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log125 3x − 2 2x + 3 = 1 3 11. − log(x) = 5.4 13. 6 − 3 log5(2x) = 0 2. log2 x3 = log2(x) 4. log5 18 − x2 = log5(6 − x) 6. log 1 2 (2x − 1) = −3 8. log(x2 − 3x) = 1 10. log x 10−3 = 4.7 12. 10 log x 10−12 = 150 14. 3 ln(x) − 2 = 1 − ln(x) 15. log3(x − 4) + log3(x + 4) = 2 16. log5(2x + 1) + log5(x + 2) = 1 17. log169(3x + 7) − log169(5x − 9) = 1 2 18. ln(x + 1) − ln(x) = 3 19. 2 log7(x) = log7(2) + log7(x + 12) 20. log(x) − log(2) = log(x + 8) − log(x + 2) 21. log3(x) = log 1 3 (x) + 8 23. (log(x))2 = 2 log(x) + 15 22. ln(ln(x)) = 3 24. ln(x2) = (ln(x))2 In Exercises 25 - 30, solve the inequality analytically. 25. 1 − ln(x) x2 < 0 27. 10 log x 10−12 ≥ 90 29. 2.3 < − log(x) < 5.4 26. x ln(x) − x > 0 28. 5.6 ≤ log x 10−3 ≤ 7.1 30. ln(x2) ≤ (ln(x))2 In Exercises 31 - 34, use your calculator to help you solve the equation or inequality. 31. ln(x) = e−x 33. ln(x2 + 1) ≥ 5 √ 32. ln(x) = 4 x 34. ln(−2x3 − x2 + 13x − 6) < 0 6.4 Logarithmic Equations and Inequalities 467 35. Since f (x) = ex is a strictly increasing
function, if a < b then ea < eb. Use this fact to solve the inequality ln(2x + 1) < 3 without a sign diagram. Use this technique to solve the inequalities in Exercises 27 - 29. (Compare this to Exercise 46 in Section 6.3.) 36. Solve ln(3 − y) − ln(y) = 2x + ln(5) for y. 37. In Example 6.4.4 we found the inverse of f (x) = log(x) 1 − log(x) to be f −1(x) = 10 x x+1. (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. (b) Find the range of f by ﬁnding the domain of f −1. x 1 − x (c) Let g(x) = and h(x) = log(x). Show that f = g ◦ h and (g ◦ h)−1 = h−1 ◦ g−1. (We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a speciﬁc example of the property.) 38. Let f (x) = 1 2 ln 1 + x 1 − x. Compute f −1(x) and ﬁnd its domain and range. 39. Explain the equation in Exercise 10 and the inequality in Exercise 28 above in terms of the Richter scale for earthquake magnitude. (See Exercise 75 in Section 6.1.) 40. Explain the equation in Exercise 12 and the inequality in Exercise 27 above in terms of sound intensity level as measured in decibels. (See Exercise 76 in Section 6.1.) 41. Explain the equation in Exercise 11 and the inequality in Exercise 29 above in terms of the pH of a solution. (See Exercise 77 in Section 6.1.) √ 42. With the help of your classmates, solve the inequality n x > ln(x) for a variety of natural numbers n. What might you conjecture about the “speed” at which f (x) = ln(x) grows versus any principal nth root function? 468 Exponential and Logarithmic Functions 6.4.
2 Answers 1. x = 5 4 4. x = −3, 4 7. x = ±10 10. x = 101.7 13. x = 25 2 16. x = 1 2 19. x = 6 22. x = ee3 25. (e, ∞) 28. 102.6, 104.1 31. x ≈ 1.3098 2. x = 1 5. x = −1 8. x = −2, 5 11. x = 10−5.4 14. x = e3/4 17. x = 2 20. x = 4 3. x = −2 6. x = 9 2 9. x = − 17 7 12. x = 103 15. x = 5 18. x = 1 e3−1 21. x = 81 23. x = 10−3, 105 24. x = 1, x = e2 26. (e, ∞) 27. 10−3, ∞ 29. 10−5.4, 10−2.3 30. (0, 1] ∪ [e2, ∞) 32. x ≈ 4.177, x ≈ 5503.665 33. ≈ (−∞, −12.1414) ∪ (12.1414, ∞) 34. ≈ (−3.0281, −3)∪(0.5, 0.5991)∪(1.9299, 2) 35. − 1 2 < x < e3 − 1 2 e2x − 1 e2x + 1 38. f −1(x) = ex − e−x ex + e−x. (To see why we rewrite this in this form, see Exercise 51 in Section 11.10.) The domain of f −1 is (−∞, ∞) and its range is the same as the domain of f, namely (−1, 1). = 36. y = 3 5e2x + 1 6.5 Applications of Exponential and Logarithmic Functions 469 6.5 Applications of Exponential and Logarithmic Functions As we mentioned in Section 6.1, exponential and logarithmic functions are used to model a wide variety of behaviors in the real world. In the examples that follow, note that while the applications are drawn from many diﬀerent disciplines, the mathematics remains essentially the same. Due to the applied nature of the problems we
will examine in this section, the calculator is often used to express our answers as decimal approximations. 6.5.1 Applications of Exponential Functions Perhaps the most well-known application of exponential functions comes from the ﬁnancial world. Suppose you have $100 to invest at your local bank and they are oﬀering a whopping 5 % annual percentage interest rate. This means that after one year, the bank will pay you 5% of that $100, or $100(0.05) = $5 in interest, so you now have $105.1 This is in accordance with the formula for simple interest which you have undoubtedly run across at some point before. Equation 6.1. Simple Interest The amount of interest I accrued at an annual rate r on an investmenta P after t years is The amount A in the account after t years is given by I = P rt aCalled the principal A = P + I = P + P rt = P (1 + rt) Suppose, however, that six months into the year, you hear of a better deal at a rival bank.2 Naturally, you withdraw your money and try to invest it at the higher rate there. Since six months is one half of a year, that initial $100 yields $100(0.05) 1 = $2.50 in interest. You take your 2 $102.50 oﬀ to the competitor and ﬁnd out that those restrictions which may apply actually do apply to you, and you return to your bank which happily accepts your $102.50 for the remaining six months of the year. To your surprise and delight, at the end of the year your statement reads $105.06, not $105 as you had expected.3 Where did those extra six cents come from? For the ﬁrst six months of the year, interest was earned on the original principal of $100, but for the second six months, interest was earned on $102.50, that is, you earned interest on your interest. This is the basic concept behind compound interest. In the previous discussion, we would say that the interest was compounded twice, or semiannually.4 If more money can be earned by earning interest on interest already earned, a natural question to ask is what happens if the interest is compounded more often, say 4 times a year, which is every three months, or ‘quarterly.’ In this case, the money
is in the account for three months, or 1 4 of a year, at a time. After the ﬁrst quarter, we = $101.25. We now invest the $101.25 for the next three have A = P (1 + rt) = $100 1 + 0.05 · 1 4 1How generous of them! 2Some restrictions may apply. 3Actually, the ﬁnal balance should be $105.0625. 4Using this convention, simple interest after one year is the same as compounding the interest only once. 470 Exponential and Logarithmic Functions months and ﬁnd that at the end of the second quarter, we have A = $101.25 1 + 0.05 · 1 ≈ $102.51. 4 Continuing in this manner, the balance at the end of the third quarter is $103.79, and, at last, we obtain $105.08. The extra two cents hardly seems worth it, but we see that we do in fact get more money the more often we compound. In order to develop a formula for this phenomenon, we need to do some abstract calculations. Suppose we wish to invest our principal P at an annual rate r and th of a year compound the interest n times per year. This means the money sits in the account 1 n between compoundings. Let Ak denote the amount in the account after the kth compounding. Then. After the second compounding, we use A1 A1 = P 1 + r 1. Continuing in as our new principal and get A2 = A1 this fashion, we get A3 = P 1 + r k. Since we compound the interest n times per year, after t years, we have nt compoundings. We have just derived the general formula for compound interest below. which simpliﬁes to A1 = P 1 + r, and so on, so that Ak =, A4 = Equation 6.2. Compounded Interest: If an initial principal P is invested at an annual rate r and the interest is compounded n times per year, the amount A in the account after t years is A(t) = P 1 + nt r n If we take P = 100, r = 0.05, and n = 4, Equation 6.2 becomes A(t) = 100 1 + 0.05 which 4 reduces to A(t) = 100(1.0125)
4t. To check this new formula against our previous calculations, we ﬁnd A 1 4 ≈ $103.79, and A(1) ≈ $105.08. ≈ $102.51, A 3 4 4 ) = 101.25, A 1 2 = 100(1.0125)4( 1 4t Example 6.5.1. Suppose $2000 is invested in an account which oﬀers 7.125% compounded monthly. 1. Express the amount A in the account as a function of the term of the investment t in years. 2. How much is in the account after 5 years? 3. How long will it take for the initial investment to double? 4. Find and interpret the average rate of change5 of the amount in the account from the end of the fourth year to the end of the ﬁfth year, and from the end of the thirty-fourth year to the end of the thirty-ﬁfth year. Solution. 1. Substituting P = 2000, r = 0.07125, and n = 12 (since interest is compounded monthly) into 12t Equation 6.2 yields A(t) = 2000 1 + 0.07125 = 2000(1.0059375)12t. 12 2. Since t represents the length of the investment in years, we substitute t = 5 into A(t) to ﬁnd A(5) = 2000(1.0059375)12(5) ≈ 2852.92. After 5 years, we have approximately $2852.92. 5See Deﬁnition 2.3 in Section 2.1. 6.5 Applications of Exponential and Logarithmic Functions 471 3. Our initial investment is $2000, so to ﬁnd the time it takes this to double, we need to ﬁnd t when A(t) = 4000. We get 2000(1.0059375)12t = 4000, or (1.0059375)12t = 2. Taking natural 12 ln(1.0059375) ≈ 9.75. Hence, it takes approximately 9 years logs as in Section 6.3, we get t = 9 months for the investment to double. ln(2) 5−4 4. To ﬁnd the average rate of change of A from the end of the fourth year to
the end of the ﬁfth year, we compute A(5)−A(4) ≈ 195.63. Similarly, the average rate of change of A from the end of the thirty-fourth year to the end of the thirty-ﬁfth year is A(35)−A(34) ≈ 1648.21. This means that the value of the investment is increasing at a rate of approximately $195.63 per year between the end of the fourth and ﬁfth years, while that rate jumps to $1648.21 per year between the end of the thirty-fourth and thirty-ﬁfth years. So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases.6 35−34 We have observed that the more times you compound the interest per year, the more money you will earn in a year. Let’s push this notion to the limit.7 Consider an investment of $1 invested at 100% interest for 1 year compounded n times a year. Equation 6.2 tells us that the amount of money in the account after 1 year is A = 1 + 1 n n. Below is a table of values relating n and A. A n 2 1 2 2.25 4 ≈ 2.4414 12 ≈ 2.6130 360 ≈ 2.7145 1000 ≈ 2.7169 10000 ≈ 2.7181 100000 ≈ 2.7182 As promised, the more compoundings per year, the more money there is in the account, but we also observe that the increase in money is greatly diminishing. We are witnessing a mathematical ‘tug of war’. While we are compounding more times per year, and hence getting interest on our interest more often, the amount of time between compoundings is getting smaller and smaller, so there is less time to build up additional interest. With Calculus, we can show8 that as n → ∞, A = 1 + 1 n → e, where e is the natural base ﬁrst presented in Section 6.1. Taking the number n of compoundings per year to inﬁnity results in what is called continuously compounded interest. Theorem 6.8. If you invest $1 at 100% interest compounded continuously, then you will have $e at the end of one year. 6
In fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really deﬁnes exponential functions and we refer the reader to a course in Calculus. 7Once you’ve had a semester of Calculus, you’ll be able to fully appreciate this very lame pun. 8Or deﬁne, depending on your point of view. 472 Exponential and Logarithmic Functions Using this deﬁnition of e and a little Calculus, we can take Equation 6.2 and produce a formula for continuously compounded interest. Equation 6.3. Continuously Compounded Interest: If an initial principal P is invested at an annual rate r and the interest is compounded continuously, the amount A in the account after t years is A(t) = P ert If we take the scenario of Example 6.5.1 and compare monthly compounding to continuous compounding over 35 years, we ﬁnd that monthly compounding yields A(35) = 2000(1.0059375)12(35) which is about $24,035.28, whereas continuously compounding gives A(35) = 2000e0.07125(35) which is about $24,213.18 - a diﬀerence of less than 1%. Equations 6.2 and 6.3 both use exponential functions to describe the growth of an investment. Curiously enough, the same principles which govern compound interest are also used to model In Biology, The Law of Uninhibited Growth states as short term growth of populations. its premise that the instantaneous rate at which a population increases at any time is directly proportional to the population at that time.9 In other words, the more organisms there are at a given moment, the faster they reproduce. Formulating the law as stated results in a diﬀerential equation, which requires Calculus to solve. Its solution is stated below. Equation 6.4. Uninhibited Growth: If a population increases according to The Law of Uninhibited Growth, the number of organisms N at time t is given by the formula N (t) = N0ekt, where N (0) = N0 (read ‘N nought’) is the initial number of organisms and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of N (t
) at time t) = k N (t) It is worth taking some time to compare Equations 6.3 and 6.4. In Equation 6.3, we use P to denote the initial investment; in Equation 6.4, we use N0 to denote the initial population. In Equation 6.3, r denotes the annual interest rate, and so it shouldn’t be too surprising that the k in Equation 6.4 corresponds to a growth rate as well. While Equations 6.3 and 6.4 look entirely diﬀerent, they both represent the same mathematical concept. Example 6.5.2. In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to ﬁve million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, ﬁnd a formula for the number of cells, N, in thousands, after t days. Solution. We begin with N (t) = N0ekt. Since N is to give the number of cells in thousands, we have N0 = 12, so N (t) = 12ekt. In order to complete the formula, we need to determine the 9The average rate of change of a function over an interval was ﬁrst introduced in Section 2.1. Instantaneous rates of change are the business of Calculus, as is mentioned on Page 161. 6.5 Applications of Exponential and Logarithmic Functions 473 growth rate k. We know that after one week, the number of cells has grown to ﬁve million. Since t measures days and the units of N are in thousands, this translates mathematically to N (7) = 5000. 3 ). Of We get the equation 12e7k = 5000 which gives k = 1 course, in practice, we would approximate k to some desired accuracy, say k ≈ 0.8618, which we can interpret as an 86.18% daily growth rate for the cells.. Hence, N (t) = 12e 7 ln 1250 7 ln( 1250 3 t Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we’ve seen already is Example 6.1.1 in Section 6
