Problem 12.9. A ship embarked on a long voyage
. At the start of the voyage, there were 500 ants in the cargo hold of the ship. One week into the voyage, there were 800 ants. Suppose the population of ants is an exponential function of time. (a) How long did it take the population to double? (b) How long did it take the population to triple? (c) When were there be 10,000 ants on board? Problem 12.12. The cities of Abnarca and Bonipto have populations that are growing exponentially. In 1980, Abnarca had a population of 25,000 people. In 1990, its population was 29,000. Bonipto had a population of 34,000 in 1980. The population of Bonipto doubles every 55 years. (a) How long does it take the population of Abnarca to double? (b) When will Abnarca’s population equal that of Bonipto? Chapter 13 Three Construction Tools Sometimes the composition of two functions can be understood by graphical manipulation. When we discussed quadratic functions and parabolas in the previous section, certain key graphical maneuvers were laid out. In this section, we extend those graphical techniques to general function graphs. 13.1 A Low-tech Exercise This section is all about building new functions from ones we already have in hand. This can be approached symbolically or graphically. Let’s begin with a simple hands-on exercise involving the curve in Figure 13.1. y = f(x) By the vertical line test, we know this represents the graph of a function y = f(x). With this picture and a piece of bendable wire we can build an INFINITE number of new functions from the original function. Begin by making a “model” of this graph by bending a piece of wire to the exact shape of the graph and place it right on top of the curve. The wire model can be manipulated in a variety of ways: slide the model back and forth horizontally, up and down vertically, expand or compress the model horizontally or vertically. Figure 13.1: Start with some curve. Another way to build new curves from old ones is to exploit the built in symmetry of the xy-coordinate system. For example, imagine reﬂecting the graph of y = f(x) across the x-axis or the y-axis. In all of the above cases, we moved from the original wire model of our function graph
to a new curve that (by the vertical line test) is the graph of a new function. The big caution in all this is that we are NOT ALLOWED to rotate or twist the curve; this kind of maneuver does lead to a new curve, but it may not be the graph of a function: See Figure 13.2. rotate NOT a function graph Figure 13.2: Rotating a curve. The pictures in Figure 13.3 highlight most of what we have to say in 165 166 CHAPTER 13. THREE CONSTRUCTION TOOLS this section; the hard work remaining is a symbolic reinterpretation of these graphical operations. left right up Horizontal shift. Vertical shift. down pull pull push push expand horizontally compress horizontally Horizontal expansion. Horizontal compression. pull pull expand vertically push push compress vertically Vertical expansion. Vertical compression. reﬂect across x-axis reﬂect across y-axis Vertical reﬂection. Horizontal reﬂection. Figure 13.3: Shifting, dilating, and reﬂecting y = f(x). 13.2 Reﬂection In order to illustrate the technique of reﬂection, we will use a concrete example: Function: y = p(x) = 2x + 2 2 Domain: 6 Range: −2 −2 x y ≤ ≤ ≤ ≤ As we know, the graph of y = p(x) on the domain −2 2 is a line of slope 2 with y-intercept 2, as pictured in Figure 13.4(a). Now, start with ≤ ≤ x 13.2. REFLECTION 167 the function equation y = p(x) = 2x + 2 and replace every occurrence of “y” by “−y.” This produces the new equation −y = 2x + 2; or, equivalently, y = q(x) = −2x − 2. x ≤ −y The domain is still −2 2, but the range will ≤ change; we obtain the new range by replacing “y” by “−y” −6. The in the original range: −2 graph of this function is a DIFFERENT line; this one has slope −2 and y-intercept −2. We contrast these two curves in Figure 13.4(b), where q(x) is graphed as the “dashed
line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of q(x) is really just the original line reﬂected across the x-axis. 6; so 2 ≥ ≤ ≥ ≤ y Next, take the original function equation y = p(x) = 2x + 2 and replace every occurrence of “x” by “−x.” This produces a new equation y = 2(−x) + 2; or, equivalently, −2 −1 y = r(x) = −2x + 2. −2 −1 y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 (a) Graph of p(x) = 2x + 2. y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 x ≥ ≥ The domain must also be checked by replacing “x” by “−x” in the original domain condition: −2 2, so −2. It just so happens in this case, the domain 2 is unchanged. This is yet another DIFFERENT line; this one has slope −2 and y-intercept 2. We contrast these two curves in Figure 13.4(c), where r(x) is graphed as the “dashed line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of r(x) is really just the original curve reﬂected across the y-axis. −x ≤ ≤ This example illustrates a general principle referred to as the reﬂection principle. Important Facts 13.2.1 (Reﬂection). Let y = f(x) be a function equation. (b) Graph of q(x) = −2x − 2. y-axis x-axis 1 2 8 6 4 2 −2 −4 −6 −8 −2 −1 (c) Graph of r(x) = −2x + 2. Figure 13.4: Reﬂecting y = p(x). (i) We can reﬂect the graph across the x-axis and the resulting curve is the graph of the new function obtained by replacing “y” by “−y” in the original equation. The domain is
the same as the domain for y = f(x). If the range for y = f(x) d. In other words, is c the reﬂection across the x-axis is the graph of y = −f(x). d, then the range of −y = f(x) is c −y ≤ ≤ ≤ ≤ y (ii) We can reﬂect the graph across the y-axis and the resulting curve is the graph of the new function obtained by replacing “x” by “−x” in the original equation. The range is the same as the range for y = f(x). If the domain for y = f(x) is a b, then the domain x ≤ ≤ 168 CHAPTER 13. THREE CONSTRUCTION TOOLS of y = f(−x) is a b. Using composition no≤ tation, the reﬂection across the y-axis is the graph of y = f(−x). −x ≤ Example 13.2.2. Consider the parallelogram-shaped region with vertices (0, 2), (0, −2), (1, 0), and (−1, 0). Use the reﬂection principle to ﬁnd functions whose graphs bound 0,2) l 2 Q = (1,0) x-axis l 1 Region R y-axis Figure 13.5: The region. R y-axis y = f(−x) y = f(x) −4 −2 2 4 x-axis −y = f(x) Figure 13.6: Reﬂecting the semicircle. R Solution. Here is a picture of the region : First off, using the two point formula for the equation of a line, we ﬁnd that the line ℓ1 passing through the points P = (0, 2) and Q = (1, 0) is the graph of the function y = f1(x) = −2x + 2. By Fact 13.2.1 (i), ℓ2 is the graph of the equation −y = −2x + 2, which we can write as the function y = f2(x) = 2x − 2. By Fact 13.2.1 (ii) applied to ℓ2, the line ℓ3 is the graph of the function y = f3(
x) = −2x − 2. Finally, by Fact 13.2.1 (i) applied to ℓ3, the line ℓ4 is the graph of the equation −y = −2x − 2, which we can write as the function y = f4(x) = 2x + 2. Figure 13.6 illustrates the fact that we need to be careful about the domain of the original function when using the reﬂection principle. For example, consider 1 − (x − 3)2. The largest possible domain of y = f(x) = 1 + x-values is 2 4 and the graph is an upper semicircle x p ≤ of radius 1 centered at the point (3, 1). ≤ Reﬂection across the x-axis gives the graph of reﬂection 1 − (x + 3)2 y = −1 − across the y-axis gives the graph of y = 1 + on the new domain −4 1 − (x − 3)2 on the same domain; −2. p x ≤ ≤ p 13.3 Shifting 2.5 2 1.5 1 0.5 −2 −1 0 y-axis x x-axis Let’s start out with the function y = f(x) = √4 − x2, which has a largest possible domain −2 2. From Chapter 6, the graph of this equation is an upper semicircle of radius 2 centered at the origin (0, 0). Sliding the graph back and forth horizontally or vertically (or both), never rotating or twisting, we are led to the “dashed curves” below (contrasted with the original graph which is plotted with a solid curve). This describes some shifted curves on a pictorial level, but what are the underlying equations? For this example, we can use the fact that all of the shifted curves are still semicircles and Chapter 6 tells us how to ﬁnd their equations. ≤ ≤ 2 1 Figure 13.7: Graph of y = √4 − x2. 13.3. SHIFTING 169 The lower right-hand dashed semicircle is of radius 2 and is centered at (3, 0), so the corresponding equation must be y = 4 − (x − 3)2. The upper left-hand dashed semicircle is of radius 2 and centered at (0, 3), so the corresponding equation must be
y = 3+√4 − x2. The upper right-hand dashed semicircle is of radius 2 and centered at (3, 3), so the corresponding equation must be y = 3 + 4 − (x − 3)2. p Keeping this same example, we can continue this kind p of shifting more generally by thinking about the effect of making the following three replacements in the equation y = √4 − x2x and y) (x − h and y − k). These substitutions lead to three new equations, each the equation of a semicircle: 5 4 3 2 1 −2 −1 1 2 3 4 5 Figure 13.8: Shifting the upper semicircle. y = 4 − (x − h)2 − x2; and, 4 − (x − h)2. ⇐ ⇐ Upper semicircle with radius 2 and center (h, 0). Upper semicircle with radius 2 and center (0, k). Upper semicircle with radius 2 and center (h, k). There are three potentially confusing points with this ⇐ y-axis original curve x-axis h negative h positive curve shifted h units right if h is positive curve shifted \|h\| units left if h is negative Figure 13.9: Potentially confusing points. example: • • ± ) of h and k. In FigBe careful with the sign (i.e., ure 13.9, if h = 1, we horizontally shift the semicircle 1 unit to the right; whereas, if h = −1, we horizontally shift the semicircle −1 units to the right. But, shifting −1 unit to the right is the same as shifting 1 unit to the left! In other words, if h is positive, then a horizontal shift by h will move the graph \|h\| units to the right; if h is negative, then a horizontal shift by h will move the graph \|h\| units to the left. If k is positive, then a vertical shift by k will move the graph \|k\| units up; if k is negative, then a vertical shift by k will move the graph \|k\| units down. These conventions insure that the “positivity” of h and k match up with “rightward” and “upward” movement of the graph. 170 CHAPTER 13. THREE CONSTRUCTION TOOLS • When shifting, the domain of allowed x-values may
change. This example illustrates an important general principle referred to as the shifting principle. Important Facts 13.3.1 (Shifting). Let y = f(x) be a function equation. (i) If we replace “x” by “x − h” in the original function equation, then the graph of the resulting new function y = f(x − h) is obtained by horizontally shifting the graph of f(x) by h. If h is positive, the picture shifts to the right h units; if h is negative, the picture shifts to the left h units. If the domain of f(x) is an interval a b, then the domain ≤ of f(x − h) is a x − h b. The range remains unchanged under ≤ horizontal shifting. ≤ ≤ x (ii) If we replace “y” by “y − k” in the original function equation, then the graph of the resulting new function y = f(x) + k is obtained by vertically shifting the graph of f(x) by k. If k is positive, the picture shifts upward k units; if k is negative, the picture shifts downward k units. If the range of f(x) is an interval c d, then the range y of f(x) + k is c d. The domain remains unchanged under ≤ vertical shifting. y − k ≤ ≤ ≤ 13.4 Dilation 1 0.75 0.5 0.25 y-axis 2 To introduce the next graphical principle we will look at the function y = f(x) = x x2 + 1. −6 −4 −2 4 6 x-axis −0.25 −0.5 −0.75 −1 Figure 13.10: Graph of y = f(x) = x x2+1. ≤ Using a graphing device, we have produced a plot of the graph on the domain −6 6. Figure 13.10 shows the x ≤ curve has a high point H (like a “mountain peak”) and a low point L (like a “valley”). Using a graphing device, we can determine that the high point is H = (1, 1 2) (it lies on −1, − 1 the line with equation y = 1 (it lies on 2 the line with equation line y = − 1 1 2. Draw two new horizontal lines with equations y = 2 1.