.1. There, the value of a car declined from its purchase price of $25,000 to nothing at all. Another real world phenomenon which follows suit is radioactive decay. There are elements which are unstable and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The assumption behind this model is that the rate of decay of an element at a particular time is directly proportional to the amount of the element present at that time. In other words, the more of the element there is, the faster the element decays. This is precisely the same kind of hypothesis which drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is hauntingly similar to Equation 6.4 with the exception that the rate constant k is negative. Equation 6.5. Radioactive Decay The amount of a radioactive element A at time t is given by the formula A(t) = A0ekt, where A(0) = A0 is the initial amount of the element and k < 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of A(t) at time t) = k A(t) Example 6.5.3. Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life10 of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, ﬁnd a function which gives the amount of Iodine-131, A, in grams, t days later. Solution. Since we start with 5 grams initially, Equation 6.5 gives A(t) = 5ekt. Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, A(8) = 2.5 8 ln 1 which means 5e8k = 2.5. Solving, we get k = 1 8 ≈ −0.08664, which we can interpret as a loss of material at a rate of 8.664% daily. Hence, A(t) = 5e− t ln(2) 8 ≈ 5e−0.08664t. = − ln(2)
2 We now turn our attention to some more mathematically sophisticated models. One such model is Newton’s Law of Cooling, which we ﬁrst encountered in Example 6.1.2 of Section 6.1. In that example we had a cup of coﬀee cooling from 160◦F to room temperature 70◦F according to the formula T (t) = 70 + 90e−0.1t, where t was measured in minutes. In this situation, we know the physical limit of the temperature of the coﬀee is room temperature,11 and the diﬀerential equation 10The time it takes for half of the substance to decay. 11The Second Law of Thermodynamics states that heat can spontaneously ﬂow from a hotter object to a colder one, but not the other way around. Thus, the coﬀee could not continue to release heat into the air so as to cool below room temperature. 474 Exponential and Logarithmic Functions which gives rise to our formula for T (t) takes this into account. Whereas the radioactive decay model had a rate of decay at time t directly proportional to the amount of the element which remained at time t, Newton’s Law of Cooling states that the rate of cooling of the coﬀee at a given time t is directly proportional to how much of a temperature gap exists between the coﬀee at time t and room temperature, not the temperature of the coﬀee itself. In other words, the coﬀee cools faster when it is ﬁrst served, and as its temperature nears room temperature, the coﬀee cools ever more slowly. Of course, if we take an item from the refrigerator and let it sit out in the kitchen, the object’s temperature will rise to room temperature, and since the physics behind warming and cooling is the same, we combine both cases in the equation below. Equation 6.6. Newton’s Law of Cooling (Warming): The temperature T of an object at time t is given by the formula T (t) = Ta + (T0 − Ta) e−kt, where T (0) = T0 is the initial temperature of the object, Ta is the ambient temperaturea and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate
of change of T (t) at time t) = k (T (t) − Ta) aThat is, the temperature of the surroundings. If we re-examine the situation in Example 6.1.2 with T0 = 160, Ta = 70, and k = 0.1, we get, according to Equation 6.6, T (t) = 70 + (160 − 70)e−0.1t which reduces to the original formula given. The rate constant k = 0.1 indicates the coﬀee is cooling at a rate equal to 10% of the diﬀerence between the temperature of the coﬀee and its surroundings. Note in Equation 6.6 that the constant k is positive for both the cooling and warming scenarios. What determines if the function T (t) is increasing or decreasing is if T0 (the initial temperature of the object) is greater than Ta (the ambient temperature) or vice-versa, as we see in our next example. Example 6.5.4. A 40◦F roast is cooked in a 350◦F oven. After 2 hours, the temperature of the roast is 125◦F. 1. Assuming the temperature of the roast follows Newton’s Law of Warming, ﬁnd a formula for the temperature of the roast T as a function of its time in the oven, t, in hours. 2. The roast is done when the internal temperature reaches 165◦F. When will the roast be done? Solution. 1. The initial temperature of the roast is 40◦F, so T0 = 40. The environment in which we are placing the roast is the 350◦F oven, so Ta = 350. Newton’s Law of Warming tells us T (t) = 350 + (40 − 350)e−kt, or T (t) = 350 − 310e−kt. To determine k, we use the fact that after 2 hours, the roast is 125◦F, which means T (2) = 125. This gives rise to the equation 350 − 310e−2k = 125 which yields k = − 1 ≈ 0.1602. The temperature function is T (t) = 350 − 310e 2 ln 45 62 2 ln( 45 62 ) ≈ 350 − 310e−0.1602t. t 6.5 Applications of Exponential and Logar
ithmic Functions 475 2. To determine when the roast is done, we set T (t) = 165. This gives 350 − 310e−0.1602t = 165 ≈ 3.22. It takes roughly 3 hours and 15 minutes to cook 0.1602 ln 37 62 whose solution is t = − 1 the roast completely. If we had taken the time to graph y = T (t) in Example 6.5.4, we would have found the horizontal asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive at this conclusion by applying a bit of ‘number sense’. As t → ∞, −0.1602t ≈ very big (−) so that e−0.1602t ≈ very small (+). The larger the value of t, the smaller e−0.1602t becomes so that T (t) ≈ 350 − very small (+), which indicates the graph of y = T (t) is approaching its horizontal asymptote y = 350 from below. Physically, this means the roast will eventually warm up to 350◦F.12 The function T is sometimes called a limited growth model, since the function T remains bounded as t → ∞. If we apply the principles behind Newton’s Law of Cooling to a biological example, it says the growth rate of a population is directly proportional to how much room the population has to grow. In other words, the more room for expansion, the faster the growth rate. The logistic growth model combines The Law of Uninhibited Growth with limited growth and states that the rate of growth of a population varies jointly with the population itself as well as the room the population has to grow. Equation 6.7. Logistic Growth: If a population behaves according to the assumptions of logistic growth, the number of organisms N at time t is given by the equation N (t) = L 1 + Ce−kLt, where N (0) = N0 is the initial population, L is the limiting population,a C is a measure of how much room there is to grow given by C = L N0 − 1. and k > 0 is the constant of proportionality which satisﬁes the equation (instantaneous rate of change of N (t) at time t) = k N (t) (L − N (t)) aThat is
, as t → ∞, N (t) → L The logistic function is used not only to model the growth of organisms, but is also often used to model the spread of disease and rumors.13 Example 6.5.5. The number of people N, in hundreds, at a local community college who have heard the rumor ‘Carl is afraid of Virginia Woolf’ can be modeled using the logistic equation N (t) = 84 1 + 2799e−t, 12at which point it would be more toast than roast. 13Which can be just as damaging as diseases. 476 Exponential and Logarithmic Functions where t ≥ 0 is the number of days after April 1, 2009. 1. Find and interpret N (0). 2. Find and interpret the end behavior of N (t). 3. How long until 4200 people have heard the rumor? 4. Check your answers to 2 and 3 using your calculator. Solution. 1. We ﬁnd N (0) = 100. Since N (t) measures the number of people who have heard the rumor in hundreds, N (0) corresponds to 3 people. Since t = 0 corresponds to April 1, 2009, we may conclude that on that day, 3 people have heard the rumor.14 1+2799e0 = 84 2800 = 3 84 2. We could simply note that N (t) is written in the form of Equation 6.7, and identify L = 84. However, to see why the answer is 84, we proceed analytically. Since the domain of N is restricted to t ≥ 0, the only end behavior of signiﬁcance is t → ∞. As we’ve seen before,15 as t → ∞, we have 1997e−t → 0+ and so N (t) ≈ ≈ 84. Hence, as t → ∞, N (t) → 84. This means that as time goes by, the number of people who will have heard the rumor approaches 8400. 84 1+very small (+) 3. To ﬁnd how long it takes until 4200 people have heard the rumor, we set N (t) = 42. Solving 1+2799e−t = 42 gives t = ln(2799) ≈ 7.937. It takes around 8 days until 4200 people have heard the rumor. 84 4. We graph y
= N (x) using the calculator and see that the line y = 84 is the horizontal asymptote of the graph, conﬁrming our answer to part 2, and the graph intersects the line y = 42 at x = ln(2799) ≈ 7.937, which conﬁrms our answer to part 3. y = f (x) = 84 1+2799e−x and y = 84 y = f (x) = 84 1+2799e−x and y = 42 14Or, more likely, three people started the rumor. I’d wager Jeﬀ, Jamie, and Jason started it. So much for telling your best friends something in conﬁdence! 15See, for example, Example 6.1.2. 6.5 Applications of Exponential and Logarithmic Functions 477 If we take the time to analyze the graph of y = N (x) above, we can see graphically how logistic growth combines features of uninhibited and limited growth. The curve seems to rise steeply, then at some point, begins to level oﬀ. The point at which this happens is called an inﬂection point or is sometimes called the ‘point of diminishing returns’. At this point, even though the function is still increasing, the rate at which it does so begins to decline. It turns out the point of diminishing returns always occurs at half the limiting population. (In our case, when y = 42.) While these concepts are more precisely quantiﬁed using Calculus, below are two views of the graph of y = N (x), one on the interval [0, 8], the other on [8, 15]. The former looks strikingly like uninhibited growth; the latter like limited growth. y = f (x) = 84 1+2799e−x for 0 ≤ x ≤ 8 y = f (x) = 84 1+2799e−x for 8 ≤ x ≤ 16 6.5.2 Applications of Logarithms Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises 75, 76 and 77 of Section 6.1, we showed that logarithms are useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acids and bases (
pH). We now present yet a diﬀerent use of the a basic logarithm function, password strength. Example 6.5.6. The information entropy H, in bits, of a randomly generated password consisting of L characters is given by H = L log2(N ), where N is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password. 1. If a 7 character case-sensitive16 password is comprised of letters and numbers only, ﬁnd the associated information entropy. 2. How many possible symbol options per character is required to produce a 7 character password with an information entropy of 50 bits? Solution. 1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as diﬀerent. There are 10 digits (0 through 9) for a total of N = 62 symbols. Since the password is to be 7 characters long, L = 7. Thus, H = 7 log2(62) = 7 ln(62) ln(2) ≈ 41.68. 16That is, upper and lower case letters are treated as diﬀerent characters. 478 Exponential and Logarithmic Functions 2. We have L = 7 and H = 50 and we need to ﬁnd N. Solving the equation 50 = 7 log2(N ) gives N = 250/7 ≈ 141.323, so we would need 142 diﬀerent symbols to choose from.17 Chemical systems known as buﬀer solutions have the ability to adjust to small changes in acidity to maintain a range of pH values. Buﬀer solutions have a wide variety of applications from maintaining a healthy ﬁsh tank to regulating the pH levels in blood. Our next example shows how the pH in a buﬀer solution is a little more complicated than the pH we ﬁrst encountered in Exercise 77 in Section 6.1. Example 6.5.7. Blood is a buﬀer solution. When carbon dioxide is absorbed into the bloodstream it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base to partially neutralize the acid. The equation18 which models blood pH in this situation is pH = 6.1 + log 800, where x is the partial pressure of carbon dioxide in arterial blood, measured x in
torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is 7.4. Solution. We set pH = 7.4 and get 7.4 = 6.1 + log 800 x x = 800 101.3 ≈ 40.09. Hence, the partial pressure of carbon dioxide in the blood is about 40 torr. = 1.3. Solving, we ﬁnd, or log 800 x Another place logarithms are used is in data analysis. Suppose, for instance, we wish to model the spread of inﬂuenza A (H1N1), the so-called ‘Swine Flu’. Below is data taken from the World Health Organization (WHO) where t represents the number of days since April 28, 2009, and N represents the number of conﬁrmed cases of H1N1 virus worldwide. 1 t 7 N 148 257 367 658 898 1085 1490 1893 2371 2500 3440 4379 4694 10 13 12 11 3 4 2 5 6 9 8 14 t N 5251 15 16 17 18 19 20 5728 6497 7520 8451 8480 8829 Making a scatter plot of the data treating t as the independent variable and N as the dependent variable gives Which models are suggested by the shape of the data? Thinking back Section 2.5, we try a Quadratic Regression, with pretty good results. 17Since there are only 94 distinct ASCII keyboard characters, to achieve this strength, the number of characters in the password should be increased. 18Derived from the Henderson-Hasselbalch Equation. See Exercise 43 in Section 6.2. Hasselbalch himself was studying carbon dioxide dissolving in blood - a process called metabolic acidosis. 6.5 Applications of Exponential and Logarithmic Functions 479 However, is there any scientiﬁc reason for the data to be quadratic? Are there other models which ﬁt the data equally well, or better? Scientists often use logarithms in an attempt to ‘linearize’ data sets - in other words, transform the data sets to produce ones which result in straight lines. To see how this could work, suppose we guessed the relationship between N and t was some kind of power function, not necessarily quadratic, say N = BtA. To try to determine the A and B, we can take the natural log of
both sides and get ln(N ) = ln BtA. Using properties of logs to expand the right hand side of this equation, we get ln(N ) = A ln(t) + ln(B). If we set X = ln(t) and Y = ln(N ), this equation becomes Y = AX + ln(B). In other words, we have a line with slope A and Y -intercept ln(B). So, instead of plotting N versus t, we plot ln(N ) versus ln(t). ln(t) ln(N ) 0 0.693 1.099 1.386 1.609 1.792 1.946 2.079 2.197 2.302 2.398 2.485 2.565 4.997 5.549 5.905 6.489 6.800 6.989 7.306 7.546 7.771 7.824 8.143 8.385 8.454 ln(t) ln(N ) 2.639 8.566 2.708 2.773 2.833 2.890 2.944 2.996 8.653 8.779 8.925 9.042 9.045 9.086 Running a linear regression on the data gives The slope of the regression line is a ≈ 1.512 which corresponds to our exponent A. The y-intercept b ≈ 4.513 corresponds to ln(B), so that B ≈ 91.201. Hence, we get the model N = 91.201t1.512, something from Section 5.3. Of course, the calculator has a built-in ‘Power Regression’ feature. If we apply this to our original data set, we get the same model we arrived at before.19 19Critics may question why the authors of the book have chosen to even discuss linearization of data when the calculator has a Power Regression built-in and ready to go. Our response: talk to your science faculty. 480 Exponential and Logarithmic Functions This is all well and good, but the quadratic model appears to ﬁt the data better, and we’ve yet to mention any scientiﬁc principle which would lead us to believe the actual spread of the ﬂu follows any kind of power function at all. If we are to attack
this data from a scientiﬁc perspective, it does seem to make sense that, at least in the early stages of the outbreak, the more people who have the ﬂu, the faster it will spread, which leads us to proposing an uninhibited growth model. If we assume N = BeAt then, taking logs as before, we get ln(N ) = At + ln(B). If we set X = t and Y = ln(N ), then, once again, we get Y = AX + ln(B), a line with slope A and Y -intercept ln(B). Plotting ln(N ) versus t gives the following linear regression. We see the slope is a ≈ 0.202 and which corresponds to A in our model, and the y-intercept is b ≈ 5.596 which corresponds to ln(B). We get B ≈ 269.414, so that our model is N = 269.414e0.202t. Of course, the calculator has a built-in ‘Exponential Regression’ feature which produces what appears to be a diﬀerent model N = 269.414(1.22333419)t. Using properties of exponents, we write e0.202t = e0.202t ≈ (1.223848)t, which, had we carried more decimal places, would have matched the base of the calculator model exactly. The exponential model didn’t ﬁt the data as well as the quadratic or power function model, but it stands to reason that, perhaps, the spread of the ﬂu is not unlike that of the spread of a rumor 6.5 Applications of Exponential and Logarithmic Functions 481 and that a logistic model can be used to model the data. The calculator does have a ‘Logistic Regression’ feature, and using it produces the model N = 10739.147 1+42.416e0.268t. This appears to be an excellent ﬁt, but there is no friendly coeﬃcient of determination, R2, by which to judge this numerically. There are good reasons for this, but they are far beyond the scope of the text. Which of the models, quadratic, power, exponential, or logistic is the ‘best model’? If by �
�best’ we mean ‘ﬁts closest to the data,’ then the quadratic and logistic models are arguably the winners with the power function model a close second. However, if we think about the science behind the spread of the ﬂu, the logistic model gets an edge. For one thing, it takes into account that only a ﬁnite number of people will ever get the ﬂu (according to our model, 10,739), whereas the quadratic model predicts no limit to the number of cases. As we have stated several times before in the text, mathematical models, regardless of their sophistication, are just that: models, and they all have their limitations.20 20Speaking of limitations, as of June 3, 2009, there were 19,273 conﬁrmed cases of inﬂuenza A (H1N1). This is well above our prediction of 10,739. Each time a new report is issued, the data set increases and the model must be recalculated. We leave this recalculation to the reader. 482 Exponential and Logarithmic Functions 6.5.3 Exercises For each of the scenarios given in Exercises 1 - 6, Find the amount A in the account as a function of the term of the investment t in years. Determine how much is in the account after 5 years, 10 years, 30 years and 35 years. Round your answers to the nearest cent. Determine how long will it take for the initial investment to double. Round your answer to the nearest year. Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the ﬁfth year, and from the end of the thirty-fourth year to the end of the thirty-ﬁfth year. Round your answer to two decimal places. 1. $500 is invested in an account which oﬀers 0.75%, compounded monthly. 2. $500 is invested in an account which oﬀers 0.75%, compounded continuously. 3. $1000 is invested in an account which oﬀers 1.25%, compounded monthly. 4. $1000 is invested in an account which oﬀers 1.25%, compounded continuously. 5. $5000 is invested in an account which oﬀers 2.125%, compounded monthly. 6
. $5000 is invested in an account which oﬀers 2.125%, compounded continuously. 7. Look back at your answers to Exercises 1 - 6. What can be said about the diﬀerence between monthly compounding and continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios where the diﬀerence between monthly and continuously compounded interest would be more dramatic. Try varying the interest rate, the term of the investment and the principal. Use computations to support your answer. 8. How much money needs to be invested now to obtain $2000 in 3 years if the interest rate in a savings account is 0.25%, compounded continuously? Round your answer to the nearest cent. 9. How much money needs to be invested now to obtain $5000 in 10 years if the interest rate in a CD is 2.25%, compounded monthly? Round your answer to the nearest cent. 10. On May, 31, 2009, the Annual Percentage Rate listed at Jeﬀ’s bank for regular savings accounts was 0.25% compounded monthly. Use Equation 6.2 to answer the following. (a) If P = 2000 what is A(8)? (b) Solve the equation A(t) = 4000 for t. (c) What principal P should be invested so that the account balance is $2000 is three years? 6.5 Applications of Exponential and Logarithmic Functions 483 11. Jeﬀ’s bank also oﬀers a 36-month Certiﬁcate of Deposit (CD) with an APR of 2.25%. (a) If P = 2000 what is A(8)? (b) Solve the equation A(t) = 4000 for t. (c) What principal P should be invested so that the account balance is $2000 in three years? (d) The Annual Percentage Yield is the simple interest rate that returns the same amount of interest after one year as the compound interest does. With the help of your classmates, compute the APY for this investment. 12. A ﬁnance company oﬀers a promotion on $5000 loans. The borrower does not have to make any payments for the ﬁrst three years, however interest will continue to be charged to the loan at 29.9% compounded continuously. What amount will be due at the end of the three year period, assuming no payments
are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due? 13. Use Equation 6.2 to show that the time it takes for an investment to double in value does not depend on the principal P, but rather, depends only on the APR and the number of compoundings per year. Let n = 12 and with the help of your classmates compute the doubling time for a variety of rates r. Then look up the Rule of 72 and compare your answers to what that rule says. If you’re really interested21 in Financial Mathematics, you could also compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69. In Exercises 14 - 18, we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula A(t) = A0ekt where A0 is the initial amount of the material and k is the decay constant. For each isotope: Find the decay constant k. Round your answer to four decimal places. Find a function which gives the amount of isotope A which remains after time t. (Keep the units of A and t the same as the given data.) Determine how long it takes for 90% of the material to decay. Round your answer to two decimal places. (HINT: If 90% of the material decays, how much is left?) 14. Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of 5.27 years. 15. Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life 14 days. 16. Chromium 51, used to track red blood cells, initial amount 75 milligrams, half-life 27.7 days. 17. Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life 432.7 years. 18. Uranium 235, used for nuclear power, initial amount 1 kg grams, half-life 704 million years. 21Awesome pun! 484 Exponential and Logarithmic Functions 19. With the help of your classmates, show that the time it takes for 90% of each isotope listed in Exercises 14 - 18 to decay does not depend on the initial amount of the substance, but rather, on only the decay constant k. Find a formula, in terms of k only, to determine how long it takes for 90%
of a radioactive isotope to decay. 20. In Example 6.1.1 in Section 6.1, the exponential function V (x) = 25 4 5 was used to model the value of a car over time. Use the properties of logs and/or exponents to rewrite the model in the form V (t) = 25ekt. x 21. The Gross Domestic Product (GDP) of the US (in billions of dollars) t years after the year 2000 can be modeled by: G(t) = 9743.77e0.0514t (a) Find and interpret G(0). (b) According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce, the 2007 GDP was $14, 369.1 billion and the 2010 GDP was $14, 657.8 billion.) 22. The diameter D of a tumor, in millimeters, t days after it is detected is given by: D(t) = 15e0.0277t (a) What was the diameter of the tumor when it was originally detected? (b) How long until the diameter of the tumor doubles? 23. Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth N (t) = N0ekt where N0 is the initial number of bacteria and t is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organism is 20 minutes. Suppose 1000 bacteria are present initially. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of bacteria N (t) after t minutes. (c) How long until there are 9000 bacteria? Round your answer to the nearest minute. 24. Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let t be the time elapsed since the ﬁrst observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth N (t) = N0ekt. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of yeast (in millions
) per cc N (t) after t hours. (c) What is the doubling time for this strain of yeast? 6.5 Applications of Exponential and Logarithmic Functions 485 25. The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, ﬁnd a function of the form N (t) = N0ekt which models the number of wolves t years after 1996. (Use t = 0 to represent the year 1996. Also, round your value of k to four decimal places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.) 26. During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to ﬁt a model of the form N (t) = N0ekt were N (t) is the number of Painesville Residents t years after 1860. (Use t = 0 to represent the year 1860. Also, round the value of k to four decimal places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the population as 19,563) What could be some causes for such a vast discrepancy? For more on this, see Exercise 37. 27. The population of Sasquatch in Bigfoot county is modeled by P (t) = 120 1 + 3.167e−0.05t where P (t) is the population of Sasquatch t years after 2010. (a) Find and interpret P (0). (b) Find the population of Sasquatch in Bigfoot county in 2013. Round your answer to the nearest Sasquatch. (c) When will the population of Sasquatch in Bigfoot county reach 60? Round your answer to the nearest year. (d) Find and interpret the end behavior of the graph of y = P (t). Check your answer using a graphing utility. 28. The half-life of the radioactive isotope Carbon-14 is about 5730 years. (a
) Use Equation 6.5 to express the amount of Carbon-14 left from an initial N milligrams as a function of time t in years. (b) What percentage of the original amount of Carbon-14 is left after 20,000 years? (c) If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42% of the original amount, approximately how old is the tool? (d) Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simpliﬁed. 486 Exponential and Logarithmic Functions 29. Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, RubidiumStrontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation 6.5 to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time t in billions of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates. 30. Use Equation 6.5 to show that k = − ln(2) h where h is the half-life of the radioactive isotope. 31. A pork roast22 was taken out of a hardwood smoker when its internal temperature had reached 180◦F and it was allowed to rest in a 75◦F house for 20 minutes after which its internal temperature had dropped to 170◦F. Assuming that the temperature of the roast follows Newton’s Law of Cooling (Equation 6.6), (a) Express the temperature T (in ◦F) as a function of time t (in minutes). (b) Find the time at which the roast would have dropped to 140◦F had it not been carved and eaten. 32. In reference to Exercise 44 in Section 5.3, if Fritzy the Fox’s speed is the same as Chewbacca the Bunny’s speed, Fritzy’s pursuit curve is given by y(x) = x2 − 1 4 1 4 ln(x) − 1 4 Use your calculator to graph