Grab the high point H on the curve and uniformly pull straight up, so that the high point now lies on the horizontal line y = 1 at (1, 1). Repeat this process by pulling L straight downward, so that the low point is now on the line y = −1 at (−1, −1). We end up with the ”stretched dashed curve” illustrated in Figure 13.11(a). In terms of the original function equation y = x x2+1, we are simply describing the graphical effect of multiplying the y-coordinate of every point on the curve by the positive number 2. In other words, the dashed curve is the graph of y = 2x x2+1 = 2 2 ) and the low point is L = 2 ), so the range is − 1 · ± x x2+1 1 2 = 2 ≤ ≤ ± y. 13.4. DILATION 171 1 2 = 4 at (1, 1 Next, draw the two horizontal lines with equations y = 1 1 4. Grab the high point H on the curve (in Figure 13.10) and uniformly push straight down, so that the high point now lies on the horizontal line y = 1 4). Repeat this process at the low point L by pushing the curve straight upward, so that the low point is now on the line y = − 1 4 at (−1, − 1 4). We end up with the new ”dashed curve” illustrated in Figure 13.11(b). In terms of the original function equation y = x x2+1, we are simply describing the graphical effect of multiplying the y-coordinate of every point on the curve by the positive number 1 2. In other words, the dashed curve is the graph of y = x 2 · ± ± 2(x2+1). We could repeat this process systematically: • • • Pick a positive number c. 1 2. If c > 1, then 2 is parallel and above the graph of 2. On the other hand, if 0 < c < 1, then this new Draw the two horizontal lines y = c the graph of y = c y = 1 line is parallel and below y = 1 2. · ± Uniformly deform the original graph (in Figure 13.10) so that the new curve has it’s high and c low points just touching y = 2. This will involve vertically stretching or compressing, depending on whether 1 < c or 0 < c < 1
, respectively. A number of possibilities are pictured in Figure 13.11(c). ± We refer to each new dashed curve as a verticaldilation of the original (solid) curve. This example illustrates an important principle. Important Facts 13.4.1 (Vertical dilation). Let c > 0 be a positive number and y = f(x) a function equation. (i) If we replace “ y ” by “ y c ” in the original equation, then the graph of the resulting new equation is obtained by vertical dilation of the graph of y = f(x). The domain of x-values is not affected. (ii) If c > 1, then the graph of y c = f(x) (i.e., y = cf(x)) is a vertically stretched version of the original graph. −6 −4 −2 y-axis 1 0.75 0.5 0.25 2 −0.25 −0.5 −0.75 −1 4 6 x-axis (a) Vertical expansion. y-axis 1 0.75 0.5 0.25 −6 −4 −2 2 −0.25 −0.5 −0.75 −1 4 6 x-axis (b) Vertical compression. y-axis 1 0.75 0.5 0.25 −6 −4 −2 4 6 x-axis 2 −0.25 −0.5 −0.75 −1 (c) Many possibilities. Figure 13.11: Dilating y = f(x). (iii) If 0 < c < 1, then the graph of y c = f(x) (i.e., y = cf(x)) is a vertically compressed version of the original graph. If we combine dilation with reﬂection across the x-axis, we can determine the graphical relationship between y = f(x) and y = cf(x), for any constant c. The key observation is that reﬂection across the x-axis corresponds to the case c = −1. 172 CHAPTER 13. THREE CONSTRUCTION TOOLS Example 13.4.2. Describe the relationship between the graphs of y = f(x) = 1 − (x + 1)2, p y = −f(x) = − 1 − (x + 1)2, and p y = −4f(x) = −4 1 − (x + 1
)2. p step 1: start with upper semicircle x-axis step 2: reﬂect across step 3: stretch curve from step 2: to get y = −4f(x) Solution. The graph of y = f(x) is an upper semicircle of radius 1 centered at the point (−1,0). To obtain the picture of the graph of y = −4f(x), we ﬁrst reﬂect y = f(x) across the x-axis; this gives us the graph of y = −f(x). Then, we vertically dilate this picture by a factor of c = 4 to get the graph of y 4 = −f(x), which is the same as the graph of the equation y = −4f(x). See Figure 13.12. Figure 13.12: Reﬂecting and dilating a lower semicircle. Let’s return to the original example y = x x2+1 and investigate a different type of dilation where the action is taking place in the horizontal direction (whereas it was in the vertical direction before). Grab the right-hand end of the graph (in Figure 13.10) and pull to the right, while at the same time pulling the left-hand end to the left. We can quantify this by stipulating that the 2, 1 high point H = 2 and the low point L = of the original curve moves to the new location moves to the new location 1, 1 2. −1, − 1 2 −2, − 1 2 y-axis 1 0.75 0.5 0.25 y-axis 1 0.75 0.5 0.25 −6 −4 −2 2 −0.25 −0.5 −0.75 −1 (a) Stretching. 4 6 x-axis −6 −4 −2 4 6 x-axis 2 −0.25 −0.5 −0.75 −1 Figure 13.13: Horizontally dilating y = x (b) Compressing. x2+1. The result will be somewhat analogous to stretching a spring. By the same token, we could push the right-hand end to the left and push the left-hand end to the right, like compressing a spring. We can quantify this by stipulating that the high point H = of the original curve moves moves to the new to the new
location location. These two situations are indicated in Figure 13.13. We refer to each of the dashed curves as a horizontal dilation of the original (solid) curve. and the low point L = −1, − 1 2 2, − 1 1, 1 2 2, 1 − 1 2 1 2 13.5. VERTEX FORM AND ORDER OF OPERATIONS 173 The tricky point is to understand what happens on the level of the original equation. In the case of the stretched graph in Figure 13.13(a), you can use a graphing device to verify that this looks like the graph of y = x/2 (x/2)2+1 ; in other words, we replaced “x” by “x/2” in the original equation. In the case of the compressed graph in Figure 13.13(b), you can use a graphing device to verify that this looks like the graph of y = 2x (2x)2+1; in other words, we replaced “x” by “ x 1/2 = 2x” in the original equation. The process just described leads to a general principle. Important Facts 13.4.3 (Horizontal dilation). Let c > 0 be a positive number and y = f(x) a function equation. (i) If we replace “ x” by “ x graph of the resulting new function y = f tal dilation of the graph of y = f(x). If the domain of f(x) is a then the domain of y = f is a c ” in the original function equation, then the is obtained by a horizonb, ≤ ≤ x x c x c b. x c ≤ is a horizontal stretch. ≤ x c (ii) If c > 1, then the graph of y = f (iii) If 0 < c < 1, then the graph of y = f is a horizontal compression. x c 13.5 Vertex Form and Order of Operations Using the language of function compositions we can clarify our discussion in Example 7.1.2. Let’s revisit that example: Example 13.5.1. The problem is to describe a sequence of geometric maneuvers that transform the graph of y = x2 into the graph of y = −3(x − 1)2 + 2. Solution. The idea is to rewrite y = −3(x−1)2 +2 as a composition of
y = x2 with four other functions, each of which corresponds to a horizontal shift, vertical shift, reﬂection or dilation. Once we have done this, we can read off the order of geometric operations using the order of composition. Along the way, pay special attention to the exact order in which we will be composing our functions; this will make a big difference. To begin with, we can isolate four key numbers in the equation: Reﬂect − z}\|{ 3 (x − Dilate by 3 Horizontal shift by h = 1 1 z}\|{ )2 + 2 Vertical shift by k = 2 \|{z} \|{z} We want to use each number to deﬁne a new function, then compose these in the correct order. We will also give our starting function y = x2 a speciﬁc name to make things deﬁnite: f(x) = x2 v(x) = x + 2 d(x) = 3x. h(x) = x − 1 r(x) = −x 174 CHAPTER 13. THREE CONSTRUCTION TOOLS Now, verify that f h(x[x − 1] d{(x − 1)2}   r(3(x − 1)2) = v i h = v[−3(x − 1)2] = −3(x − 1)2 + 2 = −3x2 + 6x − 1. 13.6 Summary of Rules 2 −2 2 Figure 13.14: A multipart function. For quick reference, we summarize the consequence of shifting and expanding symbolically and pictorially. The running example for Tables 13.1, 13.2, and 13.3 will be a multipart function y = f(x) whose graph, seen in Figure 13.14, consists of a line segment and a quarter circle on the domain −2 2: x ≤ ≤ f(x) = x + 2 √4 − x2 0 x if −2 if 0 ≤ x ≤ ≤ ≤ 2 13.6. SUMMARY OF RULES 175 Reﬂection Symbolic Change New Equation Graphical Consequence Replace x with −x. y = f(−x) Replace with −f(x). f(x) y = −f(x) A reﬂection across the y-
axis. A reﬂection across the x-axis. Picture 2 2 2 −2 −2 −2 Table 13.1: Reﬂecting y = f(x). 176 CHAPTER 13. THREE CONSTRUCTION TOOLS Shifting (Assume c > 0 ) Symbolic Change New Equation Graphical Consequence Picture Replace x with (x − c). y = f(x − c) A shift to the right c units. 2 Replace x with (x + c). y = f(x + c) A shift to the left c units. Replace with (f(x) + c). f(x) y = f(x) + c A shift up c units. 2 4 2 −4 −2 4 2 −2 2 Replace with (f(x) − c). f(x) y = f(x) − c A shift down c units. −2 2 2 Table 13.2: Shifting y = f(x). 13.6. SUMMARY OF RULES 177 Dilation Symbolic Change New Equation Graphical Consequence Picture If c > 1, replace x x with c. y = f x c A horizontal expansion. c = 2 4 3 2 1 −1−2−3−4 0 1 2 3 4 If 0 < c < 1, replace x with. x c y = f x c A horizontal compression. c = 1 2 4 3 2 1 If c > 1, replace f(x) with (cf(x)). y = cf(x) A vertical expansion. If 0 < c < 1, replace f(x) with (cf(x)). y = cf(x) A vertical compression. Table 13.3: Dilating y = f(x). −4 −1−2−1−2−3−1−2−3−4 0 1 2 3 4 178 CHAPTER 13. THREE CONSTRUCTION TOOLS 13.7 Exercises Problem 13.1. On a single set of axes, sketch a picture of the graphs of the following four equations: y = −x+√2, y = −x−√2, y = x+√2, and y = x − √2. These equations determine lines, which in turn bound a diamond shaped region in the plane. (a) Show that the unit circle sits inside this diamond tangentially; i.e. show that the
unit circle intersects each of the four lines exactly once. (b) Find the intersection points between the unit circle and each of the four lines. (c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help. y π π/2 f(x) −1 1 x Problem 13.2. The graph of a function y = f(x) is pictured with domain −2.5 3.5. x ≤ ≤ Problem 13.4. (a) Each of the six functions y = f(x) below can be written in the “standard form” y = A\|B(x − C)\| + D, for some constants A,B,C,D. Find these constants, describe the precise order of graphical operations involved in going from the graph of y = \|x\| to the graph of y = f(x) (paying close attention to the order), write out the multipart rule, sketch the graph and calculate the coordinates of the “vertex” of the graph. Sketch the graph of each of the new func- tions listed below. (a) g(x) = 2f(x + 1) (b) h(x) = 1 2 f(2x − 1) (c) j(x) = 5f( 1 3 x + 2) − 2 (a1) f(x) = \|x − 2\| (a2) f(x) = 2\|x + 3\| (a3) f(x) = \|2x − 1\| (a4) f(x) = \|2(x − 1)\| (a5) f(x) = 3\|2x − 1\| + 5 (a6) f(x) = −2\|x + 3\| − 1 Problem 13.3. The graph of a function y = f(x) is pictured with domain −1 1. Sketch the graph of the new function ≤ ≤ x y = g(x) = 1 π f(3x) − 0.5. Find the largest possible domain of the function y = g(x). p (b) Solve the following inequalities using your work in the previous part of this problem: (b1) \|x − 2\| 3 (b2) 1 ≤ ≤ 2\|x + 3\| 5 ≤ (b
3) y = 3\|2x − 1\| + 5 10 ≥ 13.7. EXERCISES 179 (c) The graphs of y = 3\|2x − 1\| + 5 and y = −\|x − 3\| + 10 intersect to form a bounded region of the plane. Find the vertices of this region and sketch a picture. (c) The graph of z = a(x) from part (a) is given below. Sketch the graph and ﬁnd the rule for the function z = 2a(3x+3)+1; make sure to specify the domain and range of this new function. Problem 13.5. Consider the function y = f(x) with multipart deﬁnition 0 2x + 2 −x + 2 0 f(x) =   0 if x ≤ if −1 if 0 if x ≤ ≥ −1 x ≤ 2 ≤ ≤ x 2 (a) Sketch the graph of y = f(x).  (b) Is y = f(x) an even function? Is y = f(x) an odd function? (A function y = f(x) is called even if f(x) = f(−x) for all x in the domain. A function y = f(x) is called odd if f(−x) = −f(x) for all x in the domain.) (c) Sketch the reﬂection of the graph across the x-axis and y-axis. Obtain the resulting multipart equations for these reﬂected curves. (d) Sketch the vertical dilations y = 2f(x) and y = 1 2 f(x). (e) Sketch the horizontal dilations y = f(2x) and y = f( 1 2 x). (f) Find a number c > 0 so that the highest point on the graph of the vertical dilation y = cf(x) has y-coordinate 11. (g) Using horizontal dilation, ﬁnd a number c > 0 so that the function values f( x c ) are non-zero for all − 5 2 < x < 5. (h) Using horizontal dilation, ﬁnd positive numbers c,d > 0 so that the function values f( 1 c (x − d)) are non-
zero precisely when 0 < x < 1. Problem 13.6. An isosceles triangle has sides of length x, x and y. In addition, assume the triangle has perimeter 12. (a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible domain of this function. (b) Assume that the maximum value of the function a(x) in (a) occurs when x = 4. Find the maximum value of z = a(x) and z = 2a(3x + 3) + 1. 10 y-axis 8 6 4 2 2 4 6 8 10 x-axis Problem 13.7. Describe how each graph differs from that of y = x2. (a) y = 2x2 (b) y = x2 − 5 (c) y = (x − 4)2 (d) y = (3x − 12)2 (e) y = 2(3x − 12)2 − 5 Problem 13.8. In each case, start with the function y = \|x\| and perform the operations described to the graph, in the order speciﬁed. Write out the resulting rule for the function and sketch the ﬁnal graph you obtain. (a) (1) horizontally compress by a factor of 2; (2) horizontally shift to the left by 2; (3) vertically stretch by a factor of 7; (4) vertically shift up 2 units. (b) (1) horizontally stretch by a factor of 2; (2) horizontally shift to the right by 2; (3) vertically compress by a factor of 7; (4) vertically shift down 2 units. (c) (1) horizontally shift to the right 2 units; (2) horizontally compress by a factor of 3. Problem 13.9. (a) Begin with the function y = f(x) = 2x. (a1) Rewrite each of the following functions in standard exponential form: f(2x), f(x − 1), f(2x − 1), f(2(x − 1)), 3f(x), 3f(2(x − 1)). 180 CHAPTER 13. THREE CONSTRUCTION TOOLS (a2) Is the function 3f(2(x−1))+1 a func- tion of exponential type? (a3) Sketch the graphs of f
(x), f(2x), f(2(x− 1)), 3f(2(x − 1)) and 3f(2(x − 1)) + 1 in the same coordinate system and explain which graphical operation(s) (vertical shifting, vertical dilation, horizontal shifting, horizontal dilation) have been carried out. (b) In general, explain what happens when you apply the four construction tools of Chapter 13 (vertical shifting, verti- cal dilation, horizontal shifting, horizontal dilation) to the standard exponential model y = Aobx. For which of the four operations is the resulting function still a standard exponential model? Problem 13.10. Begin with a sketch of the graph of the function y = 2x on the domain of all real numbers. Describe how to use the “four tools” of Chapter 13 to obtain the graphs of these functions: y = −2x, y = 2−x, y = 3(2x), 3 (2x), y = 3 + 2x, y = 2x − 2, y = 2x−2, y = 1 y = 2x+2, y = 23x, y = 2x/3. Chapter 14 Rational Functions A rationalfunction is a function of the form f(x) = p(x) are polynomials. For example, the following are all rational functions. q(x) where p(x) and q(x) f(x) = x 3x + 4 g(x) = x2 + 1 3x − 5 h(x) = 4x5 − 4x2 − 8 x3 + x2 − x + 1 j(x) = x6 x8 + 5x − 1 2 There is a huge variety of rational functions. In this course, we will concentrate our efforts exclusively toward understanding the simplest type of rational functions: linear-to-linear rational functions. Linearto-linear rational functions are rational functions in which the numerator and the denominator are both linear polynomials. The following are linear-to-linear-rational functions. k(x) = x 3x + 4 m(x) = 5x − 6 2x + 1 n(x) = 0.34x − 0.113 x − 1 p(x) = 4x + 3 4 5 8x − 1.117 The simplest example of a linear-to-linear rational function is f