this path for x > 0. Describe the behavior of y as x → 0+ and interpret this physically. 33. The current i measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by i(t) = 2 − 2e−10t where t ≥ 0 is the number of seconds after the circuit is switched on. Determine the value of i as t → ∞. (This is called the steady state current.) 34. If the voltage in the circuit in Exercise 33 above is switched oﬀ after 30 seconds, the current is given by the piecewise-deﬁned function i(t) = 2 − 2e−10t 2 − 2e−300 e−10t+300 if 0 ≤ t < 30 if t ≥ 30 With the help of your calculator, graph y = i(t) and discuss with your classmates the physical signiﬁcance of the two parts of the graph 0 ≤ t < 30 and t ≥ 30. 22This roast was enjoyed by Jeﬀ and his family on June 10, 2009. This is real data, folks! 6.5 Applications of Exponential and Logarithmic Functions 487 35. In Exercise 26 in Section 2.3, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a catenary and its basic shape is given by y = 1 2 (ex + e−x). Use your calculator to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise 47 in Section 6.3 and the one given in the answer to Exercise 38 in Section 6.4? When ﬂipped upside down, the catenary makes an arch. The Gateway Arch in St. Louis, Missouri has the shape y = 757.7 − 127.7 2 e x 127.7 + e− x 127.7 where x and y are measured in feet and −315 ≤ x ≤ 315. Find the highest point on the arch. 36. In Exercise 6a in Section 2.5, we examined the data set given below which showed how two cats and their surviving oﬀspring can produce over 80 million cats in just ten years. It is virtually impossible to see this data plotted on your calculator, so plot x versus ln
(x) as was done on page 480. Find a linear model for this new data and comment on its goodness of ﬁt. Find an exponential model for the original data and comment on its goodness of ﬁt. Year x Number of Cats N (x 10 12 66 382 2201 12680 73041 420715 2423316 13968290 80399780 37. This exercise is a follow-up to Exercise 26 which more thoroughly explores the population growth of Painesville, Ohio. According to Wikipedia, the population of Painesville, Ohio is given by Year t Population 1860 2649 1870 3728 1880 3841 1890 4755 1900 5024 1910 5501 1920 7272 1930 10944 1940 12235 1950 14432 Year t Population 1960 16116 1970 16536 1980 16351 1990 15699 2000 17503 (a) Use a graphing utility to perform an exponential regression on the data from 1860 through 1920 only, letting t = 0 represent the year 1860 as before. How does this calculator model compare with the model you found in Exercise 26? Use the calculator’s exponential model to predict the population in 2010. (The 2010 census gave the population as 19,563) (b) The logistic model ﬁt to all of the given data points for the population of Painesville t years after 1860 (again, using t = 0 as 1860) is P (t) = 18691 1 + 9.8505e−0.03617t According to this model, what should the population of Painesville have been in 2010? (The 2010 census gave the population as 19,563.) What is the population limit of Painesville? 488 Exponential and Logarithmic Functions 38. According to OhioBiz, the census data for Lake County, Ohio is as follows: Year t Population 1860 15576 1870 15935 1880 16326 1890 18235 1900 21680 1910 22927 1920 28667 1930 41674 1940 50020 1950 75979 Year t Population 1960 148700 1970 197200 1980 212801 1990 215499 2000 227511 (a) Use your calculator to ﬁt a logistic model to these data, using x = 0 to represent the year 1860. (b) Graph these data and your logistic function on your calculator to judge the reasonable- ness of the ﬁt. (c) Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the population to be
230,041.) (d) According to your model, what is the population limit of Lake County, Ohio? 39. According to facebook, the number of active users of facebook has grown signiﬁcantly since its initial launch from a Harvard dorm room in February 2004. The chart below has the approximate number U (x) of active users, in millions, x months after February 2004. For example, the ﬁrst entry (10, 1) means that there were 1 million active users in December 2004 and the last entry (77, 500) means that there were 500 million active users in July 2010. Month x Active Users in Millions U (x) 10 22 34 38 44 54 59 60 62 65 67 70 72 77 1 5.5 12 20 50 100 150 175 200 250 300 350 400 500 With the help of your classmates, ﬁnd a model for this data. 40. Each Monday during the registration period before the Fall Semester at LCCC, the Enrollment Planning Council gets a report prepared by the data analysts in Institutional Eﬀectiveness and Planning.23 While the ongoing enrollment data is analyzed in many diﬀerent ways, we shall focus only on the overall headcount. Below is a chart of the enrollment data for Fall Semester 2008. It starts 21 weeks before “Opening Day” and ends on “Day 15” of the semester, but we have relabeled the top row to be x = 1 through x = 24 so that the math is easier. (Thus, x = 22 is Opening Day.) Week x Total Headcount Week x Total Headcount 1 2 3 4 5 6 7 8 1194 1564 2001 2475 2802 3141 3527 3790 9 10 11 12 13 14 15 16 4065 4371 4611 4945 5300 5657 6056 6478 23The authors thank Dr. Wendy Marley and her staﬀ for this data and Dr. Marcia Ballinger for the permission to use it in this problem. 6.5 Applications of Exponential and Logarithmic Functions 489 Week x Total Headcount 17 18 19 20 21 22 23 24 7161 7772 8505 9256 10201 10743 11102 11181 With the help of your classmates, ﬁnd a model for this data. Unlike most of the phenomena we have studied in this section, there is no single diﬀerential equation which governs the enrollment growth
. Thus there is no scientiﬁc reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which inﬂuence enrollment at a community college and how can you take those into account mathematically? 41. When we wrote this exercise, the Enrollment Planning Report for Fall Semester 2009 had only 10 data points for the ﬁrst 10 weeks of the registration period. Those numbers are given below. Week x Total Headcount 1 2 3 4 5 6 7 8 9 10 1380 2000 2639 3153 3499 3831 4283 4742 5123 5398 With the help of your classmates, ﬁnd a model for this data and make a prediction for the Opening Day enrollment as well as the Day 15 enrollment. (WARNING: The registration period for 2009 was one week shorter than it was in 2008 so Opening Day would be x = 21 and Day 15 is x = 23.) 490 Exponential and Logarithmic Functions 6.5.4 Answers 1. 2. 3. 4. 5. 12t A(t) = 500 1 + 0.0075 12 A(5) ≈ $519.10, A(10) ≈ $538.93, A(30) ≈ $626.12, A(35) ≈ $650.03 It will take approximately 92 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 3.88. This means that the investment is growing at an average rate of $3.88 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 4.85. This means that the investment is growing at an average rate of $4.85 per year at this point. A(t) = 500e0.0075t A(5) ≈ $519.11, A(10) ≈ $538.94, A(30) ≈ $626.16, A(35) ≈ $650.09 It will take approximately 92 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 3.88. This means that the investment is growing at an average rate of $3
.88 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 4.86. This means that the investment is growing at an average rate of $4.86 per year at this point. 12t A(t) = 1000 1 + 0.0125 12 A(5) ≈ $1064.46, A(10) ≈ $1133.07, A(30) ≈ $1454.71, A(35) ≈ $1548.48 It will take approximately 55 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 13.22. This means that the investment is growing at an average rate of $13.22 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 19.23. This means that the investment is growing at an average rate of $19.23 per year at this point. A(t) = 1000e0.0125t A(5) ≈ $1064.49, A(10) ≈ $1133.15, A(30) ≈ $1454.99, A(35) ≈ $1548.83 It will take approximately 55 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 13.22. This means that the investment is growing at an average rate of $13.22 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 19.24. This means that the investment is growing at an average rate of $19.24 per year at this point. A(t) = 5000 1 + 0.02125 A(5) ≈ $5559.98, A(10) ≈ $6182.67, A(30) ≈ $9453.40, A(35) ≈ $10512.13 It will take approximately 33 years for the investment to double. 12t 12 6.5 Applications of Exponential and Logarithmic Functions 491 The average rate of change
from the end of the fourth year to the end of the ﬁfth year is approximately 116.80. This means that the investment is growing at an average rate of $116.80 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 220.83. This means that the investment is growing at an average rate of $220.83 per year at this point. 6. A(t) = 5000e0.02125t A(5) ≈ $5560.50, A(10) ≈ $6183.83, A(30) ≈ $9458.73, A(35) ≈ $10519.05 It will take approximately 33 years for the investment to double. The average rate of change from the end of the fourth year to the end of the ﬁfth year is approximately 116.91. This means that the investment is growing at an average rate of $116.91 per year at this point. The average rate of change from the end of the thirtyfourth year to the end of the thirty-ﬁfth year is approximately 221.17. This means that the investment is growing at an average rate of $221.17 per year at this point. 8. P = 2000 e0.0025·3 ≈ $1985.06 9. P = 5000 (1+ 0.0225 12 )12·10 ≈ $3993.42 (a) A(8) = 2000 1 + 0.0025 12 10. 12·8 ≈ $2040.40 (b) t = (c) P = ln(2) 12 ln 1 + 0.0025 12 2000 1 + 0.0025 12 36 ≈ $1985.06 ≈ 277.29 years 11. (a) A(8) = 2000 1 + 0.0225 12 12·8 ≈ $2394.03 (b) t = (c) P = ≈ 30.83 years ln(2) 12 ln 1 + 0.0225 12 2000 1 + 0.0225 12 12 ≈ 1.0227 so the APY is 2.27% 36 ≈ $1869.57 (d) 1 + 0.0225 12 12. A(3) = 5000e0.299·3 ≈
$12, 226.18, A(6) = 5000e0.299·6 ≈ $30, 067.29 14. 5.27 ≈ −0.1315 k = ln(1/2) A(t) = 50e−0.1315t t = ln(0.1) −0.1315 ≈ 17.51 years. 15. 14 ≈ −0.0495 k = ln(1/2) A(t) = 2e−0.0495t t = ln(0.1) −0.0495 ≈ 46.52 days. 492 16. 18. 27.7 ≈ −0.0250 k = ln(1/2) A(t) = 75e−0.0250t t = ln(0.1) −0.025 ≈ 92.10 days. Exponential and Logarithmic Functions 17. 432.7 ≈ −0.0016 k = ln(1/2) A(t) = 0.29e−0.0016t t = ln(0.1) −0.0016 ≈ 1439.11 years. 704 ≈ −0.0010 k = ln(1/2) A(t) = e−0.0010t t = ln(0.1) −0.0010 ≈ 2302.58 million years, or 2.30 billion years. 19. t = ln(0.1) k = − ln(10) k 20. V (t) = 25eln( 4 5 )t ≈ 25e−0.22314355t 21. (a) G(0) = 9743.77 This means that the GDP of the US in 2000 was $9743.77 billion dollars. (b) G(7) = 13963.24 and G(10) = 16291.25, so the model predicted a GDP of $13, 963.24 billion in 2007 and $16, 291.25 billion in 2010. (a) D(0) = 15, so the tumor was 15 millimeters in diameter when it was ﬁrst detected. (b) t = ln(2) 0.0277 ≈ 25 days. 22. 23. (a) k = ln(2) 20 �
� 0.0346 (b) N (t) = 1000e0.0346t (c) t = ln(9) 0.0346 ≈ 63 minutes 3 ln 118 52 24. ln(6) 2.5 ≈ 0.4377 (a) k = 1 2 (b) N (t) = 2.5e0.4377t (c) t = ln(2) 0.4377 ≈ 1.58 hours 25. N0 = 52, k = 1 ≈ 0.2731, N (t) = 52e0.2731t. N (6) ≈ 268. 26. N0 = 2649, k = 1 60 ln 7272 2649 ≈ 0.0168, N (t) = 2649e0.0168t. N (150) ≈ 32923, so the population of Painesville in 2010 based on this model would have been 32,923. 27. (a) P (0) = 120 (b) P (3) = 4.167 ≈ 29. There are 29 Sasquatch in Bigfoot County in 2010. 1+3.167e−0.05(3) ≈ 32 Sasquatch. 120 (c) t = 20 ln(3.167) ≈ 23 years. (d) As t → ∞, P (t) → 120. As time goes by, the Sasquatch Population in Bigfoot County will approach 120. Graphically, y = P (x) has a horizontal asymptote y = 120. 28. (a) A(t) = N e ln(2) 5730 t − ≈ N e−0.00012097t (b) A(20000) ≈ 0.088978 · N so about 8.9% remains (c) t ≈ ln(.42) −0.00012097 ≈ 7171 years old 29. A(t) = 2.3e−0.0138629t 6.5 Applications of Exponential and Logarithmic Functions 493 31. (a) T (t) = 75 + 105e−0.005005t (b) The roast would have cooled to 140◦F in about 95 minutes. 32. From the graph, it appears that as x → 0+, y → ∞. This
is due to the presence of the ln(x) term in the function. This means that Fritzy will never catch Chewbacca, which makes sense since Chewbacca has a head start and Fritzy only runs as fast as he does. y(x) = 1 4 x2 − 1 4 ln(x) − 1 4 33. The steady state current is 2 amps. 36. The linear regression on the data below is y = 1.74899x + 0.70739 with r2 ≈ 0.999995. This is an excellent ﬁt. 2 4.1897 1 2.4849 x ln(N (x)) N (x) = 2.02869(5.74879)x = 2.02869e1.74899x with r2 ≈ 0.999995. This is also an excellent ﬁt and corresponds to our linearized model because ln(2.02869) ≈ 0.70739. 9 16.4523 6 11.1988 8 14.7006 7 12.9497 10 18.2025 3 5.9454 5 9.4478 4 7.6967 37. (a) The calculator gives: y = 2895.06(1.0147)x. Graphing this along with our answer from Exercise 26 over the interval [0, 60] shows that they are pretty close. From this model, y(150) ≈ 25840 which once again overshoots the actual data value. (b) P (150) ≈ 18717, so this model predicts 17,914 people in Painesville in 2010, a more conservative number than was recorded in the 2010 census. As t → ∞, P (t) → 18691. So the limiting population of Painesville based on this model is 18,691 people. 38. (a) y = 242526 1 + 874.62e−0.07113x, where x is the number of years since 1860. (b) The plot of the data and the curve is below. (c) y(140) ≈ 232889, so this model predicts 232,889 people in Lake County in 2010. (d) As x → ∞, y → 242526, so the limiting population of Lake County based on this model is 242,526 people. 494 Exponential and Logarithmic Functions Chapter 7