(x) = 1 x whose graph is shown in Figure 14.1. This is an important example for the study of this class of functions, as we shall see. Let’s consider the graph of this function, f(x) = 1 x. We ﬁrst begin by considering the domain of f. Since the numerator of 1 x is a constant, and the denominator is just x, the only way we can run into difﬁculty when evaluating this function is if we try to divide by zero; that is, the only value of x not in the domain of this function is zero. The function is deﬁned for all x except x = 0. At x = 0, there must be a gap, or hole, in the graph. -10 -5 10 5 0 0 -5 -10 5 10 Figure 14.1: The graph of f(x) = 1 x. To get the graph started we might simply give ourselves a point on the graph. For instance, we see that f(1) = 1/1 = 1, so the point (1,1) is on the graph. Then, if we use larger values of x, we see that 1/x becomes smaller as x grows. For instance, f(2) = 1/2, f(10) = 1/10, and f(1000) = 1/1000. In addition, we see that, other than the fact that 1/x > 0 for positive x, there is nothing preventing us from 181 182 CHAPTER 14. RATIONAL FUNCTIONS making 1/x as small as we want simply by taking x big enough. Want 1/x to be less than 0.001? Just pick x bigger than 1000. Want 1/x to be less than 0.000001? Just use x bigger than 1,000,000. What this means graphically is that as x gets bigger (starting from x = 1), the curve y = 1/x gets closer and closer to the x-axis. As a result, we say that the x-axis is a horizontal asymptote for this function. We see the same behavior for negative values of x. If x is large, and negative (think -1000, or -1000000), then 1/x is very small (i.e., close to zero), and it gets smaller the larger x becomes. Graphically, this means that as x gets
large in the negative direction, the curve y = 1/x gets closer and closer to the x-axis. We say that, in both the positive and negative directions, y = 1/x is asymptotic to the x-axis. A similar thing happens when we consider x near zero. If x is a small positive number (think 1/2, or 1/10, or 1/10000), then 1/x is a large positive number. What’s more, if we think of x as getting closer to zero, 1/x gets bigger and bigger. Plus, there is no bound on how big we can make 1/x simply by taking x as close to zero as we need to. For instance, can 1/x be as big as 10000? All you need to do is pick a positive x smaller than 1/10000. Graphically, what this means is that as x approaches zero from the positive side, the y value gets larger and larger. As a result, the curve approaches the y-axis as y gets larger. We say that the y-axis is a vertical asymptote for the curve y = 1/x. We see the same phenomenon as x approaches zero from the negative side: y = 1/x gets larger in the negative direction (i.e., it gets more and more negative). The curve gets closer and closer to the negative y-axis as y becomes more and more negative. Again, we say that the y − axis is a vertical asymptote for the curve y = 1/x. It turns out that every linear-to-linear rational function has a graph that looks essentially the same as the graph of y = 1/x. Let’s see why. Consider the linear-to-linear rational function f(x) = ax+b cx+d. If we divide cx + d into ax + b, the result is f(x) = ax + b cx + d = a c + b − ad c cx + d which we can rewrite as f(x) = a c + b − ad c · 1 cx + d = a c + b − ad c · 1 c(x + d c ) = a c + bc − ad c2 1 x + d c. · If we now let A =, B = a c then we can write bc − ad c2, and C = d c, f(x) = A + B 1 x + C
. 183 If we let g(x) = 1/x then we have shown that f(x) = A + Bg(x + C), and so the graph of the function f is just a horizontally shifted, vertically shifted and vertically dilated version of the graph of g. Also, if B turned out to be negative the graph would be vertically ﬂipped, too. Why is that useful? It means that the graph of a linear-to-linear ra- tional function can only take one of two forms. Either it looks like this: Or like this: We can sketch an accurate graph of a linear-to-linear rational function by sketching the asymptotes and then sketching in just one point on the graph. That will be enough information to nail down a decent sketch. We can always plot more points to give us more precision, but one point is enough to capture the essence of the graph. Given a linear-to-linear function that we wish to graph, we must ﬁrst determine the asymptotes. There will be one horizontal asymptote and one vertical asymptote. The vertical asymptote will be a vertical line with equation x = k where k is the one x value which is not in the domain of the function. That is, ﬁnd the value of x which makes the denominator zero and that will tell you the vertical asymptote. The horizontal asymptote is a little more involved. However, we can quickly get to a shortcut. The essence of a horizontal asymptote is that it describes what value the function is approximately equal to for very large values of x. To study what a linear-to-linear function is like when x is very large, we can perform the following algebraic manipulation: f(x) = ax + b cx + d = ax + b cx + d · 1/x 1/ While dividing by x is troublesome if x equals zero, here we are assuming x is very large, so it is certainly not zero. Now, consider this last expression in the above equation. If x is very large, then b x ≈ 0 184 (where ≈ CHAPTER 14. RATIONAL FUNCTIONS means ”is approximately”) and likewise d x ≈ 0. Hence, when x is very large, f(x. ≈ We can interpret this by saying that when x is very large, the function f(
x) is is close to a constant, and that constant is a c. Thus, the horizontal asymptote of f(x) = is the horizontal line y = a c. ax + b cx + d Example 14.0.1. Sketch the graph of the function f(x) = 3x−1 2x+7. Solution. We begin by ﬁnding the asymptotes of f. The denominator is equal to zero when 2x + 7 = 0, i.e., when x = −7/2. As a result, the vertical asymptote for this function is the vertical line x = −7/2. By taking the ratio of the coefﬁcients of x in the numerator and denominator, we can ﬁnd that the horizontal asymptote is the horizontal line y = 3 2. -10 -5 10 5 0 0 -5 -10 5 10 Figure 14.2: The graph of f(x) = 3x−1 2x+7. We then sketch these two asymptotes. The last thing we need is a single point. For instance, we may evaluate f(0): f(0) = −1 7 14.1. MODELINGWITHLINEAR-TO-LINEARRATIONALFUNCTIONS185 and so the point (0, −1/7) is on the graph. With this information, we know that the curve lies below the horizontal asymptote to the right of the vertical asymptote, and consequently the curve lies above the horizontal asymptote to the left of the vertical asymptote. We graph the result in Figure 14.2. 14.1 Modeling with Linear-to-linear Rational Functions As we have done with other sorts of functions, such as linear and quadratic, we can also model using linear-to-linear rational functions. One reason for using this type of function is their asymptotic nature. Many changing quantities in the world continually increase or decrease, but with bounds on how large or small they can get. For instance, a population may steadily decrease, but a population can never be negative. Conversely, a population may steadily increase, but due to environmental and other factors we may hypothesize that the population will always stay below some upper bound. As a result, the population may ”level off”. This leveling off behavior is exempli�
��ed by the asymptotic nature of the linear-to-linear rational functions, and so this type of function provides a way to model such behavior. Given any linear-to-linear rational function, we can always divide the numerator and the denominator by the coefﬁcient of x in the denominator. In this way, we can always assume that the coefﬁcient of x in the denominator of a function we seek is 1. This is illustrated in the next example. Example 14.1.1. Let f(x) = 2x+3 5x−7. Then f(x) = = 1 5 1 5 2x + 3 5x − 7 · 5x + 3 2 5 x − 7 5. Thus, in general, when we seek a linear-to-linear rational function, we will be looking for a function of the form f(x) = ax + b x + c and thus there are three parameters we need to determine. Note that for a function of this form, the horizontal asymptote is y = a and the vertical asymptote is x = −c. Since these functions have three parameters (i.e., a, b and c), we will need three pieces of information to nail down the function. 186 CHAPTER 14. RATIONAL FUNCTIONS There are essentially three types of modeling problems that require the determination of a linear-to-linear function. The three types are based on the kind of information given about the function. The three types are: 1. You know three points the the graph of the function passes through; 2. You know one of the function’s asymptotes and two points the graph passes through; 3. You know both asymptotes and one point the graph passes through. Notice that in all cases you know three pieces of information. Since a linear-to-linear function is determined by three parameters, this is exactly the amount of information needed to determine the function. The worst case, in terms of the amount of algebra you need to do, is the ﬁrst case. Let’s look at an example of the algebra involved with this sort. Example 14.1.2. Find the linear-to-linear rational function f(x) such that f(10) = 20, f(20) = 32 and f(25) = 36. Solution. Since f(x) is a linear
-to-linear rational function, we know f(x) = ax + b x + c for constants a, b, and c. We need to ﬁnd a, b and c. We know three things. First, f(10) = 20. So f(10) = 10a + b 10 + c = 20, which we can rewrite as 10a + b = 200 + 20c. Second, f(20) = 32. So f(20) = 20a + b 20 + c = 32, which we can rewrite as 20a + b = 640 + 32c. Third, f(25) = 36. So f(25) = 25a + b 25 + c = 36, (14.1) (14.2) 14.1. MODELINGWITHLINEAR-TO-LINEARRATIONALFUNCTIONS187 which we can rewrite as 25a + b = 900 + 36c. (14.3) These three numbered equations are enough algebraic material to solve for a, b, and c. Here is one way to do that. Subtract equation 14.1 from equation 14.2 to get 10a = 440 + 12c and subtract equation 14.2 from equation 14.3 to get 5a = 260 + 4c (14.4) (14.5) Note that we’ve eliminated b. Now multiply this last equation by 2 to get 10a = 520 + 8c Subtract equation 14.4 from this to get 0 = 80 − 4c which easily give us c = 20. Plugging this value into equation 14.4, we can ﬁnd a = 68, and then we can ﬁnd b = −80. Thus, f(x) = 68x − 80 x + 20. We can check that we have done the algebra correctly by evaluating If we get f(10) = 20, f(20) = 32 and f(x) at x = 10, x = 20 and x = 25. f(25) = 36, then we’ll know our work is correct. Algebraically, this was the worst situation of the three, since it reIf, instead of knowing three points, we know quired the most algebra. one or both of the asymptotes, then we can easily ﬁnd a and/or c, and so cut down on the amount of algebra needed.
However, the method is essentially identical. Let’s now apply these ideas to a real world problem. Example 14.1.3. Clyde makes extra money selling tickets in front of the Safeco Field. The amount he charges for a ticket depends on how many he has. If he only has one ticket, he charges $100 for it. If he has 10 tickets, he charges $80 a piece. But if he has a large number of tickets, he will sell them for $50 each. How much will he charge for a ticket if he holds 20 tickets? 188 CHAPTER 14. RATIONAL FUNCTIONS Solution. We want to give a linear-to-linear rational function relating the price of a ticket y to the number of tickets x that Clyde is holding. As we saw above, we can assume the function is of the form y = ax + b x + c where a, b and c are numbers. Note that y = a is the horizontal asymptote. When x is very large, y is close to 50. This means the line y = 50 is a horizontal asymptote. Thus a = 50 and y = 50x + b x + c. Next we plug in the point (1,100) to get a linear equation in b and c. 50 1 + b 100 = · 1 + c (1 + c) = 50 + b 50 = b − 100c 100 · Similarly, plugging in (10,80) and doing a little algebra (do it now!) gives another linear equation 300 = b − 80c. Solving these two linear equations simultaneously gives c = 12.5 and b = 1300. Thus our function is y = 50x + 1300 x + 12.5 and, if Clyde holds 20 tickets, he will charge y = 50 20 + 1300 · 20 + 12.5 = $70.77 per ticket. 14.2 Summary • Every linear-to-linear rational function has a graph which is a shifted, scaled version of the curve y = 1/x. As a result, they have one vertical asymptote, and one horizontal asymptote. • Every linear-to-linear rational function f can be expressed in the form f(x) = ax + b x + c. This function has horizontal asymptote y = a and vertical asymptote x = −c. 14.3. EXERCISES 14.3 Exercises Problem 14.1. Give
the domain of each of the following functions. Find the x- and yintercepts of each function. Sketch a graph and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes. (a) f(x) = 2x x−1 (c) h(x) = x+1 x−2 (e) k(x) = 8x+16 5x− 1 2 (b) g(x) = 3x+2 2x−5 (d) j(x) = 4x−12 x+8 (f) m(x) = 9x+24 35x−100 Problem 14.2. Oscar is hunting magnetic ﬁelds with his gauss meter, a device for measuring the strength and polarity of magnetic ﬁelds. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. (a) Give a linear-to-linear rational model relating the meter reading y to how many feet x Oscar has gone into the room. (b) How far must he go for the meter to reach 10? 100? (c) Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is? Problem 14.3. In 1975 I bought an old MarIn 1995 a similar uke tin ukulele for $200. was selling for $900. In 1980 I bought a new Kamaka uke for $100. In 1990 I sold it for $400. (a) Give a linear model relating the price p of the Martin uke to the year t. Take t = 0 in 1975. (b) Give a linear model relating the price q of the Kamaka uke to the year t. Again take t = 0 in 1975. (c) When is the value of the Martin twice the value of the Kamaka? (d) Give a function f(t) which gives the ratio of the price of the Martin to the price of the Kamaka
. 189 (e) In the long run, what will be the ratio of the prices of the ukuleles? Problem 14.4. Isobel is producing and selling casette tapes of her rock band. When she had sold 10 tapes, her net proﬁt was $6. When she had sold 20 tapes, however, her net proﬁt had shrunk to $4 due to increased production expenses. But when she had sold 30 tapes, her net proﬁt had rebounded to $8. (a) Give a quadratic model relating Isobel’s net proﬁt y to the number of tapes sold x. (b) Divide the proﬁt function in part (a) by the number of tapes sold x to get a model relating average proﬁt w per tape to the number of tapes sold. (c) How many tapes must she sell in order to make $1.20 per tape in net proﬁt? Problem 14.5. Find the linear-to-linear function whose graph passes through the points (1,1), (5,2) and (20,3). What is its horizontal asymptote? Problem 14.6. Find the linear-to-linear function whose graph has y = 6 as a horizontal asymptote and passes through (0,10) and (3,7). Problem 14.7. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score is a linear-tolinear function of the number of hours that you study. If you want a score of 80%, how long do you need to study? Problem 14.8. A street light is 10 feet above a straight bike path. Olav is bicycling down the path at a rate of 15 MPH. At midnight, Olav is 33 feet from the point on the bike path directly (See the picture). The below the street light. relationship between the intensity C of light (in candlepower) and the distance d (in feet) from the light source is given by C = k d2, where k is a constant depending on the light source. 190 CHAPTER 14. RATIONAL
FUNCTIONS (a) From 20 feet away, the street light has an intensity of 1 candle. What is k? (b) Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds. (c) When will the light on Olav have maxi- mum intensity? (d) When will the intensity of the light be 2 candles? Problem 14.10. The number of customers in a local dive shop depends on the amount of money spent on advertising. If the shop spends nothing on advertising, there will be the shop spends 100 customers/day. $100, there will be 200 customers/day. As the amount spent on advertising increases, the number of customers/day increases and approaches (but never exceeds) 400 customers/day. If olav path 33ft 10ft Problem 14.9. For each of the following ﬁnd the linear to linear function f(x) satisfying the given requirements: (a) f(0) = 0, f(10) = 10, f(20) = 15 (b) f(0) = 10, f(5) = 4, f(20) = 3 (c) f(10) = 20, f(30) = 25, and the graph of f(x) has y = 30 as its horizontal asymptote (a) Find a linear to linear rational function y = f(x) that calculates the number y of customers/day if $x is spent on advertising. (b) How much must the shop spend on advertising to have 300 customers/day. (c) Sketch the graph of the function y = f(x) x on the domain 0 5000. ≤ ≤ (d) Find the rule, domain and range for the inverse function from part (c). Explain in words what the inverse function calculates. Chapter 15 Measuring an Angle So far, the equations we have studied have an algebraic character, involving the variables x and y, arithmetic operations and maybe extraction of roots. Restricting our attention to such equations would limit our ability to describe certain natural phenomena. An important example involves understanding motion around a circle, and it can be motivated by analyzing a very simple scenario: Cosmo the dog, tied by a 20 foot long tether to a post, begins walking around a circle. A number of very natural questions arise: Q S Cosmo motors around the circle P 20 feet R Figure 15.1: Cosmo the dog walking a