Hooked on Conics 7.1 Introduction to Conics In this chapter, we study the Conic Sections - literally ‘sections of a cone’. Imagine a doublenapped cone as seen below being ‘sliced’ by a plane. If we slice the cone with a horizontal plane the resulting curve is a circle. 496 Hooked on Conics Tilting the plane ever so slightly produces an ellipse. If the plane cuts parallel to the cone, we get a parabola. If we slice the cone with a vertical plane, we get a hyperbola. For a wonderful animation describing the conics as intersections of planes and cones, see Dr. Louis Talman’s Mathematics Animated Website. 7.1 Introduction to Conics 497 If the slicing plane contains the vertex of the cone, we get the so-called ‘degenerate’ conics: a point, a line, or two intersecting lines. We will focus the discussion on the non-degenerate cases: circles, parabolas, ellipses, and hyperbolas, in that order. To determine equations which describe these curves, we will make use of their deﬁnitions in terms of distances. 498 7.2 Circles Hooked on Conics Recall from Geometry that a circle can be determined by ﬁxing a point (called the center) and a positive number (called the radius) as follows. Deﬁnition 7.1. A circle with center (h, k) and radius r > 0 is the set of all points (x, y) in the plane whose distance to (h, k) is r. (x, y) r (h, k) From the picture, we see that a point (x, y) is on the circle if and only if its distance to (h, k) is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as r = (x − h)2 + (y − k)2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us the standard equation of a circle. Equation 7.1. The Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x − h)2 + (y − k)2 = r2. Example
7.2.1. Write the standard equation of the circle with center (−2, 3) and radius 5. Solution. Here, (h, k) = (−2, 3) and r = 5, so we get (x − (−2))2 + (y − 3)2 = (5)2 (x + 2)2 + (y − 3)2 = 25 Example 7.2.2. Graph (x + 2)2 + (y − 1)2 = 4. Find the center and radius. Solution. From the standard form of a circle, Equation 7.1, we have that x + 2 is x − h, so h = −2 and y − 1 is y − k so k = 1. This tells us that our center is (−2, 1). Furthermore, r2 = 4, so r = 2. Thus we have a circle centered at (−2, 1) with a radius of 2. Graphing gives us 7.2 Circles 499 y 4 3 2 1 −4 −3 −2 −1 1 x −1 If we were to expand the equation in the previous example and gather up like terms, instead of the easily recognizable (x + 2)2 + (y − 1)2 = 4, we’d be contending with x2 + 4x + y2 − 2y + 1 = 0. If we’re given such an equation, we can complete the square in each of the variables to see if it ﬁts the form given in Equation 7.1 by following the steps given below. To Write the Equation of a Circle in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square on both variables as needed. 3. Divide both sides by the coeﬃcient of the squares. (For circles, they will be the same.) Example 7.2.3. Complete the square to ﬁnd the center and radius of 3x2 − 6x + 3y2 + 4y − 4 = 0. Solution. 3x2 − 6x + 3y2 + 4y − 4 = 0 3x2 − 6x + 3y2 + 4y = 4 3 x2 − 2x + 3 y2 + 4 3 y = 4 add 4 to both sides factor out leading coeﬃcients 3 x2 − 2x
+ 1 + 3 y2 + (x − 1)2 + 3 y + (x − 1)(1) + 3 4 9 complete the square in x, y = = 25 3 25 9 factor divide both sides by 3 From Equation 7.1, we identify x − 1 as x − h, so h = 1, and y + 2 center is (h, k) = 1, − 2 3. Furthermore, we see that r2 = 25 9 so the radius is r = 5 3. 3 as y − k, so k = − 2 3. Hence, the 500 Hooked on Conics It is possible to obtain equations like (x − 3)2 + (y + 1)2 = 0 or (x − 3)2 + (y + 1)2 = −1, neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula, Equation 1.2, in conjunction with the ideas presented so far in this section. Example 7.2.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the endpoints of a diameter. Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields y r (h, k) 4 3 2 1 −2 −1 1 2 3 x Since the given points are endpoints of a diameter, we know their midpoint (h, k) is the center of the circle. Equation 1.2 gives us (h, k) = = =,, y1 + y2 2 3 + 4 2 x1 + x2 2 −1 + 2 2 7 2 1 2, The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus, (x2 − x1)2 + (y2 − y1)2 (2 − (−1))2 + (4 − 3)2 32 + 12 √ 10 2 Finally, since 2 √ 10 2 = 10 4, our answer becomes = 10 4 7.2 Circles 501 We close this section with the most important1 circle in all of mathematics: the Unit Circle. Deﬁnition 7.2. The Unit Circle is the circle centered
at (0, 0) with a radius of 1. The standard equation of the Unit Circle is x2 + y2 = 1. Example 7.2.5. Find the points on the unit circle with y-coordinate Solution. We replace y with √ 3 2 in the equation x2 + y2 = 1 to get √ 3 2. x2 + y2 = 1 √ 2 = 1 x2 + 3 2 3 4 + x2 = 1 1 4 x2 = x = ± 1 4 1 2 Our ﬁnal answers are √ 3 2 1 2, and −. √ 3 2 1 2, x = ± 1While this may seem like an opinion, it is indeed a fact. See Chapters 10 and 11 for details. 502 7.2.1 Exercises Hooked on Conics In Exercises 1 - 6, ﬁnd the standard equation of the circle and then graph it. 1. Center (−1, −5), radius 10 2. Center (4, −2), radius 3 3. Center −3, 7, radius 1 13 2 √ 2, radius π 5. Center −e, 4. Center (5, −9), radius ln(8) √ 6. Center (π, e2), radius 3 91 In Exercises 7 - 12, complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle. 7. x2 − 4x + y2 + 10y = −25 8. −2x2 − 36x − 2y2 − 112 = 0 9. x2 + y2 + 8x − 10y − 1 = 0 10. x2 + y2 + 5x − y − 1 = 0 11. 4x2 + 4y2 − 24y + 36 = 0 12. x2 + x + y2 − 6 5 y = 1 In Exercises 13 - 16, ﬁnd the standard equation of the circle which satisﬁes the given criteria. 13. center (3, 5), passes through (−1, −2) 14. center (3, 6), passes through (−1, 4) 15. endpoints of a diameter: (3, 6) and (−1, 4) 16. endpoints of a diameter: 1 2, 4, 3 2, −1 17. The Giant
Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet.2 Find an equation for the wheel assuming that its center lies on the y-axis and that the ground is the x-axis. 18. Verify that the following points lie on the Unit Circle: (±1, 0), (0, ±1), ± and √ 19. Discuss with your classmates how to obtain the standard equation of a circle, Equation 7.1, from the equation of the Unit Circle, x2 + y2 = 1 using the transformations discussed in Section 1.7. (Thus every circle is just a few transformations away from the Unit Circle.) 20. Find an equation for the function represented graphically by the top half of the Unit Circle. Explain how the transformations is Section 1.7 can be used to produce a function whose graph is either the top or bottom of an arbitrary circle. 21. Find a one-to-one function whose graph is half of a circle. (Hint: Think piecewise.) 2Source: Cedar Point’s webpage. 7.2 Circles 7.2.2 Answers 503 1. (x + 1)2 + (y + 5)2 = 100 2. (x − 4)2 + (y + 2)2 = 9 y 5 −11 −1 x 9 −5 −15 3. (x + 3)2 + y − 7 13 2 −5 4. (x − 5)2 + (y + 9)2 = (ln(8))2 y x 5 − ln(8) 5 5 + ln(8) y 27 26 7 13 1 26 x −9 + ln(8) −9 −9 − ln(8) = π2 6. (x − π)2 + y − e22 y = 91 2 3 22 y √ 2 + π − 7 2 −3 − 5 2 5. (x + e)2 + y − √ √ 2 x −e − π −e −e + π √ 2 − π e2 + 3√ 91 e2 e2 − 3√ 91 π − 3√ 91 π x π + 3√ 91 504 Hooked on Conics 7. (x − 2)2 + (y + 5)2 = 4 Center (2, −5), radius r = 2 9
. (x + 4)2 + (y − 5)2 = 42 Center (−4, 5), radius r = √ 42 11. x2 + (y − 3)2 = 0 This is not a circle. 8. (x + 9)2 + y2 = 25 Center (−9, 0), radius r = 5 2 10. x + 5 2 Center − 5 2 = 30 4, radius r = 2 12. x + 1 2 Center − 1 2 = 161 100, radius √ 30 2 √ 161 10 13. (x − 3)2 + (y − 5)2 = 65 14. (x − 3)2 + (y − 6)2 = 20 15. (x − 1)2 + (y − 5)2 = 5 17. x2 + (y − 72)2 = 4096 16. (x − 1)2 + y − 3 2 2 = 13 2 7.3 Parabolas 7.3 Parabolas 505 We have already learned that the graph of a quadratic function f (x) = ax2 + bx + c (a = 0) is called a parabola. To our surprise and delight, we may also deﬁne parabolas in terms of distance. Deﬁnition 7.3. Let F be a point in the plane and D be a line not containing F. A parabola is the set of all points equidistant from F and D. The point F is called the focus of the parabola and the line D is called the directrix of the parabola. Schematically, we have the following. F V D Each dashed line from the point F to a point on the curve has the same length as the dashed line from the point on the curve to the line D. The point suggestively labeled V is, as you should expect, the vertex. The vertex is the point on the parabola closest to the focus. We want to use only the distance deﬁnition of parabola to derive the equation of a parabola and, if all is right with the universe, we should get an expression much like those studied in Section 2.3. Let p denote the directed1 distance from the vertex to the focus, which by deﬁnition is the same as the distance from the vertex to the directrix. For simplicity, assume that the vertex is (0, 0)
and that the parabola opens upwards. Hence, the focus is (0, p) and the directrix is the line y = −p. Our picture becomes y (0, p) (0, 0) (x, y) x y = −p (x, −p) From the deﬁnition of parabola, we know the distance from (0, p) to (x, y) is the same as the distance from (x, −p) to (x, y). Using the Distance Formula, Equation 1.1, we get 1We’ll talk more about what ‘directed’ means later. 506 Hooked on Conics (x − 0)2 + (y − p)2 = x2 + (y − p)2 = x2 + (y − p)2 = (y + p)2 x2 + y2 − 2py + p2 = y2 + 2py + p2 (x − x)2 + (y − (−p))2 (y + p)2 x2 = 4py square both sides expand quantities gather like terms Solving for y yields y = x2 a = 1 4p and vertex (0, 0). 4p, which is a quadratic function of the form found in Equation 2.4 with We know from previous experience that if the coeﬃcient of x2 is negative, the parabola opens downwards. In the equation y = x2 4p this happens when p < 0. In our formulation, we say that p is a ‘directed distance’ from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0, the focus is below the vertex. The focal length of a parabola is \|p\|. If we choose to place the vertex at an arbitrary point (h, k), we arrive at the following formula using either transformations from Section 1.7 or re-deriving the formula from Deﬁnition 7.3. Equation 7.2. The Standard Equation of a Verticala Parabola: The equation of a (vertical) parabola with vertex (h, k) and focal length \|p\| is (x − h)2 = 4p(y − k) If p > 0, the parabola opens upwards; if p < 0, it opens downwards. a
That is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, only one of the variables, x, is squared. This is a quick way to distinguish an equation of a parabola from that of a circle because in the equation of a circle, both variables are squared. Example 7.3.1. Graph (x + 1)2 = −8(y − 3). Find the vertex, focus, and directrix. Solution. We recognize this as the form given in Equation 7.2. Here, x − h is x + 1 so h = −1, and y − k is y − 3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since p < 0, the focus will be below the vertex and the parabola will open downwards. y 5 4 3 2 1 −6 −5 −4 −3 −2 −1 1 2 3 4 x 7.3 Parabolas 507 The distance from the vertex to the focus is \|p\| = 2, which means the focus is 2 units below the vertex. From (−1, 3), we move down 2 units and ﬁnd the focus at (−1, 1). The directrix, then, is 2 units above the vertex, so it is the line y = 5. Of all of the information requested in the previous example, only the vertex is part of the graph of the parabola. So in order to get a sense of the actual shape of the graph, we need some more information. While we could plot a few points randomly, a more useful measure of how wide a parabola opens is the length of the parabola’s latus rectum.2 The latus rectum of a parabola is the line segment parallel to the directrix which contains the focus. The endpoints of the latus rectum are, then, two points on ‘opposite’ sides of the parabola. Graphically, we have the following. the latus rectum F V D It turns out3 that the length of the latus rectum, called the focal diameter of the parabola is \|4p\|, which, in light of Equation 7.2, is easy to ﬁnd. In our last example, for instance, when graphing (x +
1)2 = −8(y − 3), we can use the fact that the focal diameter is \| − 8\| = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the focus. Example 7.3.2. Find the standard form of the parabola with focus (2, 1) and directrix y = −4. Solution. Sketching the data yields, y 1 −1 1 2 3 x −1 −2 −3 The vertex lies on this vertical line midway between the focus and the directrix 2No, I’m not making this up. 3Consider this an exercise to show what follows. 508 Hooked on Conics From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don’t see that immediately.) Hence, the vertex lies below the focus and has an x-coordinate of 2. To ﬁnd the y-coordinate, we note that the distance from the focus to the directrix is 1 − (−4) = 5, which means the vertex lies 5 2 units (halfway) below the focus. Starting at (2, 1) and moving down 5/2 units leaves us at (2, −3/2), which is our vertex. Since the parabola opens upwards, we know p is positive. Thus p = 5/2. Plugging all of this data into Equation 7.2 give us (x − 2)2 = 4 (x − 2)2 = 10 If we interchange the roles of x and y, we can produce ‘horizontal’ parabolas: parabolas which open to the left or to the right. The directrices4 of such animals would be vertical lines and the focus would either lie to the left or to the right of the vertex, as seen below. D V F Equation 7.3. The Standard Equation of a Horizontal Parabola: The equation of a (horizontal) parabola with vertex (h, k) and focal length \|p\| is If p > 0, the parabola opens to the right; if p < 0, it opens to the left. (y − k)2 = 4p(x − h) 4plural of ‘directrix’ 7.3 Parabolas 509 Example 7