circular path. Natural Questions 15.0.1. How can we measure the angles ∠SPR, ∠QPR, and ∠QPS? How can we measure the arc lengths arc(RS), arc(SQ) and arc(RQ)? How can we measure the rate Cosmo is moving around the circle? If we know how to measure angles, can we compute the coordinates of R, S, and Q? Turning this around, if we know how to compute the coordinates of R, S, and Q, can we then measure the angles ∠SPR, ∠QPR, and ∠QPS? Finally, how can we specify the direction Cosmo is traveling? We will answer all of these questions and see how the theory which evolves can be applied to a variety of problems. The deﬁnition and basic properties of the circular functions will emerge as a central theme in this Chapter. The full problem-solving power of these functions will become apparent in our discussion of sinusoidal functions in Chapter 19. The xy-coordinate system is well equipped to study straight line motion between two locations. For problems of this sort, the important tool is the distance formula. However, as Cosmo has illustrated, not all two-dimensional motion is along a straight line. In this section, we will describe how to calculate length along a circular arc, which requires a quick review of angle measurement. 191 192 CHAPTER 15. MEASURING AN ANGLE 15.1 Standard and Central Angles An angle is the union of two rays emanating from a common point called the vertex of the angle. A typical angle can be dynamically generated by rotating a single ray from one position to another, sweeping counterclockwise or clockwise: See Figure 15.2. We often insert a curved arrow to indicate the direction in which we are sweeping out the angle. The ray ℓ1 is called the initial side and ℓ2 the terminal side of the angle ∠AOB. (terminal side) l2 B SWEEP CLOCKWISE O A l1 vertex O l1 (initial side) A vertex A O (initial side) l1 START SWEEP COUNTERCLOCKWISE (terminal side) B l2 Figure 15.2: Angle ∠AOB. Working with angles, we need to agree on a standard frame of reference for viewing them. Within the usual xy-coordinate system, imagine a model of �
�AOB in Figure 15.2 constructed from two pieces of rigid wire, welded at the vertex. Sliding this model around inside the xy-plane will not distort its shape, only its position relative to the coordinate axis. So, we can slide the angle into position so that the initial side is coincident with the positive x-axis and the vertex is the origin. Whenever we do this, we say the angle is in standard position. Once an angle is in standard position, we can construct a circle centered at the origin and view our standard angle as cutting out a particular “pie shaped wedge” of the corresponding disc. Notice, the shaded regions in Figure 15.3 depend on whether we sweep the angle counterclockwise or clockwise from the initial side. The portion of the “pie wedge” along the circle edge, which is an arc, is called the arc subtended by the angle. We can keep track of this arc using the notation arc(AB). A central angle is any angle with vertex at the center of a circle, but its initial side may or may not be the positive x-axis. For example, ∠QPS in the Figure beginning this Chapter is a central angle which is not in standard position. 15.2. AN ANALOGY 193 y-axis l2 B l2 y-axis B arc subtended arc(AB) arc(AB) vertex O x-axis l1 A vertex O x-axis l1 A COUNTERCLOCKWISE CLOCKWISE Figure 15.3: Standard angles and arcs. 15.2 An Analogy To measure the dimensions of a box you would use a ruler. In other words, you use an instrument (the ruler) as a standard against which you measure the box. The ruler would most likely be divided up into either English units (inches) or metric units (centimeters), so we could express the dimensions in a couple of different ways, depending on the units desired. By analogy, to measure the size of an angle, we need a standard against which any angle can be compared. In this section, we will describe two standards commonly used: the degree method and the radian method of angle measurement. The key idea is this: Beginning with a circular region, describe how to construct a “basic” pie shaped wedge whose interior angle becomes the standard unit of angle measurement. 15.3 Degree Method Begin by drawing a circle of radius r, call it Cr, centered
at the origin. Divide this circle into 360 equal sized pie shaped wedges, beginning with the the point (r,0) on the circle; i.e. the place where the circle crosses the x-axis. We will refer to the arcs along the outside edges of these wedges as one-degree arcs. Why 360 equal sized arcs? The reason for doing so is historically tied to the fact that the ancient Babylonians did so as they developed their study of astronomy. (There is actually an alternate system based on dividing the circular region into 400 equal sized wedges.) Any central angle which subtends one of these 360 equal sized arcs is 194 CHAPTER 15. MEASURING AN ANGLE circle Cr typical wedge (r,0) r r this angle is DEFINED to have measure 1 degree a total of 360 equal sized pie shaped wedges inside this disk etc. etc. *NOT TO SCALE* Figure 15.4: Wedges as 1◦ arcs. deﬁned to have a measure of one degree, denoted 1◦. We can now use one-degree arcs to measure any angle: Begin by sliding the angle ∠AOB into standard central position, as in Figure 15.3. Piece together consecutive one-degree arcs in a counterclockwise or clockwise direction, beginning from the initial side and working toward the terminal side, approximating the angle ∠AOB to the nearest degree. If we are allowed to divide a one-degree arc into a fractional portion, then we could precisely determine the number m of one-degree arcs which consecutively ﬁt together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, m is deﬁned to be the degree measure of the angle. If we sweep in a clockwise direction, then −m is deﬁned to be the degree measure of the angle. So, in Figure 15.3, the left-hand angle has positive degree measure while the right-hand angle has negative degree measure. Simple examples would be angles like the ones in Figure 15.5. Notice, with our conventions, the rays determining an angle with measure −135◦ sit inside the circle in the same position as those for an angle of measure 225◦; the minus sign keeps track of sweeping the positive x-axis clockwise (rather than counterclockwise). We can further divide a one-degree arc into 60 equal arcs,
each called a one minute arc. Each one-minute arc can be further divided into 60 equal arcs, each called a one second arc. This then leads to angle measures of one minute, denoted 1 ′ and one second, denoted 1 ′′: 1◦ = 60 minutes = 3600 seconds. 15.3. DEGREE METHOD 195 90◦ 180◦ 270◦ 45◦ 315◦ −135◦ Figure 15.5: Examples of common angles. For example, an angle of measure θ = 5 degrees 23 minutes 18 seconds is usually denoted 5◦ 23 ′ 18 ′′. We could express this as a decimal of degrees: 5◦ 23 ′ 18 ′′ = 5 + 23 60 = 5.3883◦. In degrees! + 18 3600 ◦ ← As another example, suppose we have an angle with measure 75.456◦ and we wish to convert this into degree/minute/second units. First, since 75.456◦ = 75◦+0.456◦, we need to write 0.456◦ in minutes by the calculation: 0.456 degree 60 × minutes degree = 27.36 ′. This tells us that 75.456◦ = 75◦27.36 ′ = 75◦27 ′ + 0.36 ′. Now we need to write 0.36 ′ in seconds via the calculation: 0.36 minutes 60 seconds/minute = 21.6 ′′. × In other words, 75.456◦ = 75◦ 27 ′ 21.6 ′′. Degree measurement of an angle is very closely tied to direction in the plane, explaining its use in map navigation. With some additional work, it is also possible to relate degree measure and lengths of circular arcs. To do this carefully, ﬁrst go back to Figure 15.3 and recall the situation 196 CHAPTER 15. MEASURING AN ANGLE where an arc arc(AB) is subtended by the central angle ∠AOB. In this situation, the arc length of arc(AB), commonly denoted by the letter s, is the distance from A to B computed along the circular arc; keep in mind, this is NOT the same as the straight line distance between the points A and B. For example, consider the six angles pictured above, of measures 90◦, 180�
�, 270◦, 45◦, −135◦, and 315◦. If the circle is of radius r and we want to compute the lengths of the arcs subtended by these six angles, then this can be done using the formula for the circumference of a circle (on the back of this text) and the following general principle: Important Fact 15.3.1. (length of a part) = (fraction of the part) (length of the whole) × For example, the circumference of the entire circle of radius r is 2πr; this is the “length of the whole” in the general principle. The arc subtended by a 90◦ angle is 90 4 of the entire circumference; this is the “fraction of the part” in the general principle. The boxed formula implies: 360 = 1 s = arc length of the 90◦ arc = 1 4 2πr = πr 2. s = distance along the arc r θ degrees r Cr Similarly, a 180◦ angle subtends an arc of length s = πr, 2πr = 7πr a 315◦ angle subtends an arc of length s = 4, etc. In general, we arrive at this formula: 315 360 Important Fact 15.3.2 (Arc length in degrees). Start with a central angle of measure θ degrees inside a circle of radius r. Then this angle will subtend an arc of length Figure 15.6: The deﬁnition of arc length. s = 2π 360 rθ 15.4 Radian Method The key to understanding degree measurement was the description of a “basic wedge” which contained an interior angle of measure 1◦; this was straightforward and familiar to all of us. Once this was done, we could proceed to measure any angle in degrees and compute arc lengths as in Fact 15.3.2. However, the formula for the length of an arc subtended by an angle measured in degrees is sort of cumbersome, involving the curious factor 2π 360. Our next goal is to introduce an alternate angle measurement scheme called radian measure that begins with a different “basic wedge”. As will become apparent, a big selling point of radian measure is that arc length calculations become easy. 15.4. RADIAN METHOD 197 r r equilateral wedge r (r,0) circle Cr r r r this angle is DEFINED to
have measure 1 radian Figure 15.7: Constructing an equilateral wedge. As before, begin with a circle Cr of radius r. Construct an equilateral wedge with all three sides of equal length r; see Figure 15.7. We deﬁne the measure of the interior angle of this wedge to be 1 radian. Once we have deﬁned an angle of measure 1 radian, we can deﬁne an angle of measure 2 radians by putting together two equilateral wedges. Likewise, an angle of measure 1 2 radian is obtained by symmetrically dividing an equilateral wedge in half, etc. Reasoning in this way, we can piece together equilateral wedges or fractions of such to compute the radian measure of any angle. It is important to notice an important relationship between the radian measure of an angle and arc length calculations. In the ﬁve angles pictured above, 1 radian, 2 radian, 3 radian, 1 2 radian and 1 4 radian, the length of the arcs 2 r, and 1 subtended by these angles θ are r, 2r, 3r, 1 4r. In other words, a pattern emerges that gives a very simple relationship between the length s of an arc and the radian measure of the subtended angle: Important Fact 15.4.1 (Arc length in radians). Start with a central angle of measure θ radians inside a circle of radius r. Then this angle will subtend an arc of length s = θr. These remarks allow us to summarize the deﬁnition of the radian measure θ of ∠AOB inside a circle of radius r by the formula: s = distance along the arc r θ radians r Cr θ = s r − s r if angle is swept counterclockwise if angle is swept clockwise Figure 15.9: Deﬁning arc length when angles are measured in radians. 198 CHAPTER 15. MEASURING AN ANGLE r r 1 radian = θ r 2r 3r r 2 radians = θ r r r 1 2 radian = θ r 2 3 radians = θ r r 1 4 radian = θ r r r 4 Figure 15.8: Measuring angles in radians. y-axis length of arc(AB) = s B θ radian O A x-axis The
units of θ are sometimes abbreviated as rad. It is important to appreciate the role of the radius of the circle Cr when using radian measure of an angle: An angle of radian measure θ will subtend an arc of length \|θ\| on the unit circle. In other words, radian measure of angles is exactly the same as arc length on the unit circle; we couldn’t hope for a better connection! Figure 15.10: Arc length after imposing a coordinate system. circle radius r The difﬁculty with radian measure versus degree measure is really one of familiarity. Let’s view a few common angles in radian measure. It is easiest to start with the case of angles in central standard position within the unit circle. Examples of basic angles would be fractional parts of one complete revolution around the unit circle; for example, 1 12 revolution, 1 4 revolution. One revolution around the unit circle describes an arc of length 2π and so the subtended angle (1 revolution) is 2π radians. We can now easily ﬁnd the radian measure of these six angles. For example, 1 12 revolution would describe an angle of measure ( 1 6 rad. Similarly, the other ﬁve angles pictured below have measures π 2 rad, π rad and 3π 2 rad. 2 revolution and 3 8 revolution, 1 4 revolution, 1 6 revolution, 1 12)2π rad= π 3 rad, π 4 rad, π All of these examples have positive radian measure. For an angle with 2 radians, we would locate B 4 revolution clockwise, etc. From these calculations and our negative radian measure, such as θ = − π by rotating 1 15.5. AREAS OF WEDGES 199 eplacements π 6 π 4 π 3 1 12 revolution 1 8 revolution 1 6 revolution π 2 π 3π 2 1 4 revolution 1 2 revolution 3 4 revolution Figure 15.11: Common angles measured in radians. previous examples of degree measure we ﬁnd that 180 degrees = π radians. (15.1) Solving this equation for degrees or radians will provide conversion formulas relating the two types of angle measurement. The formula also helps explain the origin of the curious conversion factor π 180 = 2π in Fact 15.3.2. 360 15.5 Areas of Wedges The beauty of radian measure is that
it is rigged so that we can easily compute lengths of arcs and areas of circular sectors (i.e. “pie-shaped regions”). This is a key reason why we will almost always prefer to work with radian measure. Example 15.5.1. If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice? This can be solved using a general principle: (Area of a part) = (area of the whole) (fraction of the part) × 200 CHAPTER 15. MEASURING AN ANGLE So, for our pizza: (area one slice) = (area whole pie) 1 12 = (82π) = 16π 3. (fraction of pie) × B Rθ θ O A Cr Figure 15.12: Finding the area of a “pie shaped wedge”. Let’s apply the same reasoning to ﬁnd the area of a circular sector. We know the area of the circular disc bounded by a circle of radius r is πr2. Let Rθ be the “pie shaped wedge” cut out by an angle ∠AOB with positive measure θ radians. Using the above principle area(Rθ) = (area of disc bounded by Cr) (portion of disc accounted for by Rθ) × = (πr2) θ 2π = 1 2 r2θ. For example, if r = 3 in. and θ = π the pie shaped wedge is 9 8π sq. in. 4 rad, then the area of Important Fact 15.5.2 (Wedge area). Start with a central angle with positive measure θ radians inside a circle of radius r. The area of the “pie shaped region” bounded by the angle is 1 2r2θ. Example 15.5.3. A water drip irrigation arm 1200 feet long rotates around a pivot P once every 12 hours. How much area is covered by the arm in one hour? in 37 minutes? How much time is required to drip irrigate 1000 square feet? Solution. The irrigation arm will complete one revolution in 12 hours. The angle swept out by one complete revolution is 2π radians, so after t hours the arm sweeps out an angle θ(t) given by θ(t) = 2π radians 12 hours × t hours = π 6 t rad