.3.3. Graph (y − 2)2 = 12(x + 1). Find the vertex, focus, and directrix. Solution. We recognize this as the form given in Equation 7.3. Here, x − h is x + 1 so h = −1, and y − k is y − 2 so k = 2. Hence, the vertex is (−1, 2). We also see that 4p = 12 so p = 3. Since p > 0, the focus will be the right of the vertex and the parabola will open to the right. The distance from the vertex to the focus is \|p\| = 3, which means the focus is 3 units to the right. If we start at (−1, 2) and move right 3 units, we arrive at the focus (2, 2). The directrix, then, is 3 units to the left of the vertex and if we move left 3 units from (−1, 2), we’d be on the vertical line x = −4. Since the focal diameter is \|4p\| = 12, the parabola is 12 units wide at the focus, and thus there are points 6 units above and below the focus on the parabola5 −4 −3 −2 −1 −1 1 2 3 x −2 −3 −4 If we As with circles, not all parabolas will come to us in the forms in Equations 7.2 or 7.3. encounter an equation with two variables in which exactly one variable is squared, we can attempt to put the equation into a standard form using the following steps. To Write the Equation of a Parabola in Standard Form 1. Group the variable which is squared on one side of the equation and position the non- squared variable and the constant on the other side. 2. Complete the square if necessary and divide by the coeﬃcient of the perfect square. 3. Factor out the coeﬃcient of the non-squared variable from it and the constant. Example 7.3.4. Consider the equation y2 + 4y + 8x = 4. Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. Solution. We need a perfect square (in this case, using y) on the left-hand side of the equation and factor out the coeﬃcient of the non-squared variable (in this
case, the x) on the other. 510 Hooked on Conics y2 + 4y + 8x = 4 y2 + 4y = −8x + 4 y2 + 4y + 4 = −8x + 4 + 4 complete the square in y only factor (y + 2)2 = −8x + 8 (y + 2)2 = −8(x − 1) Now that the equation is in the form given in Equation 7.3, we see that x − h is x − 1 so h = 1, and y − k is y + 2 so k = −2. Hence, the vertex is (1, −2). We also see that 4p = −8 so that p = −2. Since p < 0, the focus will be the left of the vertex and the parabola will open to the left. The distance from the vertex to the focus is \|p\| = 2, which means the focus is 2 units to the left of 1, so if we start at (1, −2) and move left 2 units, we arrive at the focus (−1, −2). The directrix, then, is 2 units to the right of the vertex, so if we move right 2 units from (1, −2), we’d be on the vertical line x = 3. Since the focal diameter is \|4p\| is 8, the parabola is 8 units wide at the focus, so there are points 4 units above and below the focus on the parabola. y 2 1 −2 −1 1 2 x −1 −2 −3 −4 −5 −6 In studying quadratic functions, we have seen parabolas used to model physical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its ‘reﬂective property’ which necessitates knowing about the focus of a parabola. For example, many satellite dishes are formed in the shape of a paraboloid of revolution as depicted below. 7.3 Parabolas 511 Every cross section through the vertex of the paraboloid is a parabola with the same focus. To see why this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reﬂect oﬀ the parabola and concentrate at the focus which then becomes the optimal place for the receiver. If, on
the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reﬂected oﬀ the parabola in a coherent fashion as in the case in a ﬂashlight. Here, the bulb is placed at the focus and the light rays are reﬂected oﬀ a parabolic mirror to give directional light. F Example 7.3.5. A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be? Solution. One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we’ll assume the vertex is (0, 0) and the parabola opens upwards. Our standard form for such a parabola is x2 = 4py. Since the focus is 2 units above the vertex, we know p = 2, so we have x2 = 8y. Visually, y 12 units wide 2 (6, y)? 6 x −6 Since the parabola is 12 feet wide, we know the edge is 6 feet from the vertex. To ﬁnd the depth, we are looking for the y value when x = 6. Substituting x = 6 into the equation of the parabola yields 62 = 8y or y = 36 2 = 4.5. Hence, the dish will be 4.5 feet deep. 8 = 9 512 7.3.1 Exercises Hooked on Conics In Exercises 1 - 8, sketch the graph of the given parabola. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch. 1. (x − 3)2 = −16y 3. (y − 2)2 = −12(x + 3) 5. (x − 1)2 = 4(y + 3) 7. (y − 4)2 = 18(x − 2) 2. (y + 4)2 = 4x 6. (x + 2)2 = −20(y − 5) 8. y + 3 2 2 = −7 x + 9 2 In Exercises 9 - 14, put the equation into standard form and identify the vertex, focus and
directrix. 9. y2 − 10y − 27x + 133 = 0 10. 25x2 + 20x + 5y − 1 = 0 11. x2 + 2x − 8y + 49 = 0 12. 2y2 + 4y + x − 8 = 0 13. x2 − 10x + 12y + 1 = 0 14. 3y2 − 27y + 4x + 211 4 = 0 In Exercises 15 - 18, ﬁnd an equation for the parabola which ﬁts the given criteria. 15. Vertex (7, 0), focus (0, 0) 16. Focus (10, 1), directrix x = 5 17. Vertex (−8, −9); (0, 0) and (−16, 0) are 18. The endpoints of latus rectum are (−2, −7) points on the curve and (4, −7) 19. The mirror in Carl’s ﬂashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror? 20. A parabolic Wi-Fi antenna is constructed by taking a ﬂat sheet of metal and bending it into a parabolic shape.5 If the cross section of the antenna is a parabola which is 45 centimeters wide and 25 centimeters deep, where should the receiver be placed to maximize reception? 21. A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch. 22. A popular novelty item is the ‘mirage bowl.’ Follow this link to see another startling appli- cation of the reﬂective property of the parabola. 23. With the help of your classmates, research spinning liquid mirrors. To get you started, check out this website. 5This shape is called a ‘parabolic cylinder.’ 7.3 Parabolas 7.3.2 Answers 513 1. (x − 3)2 = −16y Vertex (3, 0) Focus (3, −4) Directrix y = 4 Endpoints of latus rectum (−5, −4), (11, −4) y 4 3 2 1 −5−4−3−2
−1 − 10 11 x −2 −3 −4 2 Vertex − 7 3, − 5 Focus − 7 3, −2 Directrix y = −3 Endpoints of latus rectum − 10 2 3, −2, − 4 3, −2 3. (y − 2)2 = −12(x + 3) Vertex (−3, 2) Focus (−6, 2) Directrix x = 0 Endpoints of latus rectum (−6, 8), (−6, −4) y 2 1 −5 −4 −3 −2 −1 x −1 −2 −7−6−5−4−3−2−1 −1 x −2 −3 −4 514 Hooked on Conics 4. (y + 4)2 = 4x Vertex (0, −4) Focus (1, −4) Directrix x = −1 Endpoints of latus rectum (1, −2), (1, −6) 5. (x − 1)2 = 4(y + 3) Vertex (1, −3) Focus (1, −2) Directrix y = −4 Endpoints of latus rectum (3, −2), (−1, −2) 6. (x + 2)2 = −20(y − 5) Vertex (−2, 5) Focus (−2, 0) Directrix y = 10 Endpoints of latus rectum (−12, 0), (8, 0) 7. (y − 4)2 = 18(x − 2) Vertex (2, 4) Focus 13 2, 4 Directrix x = − 5 2 Endpoints of latus rectum 13 2, −5, 13 2, 13 y −1 1 2 3 4 x −1 −2 −3 −4 −5 −6 −7 −8 y −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 y 10 12 −10 −8 −6 −4 −2 2 4 6 8 x y 13 11 9 7 5 3 1 −1 −3 −5 7.3 Parabolas 515 8. y + 3 2 2 2 = − Vertex − 9 Focus − 25 Directrix x = − 11 4 Endpoints of latus rectum − 25 2 2 4, 2, − 25 4, −5 9. (y − 5)2 = 27(x − 4) Vertex (
4, 5) 4, 5 Focus 43 Directrix x = − 11 4 11. (x + 1)2 = 8(y − 6) Vertex (−1, 6) Focus (−1, 8) Directrix y = 4 13. (x − 5)2 = −12(y − 2) Vertex (5, 2) Focus (5, −1) Directrix y = 5 15. y2 = −28(x − 7) 17. (x + 8)2 = 64 9 (y + 9) −5 −4 −3 −2 −1 y x 2 1 −1 −2 −3 −4 −5 5 (y − 1) 2 10 Vertex − 2 Focus − 2 5, 19 Directrix y = 21 20 20 12. (y + 1)2 = − 1 2 (x − 10) Vertex (10, −1) Focus 79 8, −1 Directrix x = 81 8 3 (x − 2) 14. y − 9 2 = − 4 2 Vertex 2, 9 2 Focus 5 3, 9 Directrix x = 7 3 2 16. (y − 1)2 = 10 x − 15 2 or 18. (x − 1)2 = 6 y + 17 2 (x − 1)2 = −6 y + 11 2 19. The bulb should be placed 0.625 centimeters above the vertex of the mirror. (As veriﬁed by Carl himself!) 20. The receiver should be placed 5.0625 centimeters from the vertex of the cross section of the antenna. 21. The arch can be modeled by x2 = −(y − 9) or y = 9 − x2. One foot in from the base of the arch corresponds to either x = ±2, so the height is y = 9 − (±2)2 = 5 feet. 516 7.4 Ellipses Hooked on Conics In the deﬁnition of a circle, Deﬁnition 7.1, we ﬁxed a point called the center and considered all of the points which were a ﬁxed distance r from that one point. For our next conic section, the ellipse, we ﬁx two distinct points and a distance d to use in our deﬁnition. Deﬁnition 7.4. Given two distinct points F1 and F2 in the plane and
a ﬁxed distance d, an ellipse is the set of all points (x, y) in the plane such that the sum of each of the distances from F1 and F2 to (x, y) is d. The points F1 and F2 are called the focia of the ellipse. athe plural of ‘focus’ (x, y) d1 F1 d2 F2 d1 + d2 = d for all (x, y) on the ellipse We may imagine taking a length of string and anchoring it to two points on a piece of paper. The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse. The center of the ellipse is the midpoint of the line segment connecting the two foci. The major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which also contains the center and foci. The minor axis of the ellipse is the line segment connecting two opposite ends of the ellipse which contains the center but is perpendicular to the major axis. The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures we have, 7.4 Ellipses 517 s i x A r o n i M V1 Major Axis F1 C F2 V2 An ellipse with center C; foci F1, F2; and vertices V1, V2 Note that the major axis is the longer of the two axes through the center, and likewise, the minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we assume that the ellipse has its center at (0, 0), its major axis along the x-axis, and has foci (c, 0) and (−c, 0) and vertices (−a, 0) and (a, 0). We will label the y-intercepts of the ellipse as (0, b) and (0, −b) (We assume a, b, and c are all positive numbers.) Schematically, y (0, b) (−a, 0) (−c, 0) (c, 0) (x, y) x (
a, 0) (0, −b) Note that since (a, 0) is on the ellipse, it must satisfy the conditions of Deﬁnition 7.4. That is, the distance from (−c, 0) to (a, 0) plus the distance from (c, 0) to (a, 0) must equal the ﬁxed distance d. Since all of these points lie on the x-axis, we get distance from (−c, 0) to (a, 0) + distance from (c, 0) to (a, 0) = d (a + c) + (a − c) = d 2a = d 518 Hooked on Conics In other words, the ﬁxed distance d mentioned in the deﬁnition of the ellipse is none other than the length of the major axis. We now use that fact (0, b) is on the ellipse, along with the fact that d = 2a to get (0 − (−c))2 + (b − 0)2 + distance from (−c, 0) to (0, b) + distance from (c, 0) to (0, b) = 2a (0 − c)2 + (b − 0)2 = 2a b2 + c2 = 2a b2 + c2 + b2 + c2 = 2a 2 b2 + c2 = a √ √ √ √ From this, we get a2 = b2 + c2, or b2 = a2 − c2, which will prove useful later. Now consider a point (x, y) on the ellipse. Applying Deﬁnition 7.4, we get distance from (−c, 0) to (x, y) + distance from (c, 0) to (x, y) = 2a (x − c)2 + (y − 0)2 = 2a (x − c)2 + y2 = 2a (x − (−c))2 + (y − 0)2 + (x + c)2 + y2 + In order to make sense of this situation, we need to make good use of Intermediate Algebra. (x + c)2 + y2 + (x + c)2 + y22 (x − c)2 + y2 = 2a (x +
c)2 + y2 = 2a − 2a − (x + c)2 + y2 = 4a2 − 4a (x − c)2 + y2 = 4a2 + (x − c)2 − (x + c)2 (x − c)2 + y2 = 4a2 − 4cx (x − c)2 + y2 = a2 − cx = 4a 4a a (x − c)2 + y22 = a2 − cx2 (x − c)2 + y2 (x − c)2 + y22 (x − c)2 + y2 + (x − c)2 + y2 a a2 (x − c)2 + y2 = a4 − 2a2cx + c2x2 a2x2 − 2a2cx + a2c2 + a2y2 = a4 − 2a2cx + c2x2 a2x2 − c2x2 + a2y2 = a4 − a2c2 a2 − c2 x2 + a2y2 = a2 a2 − c2 We are nearly ﬁnished. Recall that b2 = a2 − c2 so that a2 − c2 x2 + a2y2 = a2 a2 − c2 b2x2 + a2y2 = a2b2 y2 b2 = 1 x2 a2 + 7.4 Ellipses 519 This equation is for an ellipse centered at the origin. To get the formula for the ellipse centered at (h, k), we could use the transformations from Section 1.7 or re-derive the equation using Deﬁnition 7.4 and the distance formula to obtain the formula below. Equation 7.4. The Standard Equation of an Ellipse: For positive unequal numbers a and b, the equation of an ellipse with center (h, k) is (x − h)2 a2 + (y − k)2 b2 = 1 Some remarks about Equation 7.4 are in order. First note that the values a and b determine how far in the x and y directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if a > b, then we have an ellipse whose major axis
is horizontal, In this case, as we’ve seen in the and hence, the foci lie to the left and right of the center. √ a2 − b2. If b > a, derivation, the distance from the center to the focus, c, can be found by c = the roles of the major and minor axes are reversed, and the foci lie above and below the center. b2 − a2. In either case, c is the distance from the center to each focus, and In this case, c = bigger denominator − smaller denominator. Finally, it is worth mentioning that if we take c = the standard equation of a circle, Equation 7.1, and divide both sides by r2, we get √ √ Equation 7.5. The Alternate Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x − h)2 r2 + (y − k)2 r2 = 1 Notice the similarity between Equation 7.4 and Equation 7.5. Both equations involve a sum of squares equal to 1; the diﬀerence is that with a circle, the denominators are the same, and with an ellipse, they are diﬀerent. If we take a transformational approach, we can consider both Equations 7.4 and 7.5 as shifts and stretches of the Unit Circle x2 + y2 = 1 in Deﬁnition 7.2. Replacing x with (x − h) and y with (y − k) causes the usual horizontal and vertical shifts. Replacing x with x a and y with y b causes the usual vertical and horizontal stretches. In other words, it is perfectly ﬁne to think of an ellipse as the deformation of a circle in which the circle is stretched farther in one direction than the other.1 Example 7.4.1. Graph (x+1)2 and minor axes, the vertices, the endpoints of the minor axis, and the foci. 9 + (y−2)2 25 = 1. Find the center, the lines which contain the major Solution. We see that this equation is in the standard form of Equation 7.4. Here x − h is x + 1 so h = −1, and y − k is y − 2 so k = 2. Hence, our
ellipse is centered at (−1, 2). We see that a2 = 9 so a = 3, and b2 = 25 so b = 5. This means that we move 3 units left and right from the center and 5 units up and down from the center to arrive at points on the ellipse. As an aid to sketching, we draw a rectangle matching this description, called a guide rectangle, and sketch the ellipse inside this rectangle as seen below on the left. 1This was foreshadowed in Exercise 19 in Section 7.2. 520 Hooked on Conics 4 −3 −2 −1 1 2 x −4 −3 −2 −1 1 2 x −1 −2 −3 −1 −2 −3 Since we moved farther in the y direction than in the x direction, the major axis will lie along the vertical line x = −1, which means the minor axis lies along the horizontal line, y = 2. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (−1, 7) and (−1, −3), and the endpoints of the minor axis are (−4, 2) and (2, 2). (Notice these points are the four points we used to draw the guide rectangle.) To ﬁnd the foci, we ﬁnd c = 16 = 4, which means the foci lie 4 units from the center. Since the major axis is vertical, the foci lie 4 units above and below the center, at (−1, −2) and (−1, 6). Plotting all this information gives the graph seen above on the right. 25 − 9 = √ √ Example 7.4.2. Find the equation of the ellipse with foci (2, 1) and (4, 1) and vertex (0, 1). Solution. Plotting the data given to us, we have y 1 1 2 3 4 5 x From this sketch, we know that the major axis is horizontal, meaning a > b. Since the center is the midpoint of the foci, we know it is (3, 1). Since one vertex is (0, 1) we have that a = 3, so a2 = 9. All that remains is to ﬁnd b2. Since the foci are 1 unit away from the center, we know c = 1. Since 9 − b2,
so b2 = 8. Substituting all of our ﬁndings into the a > b, we have c = equation (x−h)2 9 + (y−1)2 √ a2 − b2, or 1 = b2 = 1, we get our ﬁnal answer to be (x−3)2 a2 + (y−k)2 8 = 1. √ 7.4 Ellipses 521 As with circles and parabolas, an equation may be given which is an ellipse, but isn’t in the standard form of Equation 7.4. In those cases, as with circles and parabolas before, we will need to massage the given equation into the standard form. To Write the Equation of an Ellipse in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square in both variables as needed. 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1. Example 7.4.3. Graph x2 + 4y2 − 2x + 24y + 33 = 0. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. Solution. Since we have a sum of squares and the squared terms have unequal coeﬃcients, it’s a good bet we have an ellipse on our hands.2 We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1. x2 + 4y2 − 2x + 24y + 33 = 0 x2 − 2x + 4y2 + 24y = −33 x2 − 2x + 4 y2 + 6y = −33 x2 − 2x + 1 + 4 y2 + 6y + 9 = −33 + 1 + 4(9) (x − 1)2 + 4(y + 3)2 = 4 (x − 1)2 + 4(y + 3)2 4 4 4 = (x − 1)2 4 (x − 1)2 4 + (y + 3)2 = 1 + (y + 3)2 1 = 1 Now that this equation is in the standard form of Equation 7.4, we see that x −
h is x − 1 so h = 1, and y − k is y + 3 so k = −3. Hence, our ellipse is centered at (1, −3). We see that a2 = 4 so a = 2, and b2 = 1 so b = 1. This means we move 2 units left and right from the center and 1 unit up and down from the center to arrive at points on the ellipse. Since we moved farther in the x direction than in the y direction, the major axis will lie along the horizontal line y = −3, which means the minor axis lies along the vertical line x = 1. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points (−1, −3) and (3, −3), and the endpoints of the minor axis are (1, −2) and (1, −4). To ﬁnd the foci, we ﬁnd c = 3, which means 4 − 1 = √ √ 2The equation of a parabola has only one squared variable and the equation of a circle has two squared variables with identical coeﬃcients. 522 Hooked on Conics √ √ the foci lie left and right of the center, at (1 − √ 3 units from the center. Since the major axis is horizontal, the foci lie √ 3 units to the 3, −3). Plotting all of this information gives 3, −3) and (1 + y −1 1 2 3 4 x −1 −2 −3 −4 As you come across ellipses in the homework exercises and in the wild, you’ll notice they come in all shapes in sizes. Compare the two ellipses below. Certainly, one ellipse is more round than the other. This notion of ‘roundness’ is quantiﬁed below. Deﬁnition 7.5. The eccentricity of an ellipse, denoted e, is the following ratio: e = distance from the center to a focus distance from the center to a vertex In an ellipse, the foci are closer to the center than the vertices, so 0 < e < 1. The ellipse above on the left has eccentricity e ≈ 0.98; for the ellipse above on the right, e �
� 0.66. In general, the closer the eccentricity is to 0, the more ‘circular’ the ellipse; the closer the eccentricity is to 1, the more ‘eccentric’ the ellipse. Example 7.4.4. Find the equation of the ellipse whose vertices are (±5, 0) with eccentricity e = 1 4. Solution. As before, we plot the data given to us y x 7.4 Ellipses 523 From this sketch, we know that the major axis is horizontal, meaning a > b. With the vertices located at (±5, 0), we get a = 5 so a2 = 25. We also know that the center is (0, 0) because the center is the midpoint of the vertices. All that remains is to ﬁnd b2. To that end, we use the fact that the eccentricity e = 1 4 which means e = distance from the center to a focus distance from the center to a vertex = c a = c 5 = 1 4 from which we get c = 5 we get b2 = 375 ﬁnal answer x2 4. To get b2, we use the fact that c = a2 − b2, so 5 16. Substituting all of our ﬁndings into the equation (x−h)2 25 + 16y2 375 = 1. 4 = a2 + (y−k)2 25 − b2 from which b2 = 1, yields our √ √ As with parabolas, ellipses have a reﬂective property. If we imagine the dashed lines below representing sound waves, then the waves emanating from one focus reﬂect oﬀ the top of the ellipse and head towards the other focus. F1 F2 Such geometry is exploited in the construction of so-called ‘Whispering Galleries’. If a person whispers at one focus, a person standing at the other focus will hear the ﬁrst person as if they were standing right next to them. We explore the Whispering Galleries in our last example. Example 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room
is 40 feet high at the center and 100 feet wide at the ﬂoor, how far from the outer wall should each of them stand so that they will be positioned at the foci of the ellipse? Solution. Graphing the data yields 524 Hooked on Conics y 40 units tall 100 units wide x It’s most convenient to imagine this ellipse centered at (0, 0). Since the ellipse is 100 units wide and 40 units tall, we get a = 50 and b = 40. Hence, our ellipse has the equation x2 402 = 1. √ 900 = 30, so that the foci are 30 units We’re looking for the foci, and we get c = from the center. That means they are 50 − 30 = 20 units from the vertices. Hence, Jason and Jamie should stand 20 feet from opposite ends of the gallery. 502 + y2 502 − 402 = √ 7.4 Ellipses 7.4.1 Exercises 525 In Exercises 1 - 8, graph the ellipse. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity. = 1 1. 3. 5. 7. + x2 y2 169 25 (x − 2)2 4 (x − 1)2 10 (x + 2)2 16 + + + (y + 3)2 9 (y − 3)2 11 (y − 5)2 20 = 1 = 1 = 1 2. 4. 6. 8. + x2 y2 9 25 (x + 5)2 16 (x − 1)2 9 (x − 4)2 8 = 1 + + + (y − 4)2 1 (y + 3)2 4 (y − 2)2 18 = 1 = 1 = 1 In Exercises 9 - 14, put the equation in standard form. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity. 9. 9x2 + 25y2 − 54x − 50y − 119 = 0 10. 12x2 + 3y2 − 30y + 39 = 0 11. 5x2 + 18y2 − 30x + 72y + 27 = 0 12. x2 − 2x + 2
y2 − 12y + 3 = 0 13. 9x2 + 4y2 − 4y − 8 = 0 14. 6x2 + 5y2 − 24x + 20y + 14 = 0 In Exercises 15 - 20, ﬁnd the standard form of the equation of the ellipse which has the given properties. 15. Center (3, 7), Vertex (3, 2), Focus (3, 3) 16. Foci (0, ±5), Vertices (0, ±8). 17. Foci (±3, 0), length of the Minor Axis 10 18. Vertices (3, 2), (13, 2); Endpoints of the Minor Axis (8, 4), (8, 0) 19. Center (5, 2), Vertex (0, 2), eccentricity 1 2 20. All points on the ellipse are in Quadrant IV except (0, −9) and (8, 0). (One might also say that the ellipse is “tangent to the axes” at those two points.) 21. Repeat Example 7.4.5 for a whispering gallery 200 feet wide and 75 feet tall. 22. An elliptical arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch. Compare your result with your answer to Exercise 21 in Section 7.3. 526 Hooked on Conics 23. The Earth’s orbit around the sun is an ellipse with the sun at one focus and eccentricity e ≈ 0.0167. The length of the semimajor axis (that is, half of the major axis) is deﬁned to be 1 astronomical unit (AU). The vertices of the elliptical orbit are given special names: ‘aphelion’ is the vertex farthest from the sun, and ‘perihelion’ is the vertex closest to the sun. Find the distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion. 24. The graph of an ellipse clearly fails the Vertical Line Test, Theorem 1.1, so the equation of an ellipse does not deﬁne y as a function of x. However, much like with circles and horizontal parabolas, we can split an
ellipse into a top half and a bottom half, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the ellipses given in Exercises 1 - 8 above. What diﬃculties arise when you plot them on the calculator? 25. Some famous examples of whispering galleries include St. Paul’s Cathedral in London, England, National Statuary Hall in Washington, D.C., and The Cincinnati Museum Center. With the help of your classmates, research these whispering galleries. How does the whispering effect compare and contrast with the scenario in Example 7.4.5? 26. With the help of your classmates, research “extracorporeal shock-wave lithotripsy”. It uses the reﬂective property of the ellipsoid to dissolve kidney stones. 7.4 Ellipses 7.4.2 Answers 1. 2. + = 1 y2 25 x2 169 Center (0, 0) Major axis along y = 0 Minor axis along x = 0 Vertices (13, 0), (−13, 0) Endpoints of Minor Axis (0, −5), (0, 5) Foci (12, 0), (−12, 0) e = 12 13 + = 1 x2 y2 25 9 Center (0, 0) Major axis along x = 0 Minor axis along y = 0 Vertices (0, 5), (0, −5) Endpoints of Minor Axis (−3, 0), (3, 0) Foci (0, −4), (0, 4) e = 4 5 3. (x − 2)2 4 + (y + 3)2 9 = 1 Center (2, −3) Major axis along x = 2 Minor axis along y = −3 Vertices (2, 0), (2, −6) Endpoints of Minor Axis (0, −3), (4, −3) Foci (2, −3 + 5), (2, −3 − 5 e = 3 5) √ √ √ 4. (x + 5)2 16 + (y − 4)2 1 = 1 Center (−5, 4) Major axis along y = 4 Minor axis along x = −5 Vertices (−9, 4), (−1, 4) Endpoints of Minor Axis (−5, 3), (−5, 5) 15, 4), (−5 − F
oci (−5 + √ e = 15, 4) √ √ 15 4 527 x 13 −13 y 1 5 4 3 2 1 −1 −1 −2 −3 −4 −5 y 5 4 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 −4 −5 y 1 2 3 4 x −1 −2 −3 −4 −5 −6 y 5 4 3 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1 x 528 Hooked on Conics 5. (x − 1)2 10 + (y − 3)2 11 = 1 √ Center (1, 3) Major axis along x = 1 Minor axis along y = 3 11), (1, 3 − Vertices (1, 3 + Endpoints of the Minor Axis (1 − Foci (1, 2), (1, 4) e = 10, 3), (1 + 10, 3) √ √ √ 11 11 √ 11) y 6 5 4 3 2 1 6. (x − 1)2 9 + (y + 3)2 4 = 1 Center (1, −3) Major axis along y = −3 Minor axis along x = 1 Vertices (4, −3), (−2, −3) Endpoints of the Minor Axis (1, −1), (1, −5) Foci (1 + √ e = 5, −3), (1 − 5, −3) √ √ 5 3 7. (x + 2)2 16 + (y − 5)2 20 = 1 Center (−2, 5) Major axis along x = −2 Minor axis along y = 5 Vertices (−2, 5 + 2 Endpoints of the Minor Axis (−6, 5), (2, 5) Foci (−2, 7), (−2, 3) e = 5), (−2, 5 − 2 5) √ √ √ 5 5 −2 −1 1 2 3 4 y −2 −1 1 2 3 4 x x −1 −2 −3 −4 −5 y 10 6 −5 −4 −3 −2 −1 1 2 x 7.4 Ellipses 529 8. (x − 4)2 8 + (y − 2)2 18 = 1 √ Center (4, 2) Major axis along x = 4 Minor axis along y = 2 Vertices
(4, 2 + 3 Endpoints of the Minor Axis (4 − 2 2, 2), (4 + 2 √ Foci (4, 2 + √ 5 e = 3 2, 2) 10), (4, 2 − √ √ √ 2), (4, 2 − 3 10) √ 21 −2 −. 11. = 1 10. (x − 3)2 25 + (y − 1)2 9 Center (3, 1) Major Axis along y = 1 Minor Axis along x = 3 Vertices (8, 1), (−2, 1) Endpoints of Minor Axis (3, 4), (3, −2) Foci (7, 1), (−1, 1) e = 4 5 = 1 12. (x − 3)2 18 + (y + 2)2 5 √ Center (3, −2) Major axis along y = −2 Minor axis along x = 3 Vertices (3 − 3 2, −2) Endpoints of Minor Axis (3, −2 + (3, −2 − Foci (3 − √ e = √ 5) √ 13, −2), (3 + 2, −2), (3 + 3 13, −2) √ √ 26 6 √ 5), + = 1 (y − 5)2 12 x2 3 Center (0, 5) Major axis along x = 0 Minor axis along y = 5 Vertices (0, 5 − 2 Endpoints of Minor Axis (− Foci (0, 2), (0, 8) e = √ √ 3 2 3), (0, 5 + 2 √ √ 3) 3, 5), ( √ 3, 5) (x − 1)2 16 + (y − 3)2 8 = 1 Center (1, 3) Major Axis along y = 3 Minor Axis along x = 1 Vertices (5, 3), (−3, 3) Endpoints of Minor Axis (1, 3 + 2 (1, 3 − 2 Foci (1 + 2 √ 2 e = 2 2, 3), (1 − 2 2, 3) 2) √ √ √ √ 2), 530 13. + 2 = 1 4 y − 1 x2 2 9 1 Center 0, 1 2 Major Axis along x = 0 (the y-axis) Minor Axis along y = 1 2 Vertices (0, 2), (0
, −1) Endpoints of Minor Axis −1, 1 2 Foci, √ √ 5 5 0, 1− 2, 1, 1 2 √ 0, 1+ 2 5 3 e = 14. Hooked on Conics = 1 (x − 2)2 5 + (y + 2)2 6 √ Center (2, −2) Major Axis along x = 2 Minor Axis along y = −2 Vertices 2, −2 + Endpoints of Minor Axis 2 − 2 + 5, −2 Foci (2, −1), (2, −3) e = 6, (2, −2 − √ √ 6 6 √ 6) √ 5, −2, 15. 17. 19. (x − 3)2 9 + x2 y2 34 25 (x − 5)2 25 + (y − 7)2 25 = 1 = 1 + 4(y − 2)2 75 = 1 21. Jamie and Jason should stand 100 − 25 √ 16. 18. 20. + y2 x2 64 39 (x − 8)2 25 (x − 8)2 64 = 1 + + (y − 2)2 4 (y + 9)2 81 = 1 = 1 7 ≈ 33.86 feet from opposite ends of the gallery. 22. The arch can be modeled by the top half of x2 9 + y2 corresponds to either x = ±2. Plugging in x = ±2 gives y = ±3 5 ≈ 6.71 feet. height, we choose y = 3 √ 81 = 1. One foot in from the base of the arch 5 and since y represents a √ 23. Distance from the sun to aphelion ≈ 1.0167 AU. Distance from the sun to perihelion ≈ 0.9833 AU. 7.5 Hyperbolas 7.5 Hyperbolas 531 In the deﬁnition of an ellipse, Deﬁnition 7.4, we ﬁxed two points called foci and looked at points whose distances to the foci always added to a constant distance d. Those prone to syntactical tinkering may wonder what, if any, curve we’d generate if we replaced added with subtracted. The answer is a hyperbola. Deﬁnition 7.6. Given two distinct points F1 and F2 in