ians. Consequently, by Important Fact 15.5.2 the area A(t) of the irrigated region after t hours is A(t) = 1 2 (1200)2θ(t) = 1 2 (1200)2 π 6 t = 120,000πt square feet. After 1 hour, the irrigated area is A(1) = 120,000π = 376,991 sq. ft. Likewise, after 37 minutes, which is 37 60 hours, the area of the irrigated region is A( 37 60) = 232,500 square feet. To answer the ﬁnal question, we need to solve the equation A(t) = 1000; i.e., 120,000πt = 1000, so 60) = 120,000π( 37 t = 1 120π hours × 3600 seconds hour = 9.55 seconds. 15.5. AREAS OF WEDGES 201 15.5.1 Chord Approximation Our ability to compute arc lengths can be used as an estimating tool for distances between two points. Let’s return to the situation posed at the beginning of this section: Cosmo the dog, tied by a 20 foot long tether to a post in the ground, begins at location R and walks counterclockwise to location S. Furthermore, let’s suppose you are standing at the center of the circle determined by the tether and you measure the angle from R to S to be 5◦; see the left-hand ﬁgure. Because the angle is small, notice that the straight line distance d from R to S is approximately the same as the arc length s subtended by the angle ∠RPS; the right-hand picture in Figure 15.13 is a blow-up: 5◦ P 20 feet S R S s d 5◦ P 20 feet R Figure 15.13: Using the arc length s to approximate the chord d. Example 15.5.4. Estimate the distance from R to S. Solution. We ﬁrst convert the angle into radian measure via (15.1): 5◦ = 0.0873 radians. By Fact 15.3.2, the arc s has length 1.745 feet = 20.94 inches. This is approximately equal to the distance from R to S, since the angle is small. 202 CHAPTER 15. MEASURING AN ANGLE S chord s R O We
call a line segment connecting two points on a circle a chord of the circle. The above example illustrates a general principal for approximating the length of any chord. A smaller angle will improve the accuracy of the arc length approximation. Important Fact 15.5.5 (Chord Approximation). In Figure 15.14, if the central angle is small, then s \|RS\|. ≈ Figure 15.14: Chord approximation. 15.6 Great Circle Navigation A basic problem is to ﬁnd the shortest route between any two locations on the earth. We will review how to coordinatize the surface of the earth and recall the fact that the shortest path between two points is measured along a great circle. View the earth as a sphere of radius r = 3,960 miles. We could slice P0 which is both perpendicular the earth with a two-dimensional plane to a line connecting the North and South poles and passes through the center of the earth. Of course, the resulting intersection will trace out a circle of radius r = 3,960 miles on the surface of the earth, which we call the equator. We call the plane P0 the equatorial plane. Slicing the earth with any other plane P0, we can consider the right triangle pictured below and the angle θ: parallel to P North Pole line of latitude North Pole r equatorial plane P0 equator b r θ r equator b θ r 90◦ − θ◦ South Pole center of earth center of earth South Pole Figure 15.15: Measuring latitude. Essentially two cases arise, depending on whether or not the plane P slices the surface of is above or below the equatorial plane. The plane the earth in a circle, which we call a line of latitude. This terminology is somewhat incorrect, since these lines of latitude are actually circles on the surface of the earth, but the terminology is by now standard. Depending on whether this line of latitude lies above or below the equatorial plane, we refer to it as the θ◦ North line of latitude (denoted θ◦ N) or the θ◦ South line of latitude (denoted θ◦ S). Notice, the radius b of a line of latitude can vary from a maximum of 3,960 miles (in the case of θ = 0◦), P 15.6. GREAT CIRCLE NAVIGATION 203 to a minimum of 0 miles, (when θ = 90◦
). When b = 0, we are at the North or South poles on the earth. In a similar spirit, we could imagine slicing the earth with a plane Q which is perpendicular to the equatorial plane and passes through the center of the earth. The resulting intersection will trace out a circle of radius 3,960 miles on the surface of the earth, which is called a line of longitude. Half of a line of longitude from the North Pole to the South Pole is called a meridian. We distinguish one such meridian; the one which passes through Greenwich, England as the Greenwich meridian. Longitudes are measured using angles East or West of Greenwich. Pictured below, the longitude of A is θ. Because θ is east of Greenwich, θ measures longitude East, typically written θ◦ E; west longitudes would be denoted as θ◦ W. All longitudes are between 0◦ and 180◦. The meridian which is 180◦ West (and 180◦ E) is called the International Date Line. Introducing the grid of latitude and longitude lines on the earth amounts to imposing a coordinate system. In other words, any position on the earth can be determined by providing the longitude and latitude of the point. The usual convention is to list longitude ﬁrst. For example, Seattle has coordinates 122.0333◦ W, 47.6◦ N. Since the labels “N and S” are attached to latitudes and the labels “E and W” are attached to longitudes, there is no ambiguity here. This means that Seattle is on the line of longitude 122.0333◦ West of the Greenwich meridian and on the line of latitude 47.6◦ North of the equator. In the ﬁgure below, we indicate the key angles ψ = 47.6◦ and θ = 122.0333◦ by inserting the three indicated radial line segments. Greenwich meridian center of earth θ r A r Greenwich, England North Pole line of longitude International Date Line Equator South Pole Figure 15.16: The International Date Line. Seattle, WA N great circles not great circle Ψ r θ S Greenwich Meridian not great circle Figure 15.17: Distances along great circles. 204 CHAPTER 15. MEASURING AN ANGLE Now that we have imposed a coordinate system on the earth, it is natural
to study the distance between two locations. A great circle of a sphere is deﬁned to be a circle lying on the sphere with the same center as the sphere. For example, the equator and any line of longitude are great circles. However, lines of latitude are not great circles (except the special case of the equator). Great circles are very important because they are used to ﬁnd the shortest distance between two points on the earth. The important fact from geometry is summarized below. Important Fact 15.6.1 (Great Circles). The shortest distance between two points on the earth is measured along a great circle connecting them. Example 15.6.2. What is the shortest distance from the North Pole to Seattle, WA? W E N O S equator Greenwich Meridian Figure 15.18: Distance between the North Pole and Seattle, Washington. Solution. The line of longitude 122.0333◦ W is a great circle connecting the North Pole and Seattle. So, the shortest distance will be the arc length s subtended by the angle ∠NOW pictured in Figure 15.18. Since the latitude of Seattle is 47.6◦, the angle ∠EOW has measure 47.6◦. Since ∠EON is a right angle (i.e., 90◦), ∠NOW has measure 42.4◦. By Fact 15.4.1 and Equation 15.1, s = (3960 miles)(42.4◦)(0.01745 radians/degree) = 2943.7 miles, which is the shortest distance from the pole to Seattle. 15.7 Summary 360◦ = 2π radians A circular arc with radius r and angle θ has length s, with s = rθ when θ is measured in radians. A circular wedge with radius r and angle θ has area A, with • • • A = 1 2 r2θ when θ is measured in radians. 15.8. EXERCISES 15.8 Exercises Problem 15.1. Let ∠AOB be an angle of measure θ. (a) Convert θ = 13.4o into degrees/ min- utes/ seconds and into radians. (b) Convert θ = 1o4 ′44 ′′ into degrees and ra- dians. (c) Convert θ = 0.1 radian into degrees and
degrees/ minutes/ seconds. Problem 15.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is deﬁned as the length of an arc s along a great circle on the earth when the subtending angle has measure 1 ′ = “one minute” = 1/60 of one degree. Assume the radius of the earth is 3,960 miles. (a) Find the length of one nautical mile to the nearest 10 feet. (b) A vessel which travels one nautical mile in one hours time is said to have the speed of one knot; this is the usual navigational measure of speed. If a vessel is traveling 26 knots, what is the speed in mph (miles per hour)? (c) If a vessel is traveling 18 mph, what is the speed in knots? Problem 15.3. The rear window wiper blade on a station wagon has a length of 16 inches. The wiper blade is mounted on a 22 inch arm, 6 inches from the pivot point. 6" 16" 205 (c) Suppose bug A lands on the end of the blade farthest from the pivot. Assume the wiper turns through an angle of 110◦. In one cycle (back and forth) of the wiper blade, how far has the bug traveled? (d) Suppose bug B lands on the end of the wiper blade closest to the pivot. Assume the wiper turns through an angle of 110◦. In one cycle of the wiper blade, how far has the bug traveled? (e) Suppose bug C lands on an intermediate location of the wiper blade. Assume the wiper turns through an angle of 110◦. If bug C travels 28 inches after one cycle of the wiper blade, determine the location of bug C on the wiper blade. Problem 15.4. A water treatment facility operates by dripping water from a 60 foot long arm whose end is mounted to a central pivot. The water then ﬁlters through a layer of charcoal. The arm rotates once every 8 minutes. (a) Find the area of charcoal covered with water after 1 minute. (b) Find the area of charcoal covered with water after 1 second. (c) How long would it take to cover 100 square feet of charcoal with water? (d) How long would it take to cover 3245 square feet of charcoal with water? (a) If the wiper turns through an
angle of 110◦, how much area is swept clean? (b) Through how much of an angle would the wiper sweep if the area cleaned was 10 square inches? Problem 15.5. Astronomical measurements are often made by computing the small angle formed by the extremities of a distant object and using the estimating technique in 15.5.1. In the picture below, the full moon is shown to form an angle of 1 when the distance indi2 cated is 248,000 miles. Estimate the diameter of the moon. o 206 CHAPTER 15. MEASURING AN ANGLE moon (a) θ = 45◦ (b) θ = 80o (c) θ = 3 radians (d) θ = 2.46 radians (e) θ = 97o23 ′3 ′′ (f) θ = 35o24 ′2 ′′ 248,000 miles o 1/2 earth Problem 15.6. An aircraft is ﬂying at the speed of 500 mph at an elevation of 10 miles above the earth, beginning at the North pole and heading South along the Greenwich meridian. A spy satellite is orbiting the earth at an elevation of 4800 miles above the earth in a circular orbit in the same plane as the Greenwich meridian. Miraculously, the plane and satellite always lie on the same radial line from the center of the earth. Assume the radius of the earth is 3960 miles. satellite plane earth Problem 15.8. Matilda is planning a walk around the perimeter of Wedge Park, which is shaped like a circular wedge, as shown below. The walk around the park is 2.1 miles, and the park has an area of 0.25 square miles. If θ is less than 90 degrees, what is the value of the radius, r? θ r r Problem 15.9. Let C6 be the circle of radius 6 inches centered at the origin in the xycoordinate system. Compute the areas of the shaded regions in the picture below; the inner circle in the rightmost picture is the unit circle: y=x y=−x (a) When is the plane directly over a location with latitude 74◦30 ′18 ′′ N for the ﬁrst time? (b) How fast is the satellite moving? (c) When is the plane directly over the equa- tor and how far has
swept out per unit time” by the moving object, starting from some initial position. We need to somehow indicate the direction in which the angle is being swept out. This can be done by indicating “clockwise” our “counterclockwise”. Alternatively, we can adopt the convention that the positive rotational direction is counterclockwise, then insert a minus sign to indicate rotation clockwise. For example, saying that Cosmo is moving at an angular speed of ω = − π rad sec means he 2 is moving clockwise π 2 rad sec. Another way to study the rate of a circular motion is to count the number of complete circuits of the circle per unit time. This sort of rate has the form Number of Revolutions Unit of Time ; If we take “minutes” to be we will also view this as an angular speed. the preferred unit of time, we arrive at the common measurement called revolutionsper minute, usually denoted RPM or rev/min. For example, if Cosmo completes one trip around the circle every 2 minutes, then Cosmo is moving at a rate of 1 2 RPM. If instead, Cosmo completes one trip around the circle every 12 seconds, then we could ﬁrst express Cosmo’s speed in units of revolutions/second as 1 12 rev/second, then convert to RPM units: 1 12 rev sec 60 sec min = 5 RPM. As a variation, if we measure that Cosmo completed 3 2 minutes, then Cosmo’s angular speed is computed by 7 of a revolution in 3 7rev 2 min = 3 14 RPM. The only possible ambiguity involves the direction of revolution: the object can move clockwise or counterclockwise. The one shortcoming of using angular speed is that we are not directly keeping track of the distance the object is traveling. This is fairly easy to remedy. Returning to Figure 16.1, the circumference of the circle of motion is 2π(20) = 40π feet. This is the distance traveled per revolution, so we can now make conversions of angular speed into “distance traveled per unit time”; this is called the linear speed. 2 RPM, then he has a linear speed of If Cosmo is moving 1 v = 1 rev 2 min 40π ft rev = 20π ft min. Likewise, if Cosmo is moving π 7 rad sec, then v = π rad 7 sec 1 rev 2π rad 40π ft rev =
20π 7 ft sec. 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 209 Important Fact 16.1.1. This discussion is an example of what is usually called “units analysis”. The key idea we have illustrated is how to convert between two different types of units: rev min converts to − ft min 16.2 Different Ways to Measure → Circular Motion The discussion of Cosmo applies to circular motion of any object. As a matter of convention, we usually use the Greek letter ω to denote angular speed and v for linear speed. If an object is moving around a circle of radius r at a constant rate, then we can measure it’s speed in two ways: • • The angular speed ω = “revolutions” “unit time” or “degrees swept” “unit time” or “radians swept” “unit time”. The linear speed v = “distance traveled” “per unit time”. Important Facts 16.2.1 (Measuring and converting). We can convert between angular and linear speeds using these facts: 1 revolution = 360◦ = 2π radians; The circumference of a circle of radius r units is 2πr units. • • 16.2.1 Three Key Formulas If an object begins moving around a circle, there are a number of quantities we can try to relate. Some of these are “static quantities”: Take a visual “snapshot” of the situation after a certain amount of time has elapsed, then we can measure the radius, angle swept, arc length and time elapsed. Other quantities of interest are “dynamic quantities”: This means something is CHANGING with respect to time; in our case, the linear speed (which measures distance traveled per unit time) and angular speed (which measures angle swept per unit time) fall into this category. 210 CHAPTER 16. MEASURING CIRCULAR MOTION STATIC QUANTITIES DYNAMIC QUANTITIES S 1. arc length s S 5. angular speed ω 2. angle swept in time r 6. linear speed v θ P R 3. radius r 4. elapsed time t P R...take a “snapshot” after time t......see what happens per unit time... Figure 16.2: Measuring linear and angular speed. We
now know two general relationships for circular motion: (i) s = rθ, where s=arclength (a linear distance), r=radius of the circular path and θ=angle swept in RADIAN measure; this was the content of Fact 15.4.1 on page 197. (ii) θ = ωt, where θ is the measure of an angle swept, ω= angular speed and t represents time elapsed. This is really just a consequence of units manipulation. If r=20 feet and θ = 1.3 Notice how the units work in these formulas. radians, then the arc length s = 20(1.3) feet= 26 feet; this is the length If ω = 3 of the arc of radius 20 feet that is subtending the angle θ. rad/second and t = 5 seconds, then θ = 3 rad 5 seconds = 15 radians. If we replace “θ” in s = rθ of (i) with θ = ωt in (ii), then we get seconds × s = rωt. This gives us a relationship between arclength s (a distance) and time t. Plug in the fact that the linear speed is deﬁned to be v = “distance” and we get t v = s t = rωt t = rω. All of these observations are summarized below. Important Facts 16.2.2 (Three really useful formulas). If we measure angles θ in RADIANS and ω in units of radians per unit time, we have these three formulas: s = rθ θ = ωt v = rω (16.1) (16.2) (16.3) 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 211 Example 16.2.3. You are riding a stationary exercise bike and the speedometer reads a steady speed of 40 MPH (miles per hour). If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear tire. A pebble becomes stuck to the tread of the rear tire. Describe the location of the pebble after 1 second and 0.1 second. * pebble sticks to tread here Figure 16.3: Where is the pebble after t seconds? Solution. The tires will be rotating in a counterclock
wise direction and the radius r = 1 228 = 14 inches. The other given quantity, “40 MPH”, involves miles, so we need to decide which common units to work with. Either will work, but since the problem is focused on the wheel, we will utilize inches. If the speedometer reads 40 MPH, this is the linear speed of a speciﬁed location on the rear tire. We need to convert this into an angular speed, using unit conversion formulas. First, the linear speed of the wheel is v = 40 = 704 miles hr in sec. 5280 ft mile in ft 12 1 hr 60 min 1 min 60 sec Now, the angular speed ω of the wheel will be ω = 704 inches second 2(14)π inches revolution revolution second = 8 = 480 RPM It is then an easy matter to convert this to ω = 8 = 2,880 revolution second degrees second. 360 degrees revolution If the pebble begins at the “6 o’clock” position (the place the tire touches the ground on the wheel), then after 1 second the pebble will go through 8 revolutions, so will be in the “6 o’clock” position again. After 0.1 seconds, the pebble will go through rev sec 8 (0.1 sec) = 0.8 rev = (0.8 rev) 360 = 288◦. deg rev 212 CHAPTER 16. MEASURING CIRCULAR MOTION Keeping in mind that the rotation is counterclockwise, we can view the location of the pebble after 0.1 seconds as pictured below: 288◦ wise rotation counterclock- after 0.1 second * located here at time = 0.1 sec starts here * pebble sticks to tread in 6 o’clock position Figure 16.4: Computing the pebble’s position after t = 0.1 sec. We solved the previous problem using the “unit conversion method”. There is an alternate approach available, which uses one of the formulas in Fact 16.2.2. Here is how you could proceed: First, as above, we know the linear speed is v = 704 in/sec. Using the “v = ωr” formula, we have 704 in sec = ω(14 in) ω = 50.28 rad sec. Notice how the units worked out in the calculation: the
“time” unit comes from v and the “angular” unit will always be radians. As a comparison with the solution above, we can convert ω into RPM units: rad sec 1 rev 2π rad ω = 50.28 = 8 rev sec. All of the problems in this section can be worked using either the “unit conversion method” or the “v = ωr method”. 16.3 Music Listening Technology The technology of reproducing music has gone through a revolution since the early 1980’s. The “old” stereo long playing record (the LP) and the 16.3. MUSIC LISTENING TECHNOLOGY 213 “new” digital compact disc (the CD) are two methods of storing musical data for later reproduction in a home stereo system. These two technologies adopt different perspectives as to which notion of circular speed is best to work with. Long playing stereo records are thin vinyl plastic discs of radius 6 inches onto which small spiral grooves are etched into the surface; we can approximately view this groove as a circle. The LP is placed on a ﬂat 12 inch diameter platter which turns at a constant angular speed of 33 1 3 RPM. An arm on a pivot (called the tone arm) has a needle mounted on the end (called the cartridge), which is placed in the groove on the outside edge of the record. Because the grooves wobble microscopically from side-to-side, the needle will mimic this motion. In turn, this sets a magnet (mounted on the opposite end of the needle) into motion. This moving magnet sits inside a coil of wire, causing a small varying voltage; the electric signal is then fed to your stereo, ampliﬁed and passed onto your speakers, reproducing music! LP turning at 33 1 3 RPM amp speakers tonearm needle Figure 16.5: Reproducing music using analogue technology. This is known as analogue technology and is based upon the idea of maintaining a constant angular speed of 33 1 3 RPM for the storage medium (our LP). (Older analogue technologies used 45 RPM and 78 RPM records. However, 33 1 3 RPM became the consumer standard for stereo music.) With an LP, the beginning of the record (the lead-in groove) would be on the outermost edge of the record and the end of the record (the exit groove) would be close to the center. Placing the needle in the
lead-in groove, the needle gradually works its way to the exit groove. However, whereas the angular speed of the LP is a constant 33 1 3 RPM, the linear speed at the needle can vary quite a bit, depending on the needle location. 214 CHAPTER 16. MEASURING CIRCULAR MOTION 6 1 Figure 16.6: Lead-in and exit grooves. Example 16.3.1 (Analogue LP’s). The “lead-in groove” is 6 inches from the center of an LP, while the “exit groove” is 1 inch from the center. What is the linear speed (MPH) of the needle in the “lead-in groove”? What is the linear speed (MPH) of the needle in the “exit groove”? Find the location of the needle if the linear speed is 1 MPH. Solution. This is a straightforward application of Fact 16.2.1. Let v6 (resp. v1) be the linear speed at the lead in groove (resp. exit groove); the subscript keeps track of the needle radial location. Since the groove is approximately a circle, 2(6)π inches rev v6 = 33 1 3 = 1257 rev min in min 1257 in min 5280 ft mile = 1.19 MPH = 60 min hour 12 in ft Similarly, v1 = 0.2 MPH. To answer the remaining question, let r be the radial distance from the center of the LP to the needle location on the record. If vr = 1 MPH: 1 mile hour = vr = 33 1 3 rev min 2rπ in rev 60 min hr 1 ft 12 in 1 mile 5280 ft So, when the needle is r = 5.04 inches from the center, the linear speed is 1 MPH. laser laser support arm moves back and forth spinning CD Figure 16.7: Reproducing music using digital technology. In the early 1980’s, a new method of storing and reproducing music was introduced; this medium is called the digital compact disc, referred to as a CD for short. This is a thin plastic disc of diameter 4.5 inches, which appears to the naked eye to have a shiny silver coating on one side. Upon microscopic examination one would ﬁnd concentric circles of pits in the silver coating. This disc is placed in a CD player, which spins the disc. A laser located above the spinning disc will project onto the spinning
disc. The pits in the silver coating will cause the reﬂected laser light to vary in intensity. A sensor detects this variation, converting it to a digital signal (the analogue to digital or AD conversion). This is fed into a digital to analogue or DA conversion device, which sends a signal to your stereo, again producing music. 16.4. BELT AND WHEEL PROBLEMS 215 The technology of CD ′s differs from that of LP ′s in two crucial ways. First, the circular motion of the spinning CD is controlled so that the target on the disc below the laser is always moving at a constant linear speed of 1.2 meters minute. Secondly, the beginning location of the laser will be on the inside portion of the disc, working its way outward to the end. In this context, it makes sense to study how the angular speed of the CD is changing, as the laser position changes. sec = 2,835 inches Example 16.3.2 (Digital CD’s). What is the angular speed (in RPM) of a CD if the laser is at the beginning, located 3 4 inches from the center of the disc? What is the angular speed (in RPM) of a CD if the laser is at the end, located 2 inches from the center of the disc? Find the location of the laser if the angular speed is 350 RPM. Solution. This is an application of Fact 16.2.1. Let ω3/4 be the angular speed at the start and ω2 the angular speed at the end of the CD; the subscript is keeping track of the laser distance from the CD center. ω2 = (2835 inches/min) (2(2)π inches/rev) = 225.6 RPM ω3/4 = 2835 inches/min) (2(0.75)π inches/rev) = 601.6 RPM ′′ 2 ′′ 3 4 Start of CD End of CD Figure 16.8: Computing the angular speed of a CD. To answer the remaining question, let r be the radial distance from the center of the CD to the laser location on the CD. If the angular speed ωr at this location is 350 RPM, we have the equation 350 RPM = ωr 2,835 inches minute 2rπ inches revolution 1.289 inches = r. = So, when the laser is 1.289 inches from the center, the CD is moving
350 RPM. 16.4 Belt and Wheel Problems The industrial revolution spawned a number of elaborate machines involving systems of belts and wheels. Computing the speed of various belts and wheels in such a system may seem complicated at ﬁrst glance. The situation can range from a simple system of two wheels with a belt connecting them, to more elaborate designs. We call problems of this sort belt and wheel problems, or more generally, connected wheel problems. Solving problems of this type always uses the same strategy, which we will ﬁrst highlight by way of an example. 216 CHAPTER 16. MEASURING CIRCULAR MOTION Figure 16.9: Two typical connected wheel scenarios. front sprocket radius = 5 inches rear sprocket radius = 2 inches (a) A stationary exercise bike. radius A = 14 inches A B radius B = 2 inches C radius C = 5 inches (b) A model of the bike’s connected wheels. Figure 16.10: Visualizing the connected wheels of an exercise bike. Example 16.4.1. You are riding a stationary exercise bike. Assume the rear wheel is 28 inches in diameter, the rear sprocket has radius 2 inches and the front sprocket has radius 5 inches. How many revolutions per minute of the front sprocket produces a forward speed of 40 MPH on the bike (miles per hour)? Solution. There are 3 wheels involved with a belt (the bicycle chain) connecting two of the wheels. In this problem, we are provided with the linear speed of wheel A (which is 40 MPH) and we need to ﬁnd the angular speed of wheel C=front sprocket. Denote by vA, vB, and vC the linear speeds of each of the wheels A, B, and C, respectively. Likewise, let ωA, ωB, and ωC denote the angular speeds of each of the wheels A, B, and C, respectively. In addition, the chain connecting the wheels B and C will have a linear speed, which we will denote by vchain. The strategy is broken into a sequence of steps which leads us from the known linear speed vA to the angular speed ωC of wheel C: • • • • Step 1: Given vA, ﬁnd ωA. Use the fact ωA = vA rA. Step 2: Observe ωA = ωB; this is because the wheel and rear
sprocket are both rigidly mounted on a common axis of rotation. Step 3: Given ωB, ﬁnd vB. Use the fact vB = rBωB = rBωA = vA. rB rA Step 4: Observe vB = vchain = vC; this is because the chain is directly connecting the two sprockets and assumed not to slip. 16.4. BELT AND WHEEL PROBLEMS 217 • Step 5: Given vC, ﬁnd ωC. Use the fact ωC = vC rC vB rC rB rArC vA. = = Saying that the speedometer reads 40 MPH is the same as saying that the linear speed of a location on the rear wheel is vA = 40 MPH. Converting this into angular speed was carried out in our solution to Example 16.2.3 above; we found that ωA = 480 RPM. This completes Step 1 and so by Step 2, ωA = ωB = 480 RPM. For Step 3, we convert ωB = 480 RPM into linear speed following Fact 16.2.1: vB = 480 = 6,032 revolution minute inches minute. (2(2)π) inches revolution By Step 4, conclude that the linear speed of wheel C is vC = 6,032 inches/min. Finally, to carry out Step 5, we convert the linear speed into angular speed: ωC = 6,032 inches min 2(5)π inches rev = 192 RPM = 3.2. rev sec In conclusion, the bike rider must pedal the front sprocket at the rate of 3.2 rev sec. This example indicates the basic strategy used in all belt/wheel prob- lems. Important Facts 16.4.2 (Belt and Wheel Strategy). Three basic facts are used in all such problems: • • Using “unit conversion” or Fact 16.2.2 allows us to go from linear speed v to angular speed ω, and vice versa. If two wheels are fastened rigidly to a common axle, then they have the same angular speed. (Caution: two wheels fastened to a common axle typically do not have the same linear speed!) • If two wheels are connected by a belt (or chain), the linear speed of the belt coincides with the linear speed of each wheel. 218 CHAPTER 16. MEASURING
CIRCULAR MOTION 16.5 Exercises Problem 16.1. The restaurant in the Space Needle in Seattle rotates at the rate of one revolution per hour. (a) Through how many radians does it turn in 100 minutes? (b) How long does it take the restaurant to rotate through 4 radians? (c) How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters? Problem 16.2. You are riding a bicycle along a level road. Assume each wheel is 26 inches in diameter, the rear sprocket has a radius of 3 inches and the front sprocket has a radius of 7 inches. How fast do you need to pedal (in revolutions per minute) to achieve a speed of 35 mph? front wheel rear wheel rear sprocket front sprocket Problem 16.3. Answer the following angular speed questions. (a) A wheel of radius 22 ft. is rotating 11 RPM counterclockwise. Considering a point on the rim of the rotating wheel, what is the angular speed ω in rad/sec and the linear speed v in ft/sec? (b) A wheel of radius 8 in. is rotating 15o/sec. What is the linear speed v, the angular speed in RPM and the angular speed in rad/sec? (c) You are standing on the equator of the earth (radius 3960 miles). What is your linear and angular speed? Problem 16.4. Lee is running around the perimeter of a circular track at a rate of 10 ft/sec. The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns. (a) Sketch a picture of the situation. (b) How far has Lee traveled once he returns to his starting position? (c) How much time will elapse during Lee’s circuit? (d) Find the area of the pie shaped sector enclosed by Lee’s path. Problem 16.5. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the worlds greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel
should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. 12 RPM θ P 4 feet (a) Find the radius of the ferris wheel. (b) Once the wheel is built, John suggests that Tiff should take the ﬁrst ride. The wheel starts turning when Tiff is at the location P, which makes an angle θ with the horizontal, as pictured. It takes her 1.3 seconds to reach the top of the ride. Find the angle θ. (d) An auto tire has radius 12 inches. If you are driving 65 mph, what is the angular speed in rad/sec and the angular speed in RPM? (c) Poor engineering causes Tiff’s seat to ﬂy off in 6 seconds. Describe where Tiff is located (an angle description) the instant she becomes a human missile. 16.5. EXERCISES 219 Problem 16.6. Michael and Aaron are on the “UL-Tossum” ride at Funworld. This is a merry-go-round of radius 20 feet which spins counterclockwise 60 RPM. The ride is driven by a belt connecting the outer edge of the ride to a drive wheel of radius 3 feet: Drive wheel radius 3 ft drive belt Aaron O P Michael main ride radius 20 ft (a) Assume Michael is seated on the edge of the ride, as pictured. What is Michael’s linear speed in mph and ft/sec? (b) What is the angular speed of the drive wheel in RPM? (c) Suppose Aaron is seated 16 feet from the center of the ride. What is the angular speed of Aaron in RPM? What is the linear speed of Aaron in ft/sec? (d) After 0.23 seconds Michael will be located at S as pictured. What is the angle ∠POS in degrees? What is the angle ∠POS in radians? How many feet has Michael traveled? S θ (e) Assume Michael has traveled 88 feet from the position P to a new position Q. How many seconds will this take? What will be the angle swept out by Michael? length of arc (PQ) is 88 ft O P Q Problem 16.7. You are riding a bicycle along a level road. Assume each wheel is 28 inches in diameter, the rear
sprocket has radius 3 inches and the front sprocket has radius r inches. Suppose you are pedaling the front sprocket at the rate of 1.5 rev sec and your forward speed is 11 mph on the bike. What is the radius of the front sprocket? Problem 16.8. You are designing a system of wheels and belts as pictured below. You want wheel A to rotate 20 RPM while wheel B rotates 42 RPM. Wheel A has a radius of 6 inches, wheel B has a radius of 7 inches and wheel C has a radius of 1 inch. Assume wheels C and D are rigidly fastened to the same axle. What is the radius r of wheel D? O P A D C B 220 CHAPTER 16. MEASURING CIRCULAR MOTION Chapter 17 The Circular Functions SupposeCosmobeginsatlocation R andwalksinacounterclockwise direction, always maintaining a tight 20 ft long tether. As Cosmo moves around the circle, how can we describe his location at any given instant? In one sense, we have already answered this question: The measure of ∠RPS1 exactly pins down a location on the circle of radius 20 feet. But, we really might prefer a description of the horizontal and vertical coordinates of Cosmo; this would tie in better with the coordinate system we typically use. Solving this problem will require NEW functions, called the circular functions. S2 S3 S1 P 20 feet R S4 Figure 17.1: Cosmo moves counterclockwise maintaining a tight tether. Where’s Cosmo? 17.1 Sides and Angles of a Right Triangle Example 17.1.1. You are preparing to make your ﬁnal shot at the British Pocket Billiard World Championships. The position of your ball is as in Figure 17.2, and you must play the ball off the left cushion into the lower-right corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the cushion? Solution. This problem depends on two basic facts. First, the angles of entry and exit between the path the cushion will be equal. Secondly, the two obvious right triangles in this picture are similar triangles. Let x represent the distance from the bottom left corner to the impact point of the ball’s path: Properties of similar triangles tell us that the ratios of 5−x = 12 x. If we solve this equation 4 = 3.75 feet. common sides
are equal: for x, we obtain x = 15 4 221 The billiard table layout. 4 ft 5 ft ﬁnd this location this pocket for the big money 6 ft 12 ft 4 θ θ 5 − x x Mathmatically modeling the bank shot. Figure 17.2: A pocket billiard banking problem. 222 CHAPTER 17. THE CIRCULAR FUNCTIONS This discussion is enough to win the tourney. But, of course, there are still other questions we can ask about this simple example: What is the angle θ? That is going to require substantially more work; indeed the bulk of this Chapter! It turns out, there is a lot of mathematical mileage in the idea of studying ratios of sides of right triangles. The ﬁrst step, which will get the ball rolling, is to introduce new functions whose very deﬁnition involves relating sides and angles of right triangles. 17.2 The Trigonometric Ratios B hypotenuse side opposite θ A θ side adjacent θ C Figure 17.3: Labeling the sides of a right triangle. From elementary geometry, the sum of the angles of any triangle will equal 180◦. Given a right triangle ABC, since one of the angles is 90◦, the remaining two angles must be acute angles; i.e., angles of measure between 0◦ and 90◦. If we specify one of the acute angles in a right triangle ABC, say angle θ, we can label the three sides using △ this terminology. We then consider the following three ratios of side lengths, referred to as trigonometric ratios: △ sin(θ) def= length of side opposite θ length of hypotenuse cos(θ) def= length of side adjacent θ length of hypotenuse tan(θ) def= length of side opposite θ length of side adjacent to θ. (17.1) (17.2) (17.3) 2, cos(θ) = √3 13, tan(θ) = 5 12., cos(θ) = 1 √2 2, tan(θ) = 1 √3 For example, we have three right triangles in Figure 17.4; you can verify that the Pythagorean Theorem holds in each of the cases. In the left-hand triangle, sin(θ) = 5 13, cos(θ) = 12
In the middle triangle, sin(θ) = 1, tan(θ) = 1. In the right-hand triangle, √2 sin(θ) = 1. The symbols “sin”, “cos”, and “tan” are abbreviations for the words sine, cosine and tangent, respectively. As we have deﬁned them, the trigonometric ratios depend on the dimensions of the triangle. However, the same ratios are obtained for any right triangle with acute angle θ. This follows from the properties of similar ADE are similar. If triangles. Consider Figure 17.5. Notice \|AB\|. On the other we use \|AD\|. Since the ratios of com= \|AE\| \|AD\|, △ ADE, we obtain cos(θ) = \|AE\| hand, if we use mon sides of similar triangles must agree, we have cos(θ) = \|AC\| \|AB\| ABC to compute cos(θ), then we ﬁnd cos(θ) = \|AC\| ABC and △ △ △ 17.2. THE TRIGONOMETRIC RATIOS 223 13 θ 12 5 √2 θ 1 1 2 θ √3 1 Figure 17.4: Computing trigonometric ratios for selected right triangles. which is what we wanted to be true. The same argument can be used to show that sin(θ) and tan(θ) can be computed using any right triangle with acute angle θ. Except for some “rigged” right triangles, it is not easy to calculate the trigonometric ratios. Before the 1970’s, approximate values of sin(θ), cos(θ), tan(θ) were listed in long tables or calculated using a slide rule. Today, a scientiﬁc calculator saves the day on these computations. Most scientiﬁc calculators will give an approximation for the values of the trigonometric ratios. However, it is good to keep in mind we can compute the EXACT values of the trigonometric ratios when θ = 0, π 4, π 2 radians or, equivalently, when θ = 0◦, 30◦, 45◦, 60◦, 90◦. 17.5: Applying Figure trigonometric ratios to any
right triangle. Angle θ Trigonometric Ratio Deg Rad sin(θ) cos(θ) tan(θ) 0◦ 30◦ 45◦ 60◦ 902 2 √3 2 1 1 √3 2 √2 2 1 2 0 0 1 √3 1 √3 Undeﬁned Table 17.1: Exact Trigonometric Ratios 224 CHAPTER 17. THE CIRCULAR FUNCTIONS!!! CAUTION!!! Some people make a big deal of “approximate” vs. “exact” answers; we won’t worry about it here, unless we are speciﬁcally asked for an exact answer. However, here is something we will make a big deal about: When computing values of cos(θ), sin(θ), and tan(θ) on your calculator, make sure you are using the correct “angle mode” when entering θ; i.e. “degrees” or “radians”. For example, if θ = 1◦, then cos(1◦) = 0.9998, sin(1◦) = 0.0175, and In contrast, if θ = 1 radians, then cos(1) = 0.5403, tan(1◦) = 0.0175. sin(1) = 0.8415, and tan(1) = 1.5574. 17.3 Applications h θ h cos(θ) h sin(θ) a tan(θ) θ a Figure 17.6: What do these ratios mean? When confronted with a situation involving a right triangle where the measure of one acute angle θ and one side are known, we can solve for the remaining sides using the appropriate trigonometric ratios. Here is the key picture to keep in mind: Important Facts 17.3.1 (Trigonometric ratios). Given a right triangle, the trigonometric ratios relate the lengths of the sides as shown in Figure 17.6. Example 17.3.2. To measure the distance across a river for a new bridge, surveyors placed poles at locations A, B and C. The length \|AB\| = 100 feet and the measure of the angle ∠ABC is 31◦18 ′. Find the distance to span the river. If the measurement
of the angle ∠ABC is only accurate within 2 ′, ﬁnd the possible error in \|AC\|. ± C d A B 100 310 18 ′ Solution. The trigonometric ratio relating these two sides would be the tangent and we can convert θ into decimal form, arriving at: tan(31◦18 ′) = tan(31.3◦) = \|AC\| \|BA\| = d 100 Figure 17.7: The distance spanning a river. therefore d = 60.8 feet. This tells us that the bridge needs to span a gap of If the measurement of the angle was in error by +2 ′, then 60.8 feet. tan(31◦20 ′) = tan(31.3333◦) = 0.6088 and the span is 60.88 ft. On the other hand, if the measurement of the angle was in error by −2 ′, then tan(31◦16 ′) = tan(31.2667◦) = 0.6072 and the span is 60.72 ft. 17.3. APPLICATIONS 225 Example 17.3.3. A plane is ﬂying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be 18◦ above the horizontal, then immediately ﬂies the plane at an angle of 20◦ above horizontal. The airspeed of the plane is 100 mph. After 5 minutes, the plane is directly above the top of the mountain. How high is the plane above the top of the mountain (when it passes over)? What is the height of the mountain? T E L P 2000 ft S sealevel Figure 17.8: Flying toward a mountain. Solution. We can compute the hypotenuse of ing the speed and time information about the plane: △ LPT by us- \|PT \| = (100 mph)(5 minutes)(1 hour/60 minutes) = 25 3 miles. The deﬁnitions of the trigonometric ratios show: \|TL\| = \|PL\| = 25 3 25 3 sin(20◦) = 2.850 miles, and cos(20◦) = 7.831 miles. With this data, we can now ﬁnd \|EL\|: \|EL\| = \|PL\| tan(18◦) = 2.544 miles. The height of the plane above the peak is
\|TE\| = \|TL\| − \|EL\| = 2.850 − 2.544 = 0.306 miles = 1,616 feet. The elevation of the peak above sea level is given by: Peak elevation = plane altitude + \|EL\| = \|SP\| + \|EL\| = 2,000 + (2.544)(5,280) = 15,432 feet. Example 17.3.4. A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they measure the angle γ = 48◦ at position B (see picture), then descend 400 feet to position A and make two measurements of α = 13◦ (the measure of ∠EAD), β = 53◦ (the measure of ∠CAD). Determine the width of the canyon to the nearest foot. 400 ft B γ A α β C E canyon D Figure 17.9: Finding the width of a canyon. Solution. We will need to exploit three right triangles in ACE. Our goal is to compute \|ED\| = ACD, and the picture: \|CD\| − \|CE\|, which suggests more than one right triangle will come into play. BCD, △ △ △ 226 CHAPTER 17. THE CIRCULAR FUNCTIONS The ﬁrst step is to use ACD to obtain a system of two equations and two unknowns involving some of the side lengths; we will then solve the system. From the deﬁnitions of the trigonometric ratios, BCD and △ △ \|CD\| = (400 + \|AC\|) tan(48◦) \|CD\| = \|AC\| tan(53◦). Plugging the second equation into the ﬁrst and rearranging we get \|AC\| = 400 tan(48◦) tan(53◦) − tan(48◦) = 2,053 feet. Plugging this back into the second equation of the system gives \|CD\| = (2053) tan(53◦) = 2724 feet. The next step is to relate ACE, which can now be done ACD and △ in an effective way using the calculations above. Notice that the measure of ∠CAE is β − α = 40◦. We have △ \|CE\| = \|AC\| tan(40◦) = (2053) tan(40◦) = 1,7
23 feet. As noted above, \|ED\| = \|CD\| − \|CE\| = 2,724 − 1,723 = 1,001 feet is the width of the canyon. 17.4 Circular Functions S = (x,y) y 20 θ x P R If Cosmo is located somewhere in the ﬁrst quadrant of Figure 17.1, represented by the location S, we can use the trigonometric ratios to describe his coordinates. Impose the indicated xy-coordinate system with origin at P and extract the pictured right triangle with vertices at P and S. The radius is 20 ft. and applying Fact 17.3.1 gives S = (x, y) = (20 cos(θ), 20 sin(θ)). Figure 17.10: Cosmo on a circular path. Unfortunately, we run into a snag if we allow Cosmo to wander into the second, third or fourth quadrant, since then the angle θ is no longer acute. 17.4. CIRCULAR FUNCTIONS 227 17.4.1 Are the trigonometric ratios functions? Recall that sin(θ), cos(θ), and tan(θ) are deﬁned for acute angles θ inside a right triangle. We would like to say that these three equations actually deﬁne functions where the variable is an angle θ. Having said this, it is natural to ask if these three equations can be extended to be deﬁned for ANY angle θ. For example, we need to explain how sin is deﬁned. 2π 3 To start, we begin with the unit circle pictured in the xy-coordinate system. Let θ = ∠ROP be the angle in standard central position shown in Figure 17.11. If θ is positive (resp. negative), we adopt the convention that θ is swept out by counterclockwise (resp. clockwise) rotation of the initial side OR. The objective is to ﬁnd the coordinates of the point P in this ﬁgure. Notice that each coordinate of P (the x-coordinate and the y-coordinate) will depend on the given angle θ. For this reason, we need to introduce two new functions involving the variable θ. P 1 A unit circle with radius = 1. θ O 1 R Figure 17.11: Coordinates of points
on the unit circle. Deﬁnition 17.4.1. Let θ be an angle in standard central position inside the unit circle, as in Figure 17.11. This angle determines a point P on the unit circle. Deﬁne two new functions, cos(θ) and sin(θ), on the domain of all θ values as follows: cos(θ) sin(θ) def= horizontal x-coordinate of P on unit circle def= vertical y-coordinate of P on unit circle. r = 1 kilometer We refer to sin(θ) and cos(θ) as the basiccircularfunctions. Keep in mind that these functions have variables which are angles (either in degree or radian measure). These functions will be on your calculator. Again, BE CAREFUL to check the angle mode setting on your calculator (“degrees” or “radians”) before doing a calculation. Figure 17.12: driving track. 0.025 rad sec Michael starts here A circular Example 17.4.2. Michael is test driving a vehicle counterclockwise around a desert test track which is circular of radius 1 kilometer. He starts at the location pictured, traveling 0.025 rad Impose coordinates as pictured. Where is sec. Michael located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds and θ(t) the angle swept out the by Michael after t seconds. Since we are given the angular speed, we get θ(t) = 0.025t radians. y-axis M(t) = (x(t),y(t)) θ(t) 0.025 rad sec x-axis Michael starts here Figure 17.13: Modeling Michael’s location. 228 CHAPTER 17. THE CIRCULAR FUNCTIONS Since the angle θ(t) is in central standard position, we get M(t) = (cos(θ(t)), sin(θ(t))) = (cos(0.025t), sin(0.025t)). So, after 18 seconds Michael’s location will be M(18) = (0.9004, 0.4350).!!! CAUTION!!! Interpreting the coordinates of the point P = (cos(
θ), sin(θ)) in Figure 17.11 only works if the angle θ is viewed in central standard position. You must do some additional work if the angle is placed in a different position; see the next Example. y-axis r = 1 kilometer 0.025 rad sec x-axis Michael starts here Angela starts here 0.03 rad sec (a) Angela and Michael on the same test track. y-axis M(t) Angela starts here 0.03 rad sec β(t) θ(t) α(t) A(t) 0.025 rad sec x-axis Michael starts here (b) Modeling the motion of Angela and Michael. Example 17.4.3. Both Angela and Michael are test driving vehicles counterclockwise around a desert test track which is circular of radius 1 kilometer. They start at the locations shown in Figure 17.14(a). Michael is traveling 0.025 rad/sec and Angela is traveling 0.03 rad/sec. Impose coordinates as pictured. Where are the drivers located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds. Likewise, let A(t) be the point on the circle of motion representing Angela’s location after t seconds. Let θ(t) be the angle swept out the by Michael and α(t) the angle swept out by Angela after t seconds. Since we are given the angular speeds, we get θ(t) = 0.025t radians, and α(t) = 0.03t radians. From the previous Example 17.4.2, Figure 17.14: Visualizing motion on a circular track. M(t) = (cos(0.025t), sin(0.025t)), and M(18) = (0.9004, 0.4350). Angela’s angle α(t) is NOT in central standard position, so we must observe that α(t) + π = β(t), where β(t) is in central standard position: See Figure 17.14(b). We conclude that A(t) = (cos(β(t)), sin(β(t))) = (cos(π + 0.03t), sin(π + 0.03t)). (−0.8577, −0.5141).
So, after 18 seconds Angela’s location will be A(18) = 17.4. CIRCULAR FUNCTIONS 229 17.4.2 Relating circular functions and right triangles If the point P on the unit circle is located in the ﬁrst quadrant, then we can compute cos(θ) and sin(θ) using trigonometric ratios. In general, it’s useful to relate right triangles, the unit circle and the circular functions. To describe this connection, given θ we place it in central standard position in the unit circle, where ∠ROP = θ. Draw a line through P perpendicular to the x-axis, obtaining an inscribed right triangle. Such a right triangle has hypotenuse of length 1, vertical side of length labeled b and horizontal side of length labeled a. There are four cases: See Figure 17.16. unit circle (radius = 1) P θ sin(θ) O R cos(θ) Figure 17.15: The point P in the ﬁrst quadrant CASE I CASE II CASE III CASE IV Figure 17.16: Possible positions of θ on the unit circle. Case I has already been discussed, arriving at cos(θ) = a and sin(θ) = b. In Case II, we can interpret cos(θ) = −a, sin(θ) = b. We can reason similarly in the other Cases III and IV, using Figure 17.16, and we arrive at this conclusion: Important Facts 17.4.4 (Circular functions and triangles). View θ as in Figure 17.16 and form the pictured inscribed right triangles. Then we can interpret cos(θ) and sin(θ) in terms of these right triangles as follows: sin(θ) = b Case I : cos(θ) = a, Case II : cos(θ) = −a, sin(θ) = b Case III : cos(θ) = −a, sin(θ) = −b sin(θ) = −b Case IV : cos(θ) = a, 230 CHAPTER 17. THE CIRCULAR FUNCTIONS 17.5 What About Other Circles? T P θ O R S Cr unit circle Figure 17.17: Points on other circles. What happens if we begin with a circle Cr with radius r (possibly different than 1
) and want to compute the coordinates of points on this circle? The circular functions can be used to answer this more general question. Picture our circle Cr centered at the origin in the same picture with unit circle C1 and the angle θ in standard central position for each circle. As pictured, we can view θ = ∠ROP = ∠SOT. If P = (x,y) is our point on the unit circle corresponding to the angle θ, then the calculation below shows how to compute coordinates on general circles: P = (x,y) = (cos(θ), sin(θ)) C1 ∈ x2 + y2 = 1 r2x2 + r2y2 = r2 (rx)2 + (ry)2 = r2 T = (rx, ry) = (r cos(θ), r sin(θ)) Cr. ∈ ⇔ ⇔ ⇔ ⇔ Important Fact 17.5.1. Let Cr be a circle of radius r centered at the origin and θ = ∠SOT an angle in standard central position for this circle, as in Figure 17.17. Then the coordinates of T = (r cos(θ), r sin(θ)). y-axis β = π − α = 2.9416 -axis circle radius = 1 circle radius = 2 circle radius = 3 α B Figure 17.18: Coordinates of points on circles. Examples 17.5.2. Consider the picture below, with θ = 0.8 radians and α = 0.2 radians. What are the coordinates of the labeled points? Solution. The angle θ is in standard central position; α is a central angle, but it is not in standard position. Notice, β = π − α = 2.9416 is an angle in standard central position which locates the same points U, T, S as the angle α. Applying Deﬁnition 17.4.1 on page 227: = (0.6967, 0.7174) P = (cos(0.8), sin(0.8)) = (1.3934, 1.4347) Q = (2 cos(0.8), 2 sin(0.8)) = (2.0901, 2.1521) R = (3 cos(0.8), 3 sin(0.8)) S = (cos
(2.9416), sin(2.9416)) = (−0.9801, 0.1987) T = (2 cos(2.9416), 2 sin(2.9416)) = (−1.9602, 0.3973) U = (3 cos(2.9416), 3 sin(2.9416)) = (−2.9403, 0.5961). 17.6. OTHER BASIC CIRCULAR FUNCTION 231 Example 17.5.3. Suppose Cosmo begins at the position R in the ﬁgure, walking around the circle of radius 20 feet with an angular speed of 4 5 RPM counterclockwise. After 3 minutes have elapsed, describe Cosmo’s precise location. S 5 = 12 Solution. Cosmo has traveled 3 4 5 revolutions. If θ is = the angle traveled after 3 minutes, θ = 24π 5 radians = 15.08 radians. By (15.5.1), we have x = = 20 cos 11.76 feet. Conclude that Cosmo is located at the point S = (−16.18, 11.76). Using (15.1), θ = 864◦ = 2(360◦) + 144◦; this means that Cosmo walks counterclockwise around the circle two complete revolutions, plus 144◦. = −16.18 feet and y = 20 sin 2π radians 24π 5 rad 24π 5 rad 12 5 rev rev P 20 feet R Figure 17.19: Where is Cosmo after 3 minutes? 17.6 Other Basic Circular Function Given any angle θ, our constructions offer a concrete link between the cosine and sine functions and right triangles inscribed inside the unit circle: See Figure 17.20. P P θ O R θ O R CASE I CASE II θ O P CASE III R O θ R P CASE IV Figure 17.20: Computing the slope of a line using the function tan(θ). The slope of the hypotenuse of these inscribed triangles is just the slope of the line through OP. Since P = (cos(θ), sin(θ)) and O = (0, 0): Slope = ∆y ∆x = sin(θ) cos(θ) ; this would be valid as long as cos(θ) new circular function called the tangent of θ by the rule
= 0. This calculation motivates a tan(θ) = sin(θ) cos(θ), provided cos(θ) = 0. 6 6 232 CHAPTER 17. THE CIRCULAR FUNCTIONS The only time cos(θ) = 0 is when the corresponding point P on the unit circle has x-coordinate 0. But, this only happens at the positions (0, 1) and (0, −1) on the unit circle, corresponding to angles of the form θ =. These are the cases when the inscribed right triangle would “degenerate” to having zero width and the line segment OP becomes vertical. In summary, we then have this general idea to keep in mind: 5π 2, 3π 2, · · · π 2, ± ± ± Important Fact 17.6.1. The slope of a line = tan(θ), where θ is the angle the line makes with the x-axis (or any other horizontal line) Three other commonly used circular functions come up from time to time. The cotangent function y = cot(θ), the secant function y = sec(θ) and the cosecant function y = csc(θ) are deﬁned by the formulas: sec(θ) def= 1 cos(θ), csc(θ) def= 1 sin(θ), cot(θ) def= 1 tan(θ). Just as with the tangent function, one needs to worry about the values of θ for which these functions are undeﬁned (due to division by zero). We will not need these functions in this text. North Alaska Northwest West SeaTac 0 115 South 0 50 East 0 20 Delta (a) The ﬂight paths of three airplanes. Alaska x = 30 Q N Northwest P y = 20 E R Delta W x = −50 S (b) Modeling the paths of each ﬂight. Figure 17.21: Visualizing and modeling departing airplanes. Example 17.6.2. Three airplanes depart SeaTac Airport. A NorthWest ﬂight is heading in a direction 50◦ counterclockwise from East, an Alaska ﬂight is heading 115◦ counterclockwise from East and a Delta ﬂight is heading 20◦ clockwise from East. Find the location of
the Northwest ﬂight when it is 20 miles North of SeaTac. Find the location of the Alaska ﬂight when it is 50 miles West of SeaTac. Find the location of the Delta ﬂight when it is 30 miles East of SeaTac. impose a coordinate Solution. We system in Figure 17.21(a), where “East” (resp. “North”) points along the positive x-axis (resp. positive y-axis). To solve the problem, we will ﬁnd the equation of the three lines representing the ﬂight paths, then determine where they intersect the appropriate horizontal or vertical line. The Northwest and Alaska directions of ﬂight are angles in standard central position; the Delta ﬂight direction will be −20◦. We can imagine right triangles with their hypotenuses along the directions of ﬂight, then using the tangent function, we have these three immediate conclusions: slope NW line = tan(50◦) = 1.19, slope Alaska line = tan(115◦) = −2.14, and slope Delta line = tan(−20◦) = −0.364. 17.6. OTHER BASIC CIRCULAR FUNCTION 233 All three ﬂight paths pass through the origin (0,0) of our coordinate system, so the equations of the lines through the ﬂight paths will be: NW ﬂight : y = 1.19x, Alaska ﬂight : y = −2.14x, Delta ﬂight : y = −0.364x. The Northwest ﬂight is 20 miles North of SeaTac when y = 20; plugging into the equation of the line of ﬂight gives 20 = 1.19x, so x = 16.81 and the plane location will be P = (16.81, 20). Similarly, the Alaska ﬂight is 50 miles West of SeaTac when x = −50; plugging into the equation of the line of ﬂight gives y = (−2.14)(−50) = 107 and the plane location will be Q = (−50, 107). Finally, check that the Delta ﬂight is at R = (30, −10.92) when it is 30 miles East of SeaTac. 234 CHAPTER 17. THE CIRCULAR
FUNCTIONS 17.7 Exercises Problem 17.1. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the world’s greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. Recall, we worked on this in Exercise 16.5. 12 RPM θ P 4 feet (a) Impose a coordinate system and ﬁnd the coordinates T (t) = (x(t),y(t)) of Tiff at time t seconds after she starts the ride. (b) Tiff becomes a human missile after 6 seconds on the ride. Find Tiff’s coordinates the instant she becomes a human missile. (c) Find the equation of the tangential line along which Tiff travels the instant she becomes a human missile. Sketch a picture indicating this line and her initial direction of motion along it when the seat detaches. Problem 17.2. (a) Find the equation of a line passing through the point (-1,2) and making an angle of 13o with the x-axis. (Note: There are two answers; ﬁnd them both.) (b) Find the equation of a line making an angle of 8o with the y-axis and passing through the point (1,1). (Note: There are two answers; ﬁnd them both.) Problem 17.3. The crew of a helicopter needs to land temporarily in a forest and spot a ﬂat horizontal piece of ground (a clearing in the forest) as a potential landing site, but are uncertain whether it is wide enough. They make two measurements from A (see picture) ﬁnding α = 25o and β = 54o. They rise vertically 100 feet to B and measure γ = 47o. Determine the width of the clearing to the nearest foot. B 100 feet γ A β α C E clearing D Problem 17.4. Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12
seconds to reach the northermost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive y-axis. (a) Give Marla’s coordinates at her starting point. (b) Give Marla’s coordinates when she has been running for 10 seconds. (c) Give Marla’s coordinates when she has been running for 901.3 seconds. Problem 17.5. A merry-go-round is rotating at the constant angular speed of 3 RPM counterclockwise. The platform of this ride is a circular disc of radius 24 feet. You jump onto the ride at the location pictured below. rotating 3 RPM jump on here θ 17.7. EXERCISES 235 (a) If θ = 34o, then what are your xy- coordinates after 4 minutes? (b) If θ = 20o, then what are your xy- coordinates after 45 minutes? (c) If θ = −14o, then what are your xycoordinates after 6 seconds? Draw an accurate picture of the situation. (d) If θ = −2.1 rad, then what are your xy-coordinates after 2 hours and 7 seconds? Draw an accurate picture of the situation. (e) If θ = 2.1 rad, then what are your xycoordinates after 5 seconds? Draw an accurate picture of the situation. Problem 17.6. Shirley is on a ferris wheel which spins at the rate of 3.2 revolutions per minute. The wheel has a radius of 45 feet, and the center of the wheel is 59 feet above the ground. After the wheel starts moving, Shirley takes 16 seconds to reach the top of the wheel. How high above the ground is she when the wheel has been moving for 9 minutes? Problem 17.7. The top of the Boulder Dam has an angle of elevation of 1.2 radians from a point on the Colorado River. Measuring the angle of elevation to the top of the dam from a point 155 feet farther down river is 0.9 radians; assume the two angle measurements are taken at the same elevation above sea level. How high is the dam? downriver dam 0.9 1.2 155 ft a Problem 17.8. A radio station obtains a permit to increase the height of their radio tower on Queen Anne Hill by no more than 100 feet. You are
the head of the Queen Anne Community Group and one of your members asks you to make sure that the radio station does not exceed the limits of the permit. After ﬁnding a relatively ﬂat area nearby the tower (not necessarily the same altitude as the bottom of the tower), and standing some unknown distance away from the tower, you make three measurements all at the same height above sea level. You observe that the top of the old tower makes an angle of 39◦ above level. You move 110 feet away from the original measurement and observe that the old top of the tower now makes an angle of 34◦ above level. Finally, after the new construction is complete, you observe that the new top of the tower, from the same point as the second measurement was made, makes an angle of 40◦ above the horizontal. All three measurements are made at the same height above sea level and are in line with the tower. Find the height of the addition to the tower, to the nearest foot. Problem 17.9. Charlie and Alexandra are running around a circular track with radius 60 meters. Charlie started at the westernmost point of the track, and, at the same time, Alexandra started at the northernmost point. They both run counterclockwise. Alexandra runs at 4 meters per second, and will take exactly 2 minutes to catch up to Charlie. Impose a coordinate system, and give the x- and y-coordinates of Charlie after one minute of running. Problem 17.10. George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at the easternmost point. The illustration below shows their starting positions and running directions. They start running toward each other at constant speeds. George runs at 9 feet per second. Paula takes 50 seconds to run a lap of the track. George and Paula pass each other after 11 seconds. N George Paula After running for 3 minutes, how far east of his starting point is George? 236 CHAPTER 17. THE CIRCULAR FUNCTIONS Problem 17.11. A kite is attached to 300 feet of string, which makes a 42 degree angle with the level ground. The kite pilot is holding the string 4 feet above the ground. kite o 42 4 feet ground level (a) How high above the ground is the kite? (b) Suppose that power lines are located Is 250 feet in front of the kite ﬂyer. any portion of the kite
or string over the power lines? Problem 17.12. In the pictures below, a bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle of motion. In each of the cases, answer these questions: (a) Find an angle θ0 in standard central position that gives the bugs initial location. (In some cases, this is the angle given in the picture; in other cases, you will need to do something.) (b) The location angle of the bug at time t is given by the formula θ(t) = θ0 + ωt. Plug in the values for θ0 and ω to explicitly obtain a formula for θ(t). (c) Find the coordinates of the bug at time t. (d) What are the coordinates of the bug after 1 second? After 0 seconds? After 3 seconds? After 22 seconds? ω=4π/9rad/sec bug lands here 1.2 rad 2 in bug lands here ω=4π/9 rad/sec ω= 4π/9rad/sec bug lands here 0.5 rad 2 in 2 in Chapter 18 Trigonometric Functions Our deﬁnitions of the circular functions are based upon the unit circle. This makes it easy to visualize many of their properties. 18.1 Easy Properties of Circular Functions How can we determine the range of function values for cos(θ) and sin(θ)? To begin with, recall the abstract deﬁnition for the range of a function f(θ): Range of f = {f(θ) : θ is in the domain}. Using the unit circle constructions of the basic circular functions, it is easy to visualize the range of cos(θ) and sin(θ). Beginning at the position (1, 0), imagine a ball If we moving counterclockwise around the unit circle. “freeze” the motion at any point in time, we will have swept out an angle θ and the corresponding position P(θ) on the circle will have coordinates P(θ) = (cos(θ), sin(θ)). ball moves from 0 to π 2 radians around unit circle light source x-axis y-axis y-axis light source ball
moves from 0 π 2 radians around unit circle to x-axis (a) What do you see on the y-axis? (b) What do you see on the x-axis? Figure 18.2: Projecting the coordinates of points onto the y-axis and the x-axis. By studying the coordinates of the ball as it moves in the ﬁrst quadπ/2 radians. rant, we will be studying cos(θ) and sin(θ), for 0 θ 237 ≤ ≤ ball moves counterclockwise y-axis (0,1) P(θ) sin(θ) 1 (−1,0) θ (1,0) cos(θ) x-axis UNIT CIRCLE (0, − 1) Figure 18.1: Visualizing the range of sin(θ) and cos(θ). 238 CHAPTER 18. TRIGONOMETRIC FUNCTIONS We can visualize this very concretely. Imagine a light source as in Figure 18.2(a); then a shadow projects onto the vertical y-axis. The shadow locations you would see on the y-axis are precisely the values sin(θ), for π/2 radians. Similarly, imagine a light source as in Figure 18.2(b); 0 then a shadow projects onto the horizontal x-axis. The shadow locations you would see on the x-axis are precisely the values cos(θ), for 0 π/2 radians. ≤ ≤ ≤ ≤ θ θ There are two visual conclusions: First, the function values of sin(θ) vary from 0 to 1 as θ varies from 0 to π/2. Secondly, the function values of cos(θ) vary from 1 to 0 as θ varies from 0 to π/2. Of course, we can go ahead and continue analyzing the motion as the ball moves into the second, third and fourth quadrant, ending up back at the starting position (1, 0). See Figure 18.3. y-axis #2 #3 #1 #4 ball moves from 0 to 2π radians around unit circle x-axis light source light source #2 #1 #3 #4 ball moves from 0 to 2π radians around nit circle x-axis (a) What do you see on the y-axis? (b) What do you see on the x-axis? Figure

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.