the plane and a ﬁxed distance d, a hyperbola is the set of all points (x, y) in the plane such that the absolute value of the diﬀerence of each of the distances from F1 and F2 to (x, y) is d. The points F1 and F2 are called the foci of the hyperbola. (x1, y1) F2 F1 (x2, y2) In the ﬁgure above: the distance from F1 to (x1, y1) − the distance from F2 to (x1, y1) = d and the distance from F2 to (x2, y2) − the distance from F1 to (x2, y2) = d Note that the hyperbola has two parts, called branches. The center of the hyperbola is the midpoint of the line segment connecting the two foci. The transverse axis of the hyperbola is the line segment connecting two opposite ends of the hyperbola which also contains the center and foci. The vertices of a hyperbola are the points of the hyperbola which lie on the transverse axis. In addition, we will show momentarily that there are lines called asymptotes which the branches of the hyperbola approach for large x and y values. They serve as guides to the graph. In pictures, 532 Hooked on Conics Transverse Axis F1 V1 C V2 F2 A hyperbola with center C; foci F1, F2; and vertices V1, V2 and asymptotes (dashed) Before we derive the standard equation of the hyperbola, we need to discuss one further parameter, the conjugate axis of the hyperbola. The conjugate axis of a hyperbola is the line segment through the center which is perpendicular to the transverse axis and has the same length as the line segment through a vertex which connects the asymptotes. In pictures we have V1 V2 Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse and conjugate axes whose center is the center of the hyperbola and whose diagonals are contained in the asymptotes. This guide rectangle, much akin to the one we saw Section 7.4 to help us graph ellipses, will
aid us in graphing hyperbolas. Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall assume that the center is (0, 0), the vertices are (a, 0) and (−a, 0) and the foci are (c, 0) and (−c, 0). We label the 7.5 Hyperbolas 533 endpoints of the conjugate axis (0, b) and (0, −b). (Although b does not enter into our derivation, we will have to justify this choice as you shall see later.) As before, we assume a, b, and c are all positive numbers. Schematically we have y (0, b) (x, y) (−c, 0) (−a, 0) (a, 0) x (c, 0) (0, −b) Since (a, 0) is on the hyperbola, it must satisfy the conditions of Deﬁnition 7.6. That is, the distance from (−c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal the ﬁxed distance d. Since all these points lie on the x-axis, we get distance from (−c, 0) to (a, 0) − distance from (c, 0) to (a, 0) = d (a + c) − (c − a) = d 2a = d In other words, the ﬁxed distance d from the deﬁnition of the hyperbola is actually the length of the transverse axis! (Where have we seen that type of coincidence before?) Now consider a point (x, y) on the hyperbola. Applying Deﬁnition 7.6, we get distance from (−c, 0) to (x, y) − distance from (c, 0) to (x, y) = 2a (x − c)2 + (y − 0)2 = 2a (x − c)2 + y2 = 2a (x − (−c))2 + (y − 0)2 − (x + c)2 + y2 − Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation 7.4, we arrive at the following.1 1It is a good
exercise to actually work this out. 534 Hooked on Conics a2 − c2 x2 + a2y2 = a2 a2 − c2 What remains is to determine the relationship between a, b and c. To that end, we note that since a and c are both positive numbers with a < c, we get a2 < c2 so that a2 − c2 is a negative number. Hence, c2 − a2 is a positive number. For reasons which will become clear soon, we re-write the equation by solving for y2/x2 to get a2 − c2 x2 + a2y2 = a2 a2 − c2 − c2 − a2 x2 + a2y2 = −a2 c2 − a2 a2y2 = c2 − a2 x2 − a2 c2 − a2 y2 x2 = − c2 − a2 c2 − a2 a2 x2 (c2−a2) x2 → 0 so that y2 a x as \|x\| grows large. Thus y = ± b (c2−a2) a2 x2 → As x and y attain very large values, the quantity. By setting b2 = c2 − a2 we get y2 x2 → b2 a x are the asymptotes to the graph as predicted and our choice of labels for the endpoints of the conjugate axis is justiﬁed. In our equation of the hyperbola we can substitute a2 − c2 = −b2 which yields a2. This shows that y → ± b a2 − c2 x2 + a2y2 = a2 a2 − c2 −b2x2 + a2y2 = −a2b2 y2 b2 = 1 x2 a2 − The equation above is for a hyperbola whose center is the origin and which opens to the left and right. If the hyperbola were centered at a point (h, k), we would get the following. Equation 7.6. The Standard Equation of a Horizontala Hyperbola For positive numbers a and b, the equation of a horizontal hyperbola with center (h, k) is aThat is, a hyperbola whose branches open to the left and right (x − h)2 a2 − (y − k)2 b2 =
1 If the roles of x and y were interchanged, then the hyperbola’s branches would open upwards and downwards and we would get a ‘vertical’ hyperbola. Equation 7.7. The Standard Equation of a Vertical Hyperbola For positive numbers a and b, the equation of a vertical hyperbola with center (h, k) is: (y − k)2 b2 − (x − h)2 a2 = 1 The values of a and b determine how far in the x and y directions, respectively, one counts from the center to determine the rectangle through which the asymptotes pass. In both cases, the distance 7.5 Hyperbolas 535 a2 + b2. from the center to the foci, c, as seen in the derivation, can be found by the formula c = Lastly, note that we can quickly distinguish the equation of a hyperbola from that of a circle or ellipse because the hyperbola formula involves a diﬀerence of squares where the circle and ellipse formulas both involve the sum of squares. √ Example 7.5.1. Graph the equation (x−2)2 4 − y2 transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 25 = 1. Find the center, the lines which contain the Solution. We ﬁrst see that this equation is given to us in the standard form of Equation 7.6. Here x − h is x − 2 so h = 2, and y − k is y so k = 0. Hence, our hyperbola is centered at (2, 0). We see that a2 = 4 so a = 2, and b2 = 25 so b = 5. This means we move 2 units to the left and right of the center and 5 units up and down from the center to arrive at points on the guide rectangle. The asymptotes pass through the center of the hyperbola as well as the corners of the rectangle. This yields the following set up2 −1 1 2 3 4 5 6 x −1 −2 −3 −4 −5 −6 −7 Since the y2 term is being subtracted from the x2 term, we know that the branches of the hyperbola open to the left and right. This means that the transverse axis lies along the x-
axis. Hence, the conjugate axis lies along the vertical line x = 2. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are 2 units to the left and right of (2, 0) at (0, 0) and (4, 0). To ﬁnd the foci, we need c = 29. Since the foci 29, 0) lie on the transverse axis, we move (approximately (−3.39, 0)) and (2 + 29, 0) (approximately (7.39, 0)). To determine the equations of the asymptotes, recall that the asymptotes go through the center of the hyperbola, (2, 0), as well as the corners of guide rectangle, so they have slopes of ± b 2. Using the point-slope equation 29 units to the left and right of (2, 0) to arrive at (2 − √ a2 + b2 = a = ± 5 4 + 25 = √ √ √ √ √ 536 Hooked on Conics of a line, Equation 2.2, yields y − 0 = ± 5 it all together, we get 2 (x − 2), so we get y = 5 2 x − 5 and y = − 5 2 x + 5. Putting y 7 6 5 4 3 2 1 −3 −2 −1 −2 −3 −4 −5 −6 −7 Example 7.5.2. Find the equation of the hyperbola with asymptotes y = ±2x and vertices (±5, 0). Solution. Plotting the data given to us, we have y 5 −5 x 5 −5 This graph not only tells us that the branches of the hyperbola open to the left and to the right, it also tells us that the center is (0, 0). Hence, our standard form is x2 b2 = 1. Since the vertices are (±5, 0), we have a = 5 so a2 = 25. In order to determine b2, we recall that the slopes of the asymptotes are ± b 5 = 2, so b = 10. Hence, b2 = 100 and our ﬁnal answer is x2 a. Since a = 5 and the slope of the line y = 2x is 2,
we have that b 25 − y2 a2 − y2 100 = 1. 7.5 Hyperbolas 537 As with the other conic sections, an equation whose graph is a hyperbola may not be given in either of the standard forms. To rectify that, we have the following. To Write the Equation of a Hyperbola in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side 2. Complete the square in both variables as needed 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes 1 Example 7.5.3. Consider the equation 9y2 − x2 − 6x = 10. Put this equation in to standard form and graph. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci, and the equations of the asymptotes. Solution. We need only complete the square on x: 9y2 − x2 − 6x = 10 9y2 − 1 x2 + 6x = 10 9y2 − x2 + 6x + 9 = 10 − 1(9) 9y2 − (x + 3)2 = 1 y2 (x + 3)2 1 1 9 = 1 − 9 so b = 1 Now that this equation is in the standard form of Equation 7.7, we see that x−h is x+3 so h = −3, and y − k is y so k = 0. Hence, our hyperbola is centered at (−3, 0). We ﬁnd that a2 = 1 so a = 1, and b2 = 1 3. This means that we move 1 unit to the left and right of the center and 1 3 units up and down from the center to arrive at points on the guide rectangle. Since the x2 term is being subtracted from the y2 term, we know the branches of the hyperbola open upwards and downwards. This means the transverse axis lies along the vertical line x = −3 and the conjugate axis lies along the x-axis. Since the vertices of the hyperbola are where the hyperbola intersects and the transverse axis, we get that the vertices are 1 −3, − 1 3 3 of a unit above and below (−3, 0) at −3, 1. To ﬁnd the f
oci, we use 3 √ √ c = a2 + b2 = Since the foci lie on the transverse axis, we move 10 3 1 9 √ 10 3 units above and below (−3, 0) to arrive at. To determine the asymptotes, recall that the asymptotes go through the center of the hyperbola, (−3, 0), as well as the corners of guide rectangle, so they have slopes 3. Using the point-slope equation of a line, Equation 2.2, we get y = 1 of ± b 3 x + 1 and. Putting it all together, we get 10 3 −3, − + 1 = −3, and 10 3 √ 538 Hooked on Conics −6 −1 y 1 −1 x Hyperbolas can be used in so-called ‘trilateration,’ or ‘positioning’ problems. The procedure outlined in the next example is the basis of the (now virtually defunct) LOng Range Aid to Navigation (LORAN for short) system.2 Example 7.5.4. Jeﬀ is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeﬀ records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be located. Solution. Since Jeﬀ hears Sasquatch sooner, it is closer to Jeﬀ than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeﬀ by multiplying 760 miles hour × 1 hour 3600 seconds × 9 seconds = 1.9 miles This means that Sasquatch is 1.9 miles closer to Jeﬀ than it is to Carl. In other words, Sasquatch must lie on a path where (the distance to Carl) − (the distance to Jeﬀ) = 1.9 This is exactly the situation in the deﬁnition of a hyperbola, Deﬁnition 7.6. In this case, Jeﬀ and Carl are located at the foci,3 and our ﬁxed distance d is 1.9. For simplicity, we assume the hyperbola is
centered at (0, 0) with its foci at (−5, 0) and (5, 0). Schematically, we have 2GPS now rules the positioning kingdom. Is there still a place for LORAN and other land-based systems? Do satellites ever malfunction? 3We usually like to be the center of attention, but being the focus of attention works equally well. 7.5 Hyperbolas 539 y 6 5 4 3 2 1 Jeﬀ −6 −5 −4 −3 −2 −1 1 2 3 4 Carl 5 6 x −1 −2 −3 −4 −5 −6 We are seeking a curve of the form x2 b2 = 1 in which the distance from the center to each focus is c = 5. As we saw in the derivation of the standard equation of the hyperbola, Equation 7.6, d = 2a, so that 2a = 1.9, or a = 0.95 and a2 = 0.9025. All that remains is to ﬁnd b2. To that end, we recall that a2 + b2 = c2 so b2 = c2 − a2 = 25 − 0.9025 = 24.0975. Since Sasquatch is closer to Jeﬀ than it is to Carl, it must be on the western (left hand) branch of a2 − y2 0.9025 − y2 x2 24.0975 = 1. In our previous example, we did not have enough information to pin down the exact location of Sasquatch. To accomplish this, we would need a third observer. Example 7.5.5. By a stroke of luck, Kai was also camping in the woods during the events of the previous example. He was located 6 miles due north of Jeﬀ and heard the Sasquatch call 18 seconds after Jeﬀ did. Use this added information to locate Sasquatch. Solution. Kai and Jeﬀ are now the foci of a second hyperbola where the ﬁxed distance d can be determined as before 760 miles hour × 1 hour 3600 seconds × 18 seconds = 3.8 miles Since Jeﬀ was positioned at (−5, 0), we place Kai at (−5, 6). This puts the center of the new hyperbola at (−5, 3). Plotting Kai’s position and the new center gives us the
diagram below on the left. The second hyperbola is vertical, so it must be of the form (y−3)2 a2 = 1. As before, the distance d is the length of the major axis, which in this case is 2b. We get 2b = 3.8 so that b = 1.9 and b2 = 3.61. With Kai 6 miles due North of Jeﬀ, we have that the distance from the center to the focus is c = 3. Since a2 + b2 = c2, we get a2 = c2 − b2 = 9 − 3.61 = 5.39. Kai heard the Sasquatch call after Jeﬀ, so Kai is farther from Sasquatch than Jeﬀ. Thus Sasquatch must lie 3.61 − (x+5)2 on the southern branch of the hyperbola (y−3)2 5.39 = 1. Looking at the western branch of the b2 − (x+5)2 540 Hooked on Conics hyperbola determined by Jeﬀ and Carl along with the southern branch of the hyperbola determined by Kai and Jeﬀ, we see that there is exactly one point in common, and this is where Sasquatch must have been when it called. Kai Jeﬀ −9−8−7−6−5−4−3−2−1 y Carl 1 2 3 4 5 6 x y Carl 1 2 3 4 5 6 x Kai Jeﬀ 6 5 4 3 2 1 −9−8−7−6−5−4−3−2 Sasquatch −2 −3 −4 −5 −6 6 5 4 3 2 1 −1 −2 −3 −4 −5 −6 To determine the coordinates of this point of intersection exactly, we would need techniques for solving systems of non-linear equations (which we won’t see until Section 8.7), so we use the calculator4 Doing so, we get Sasquatch is approximately at (−0.9629, −0.8113). Each of the conic sections we have studied in this chapter result from graphing equations of the form Ax2 + Cy2 + Dx + Ey + F = 0 for diﬀerent choices of A, C, D, E, and5 F. While we’ve seen examples6 demonstrate how to convert an equation from
this general form to one of the standard forms, we close this chapter with some advice about which standard form to choose.7 Strategies for Identifying Conic Sections Suppose the graph of equation Ax2 + Cy2 + Dx + Ey + F = 0 is a non-degenerate conic section.a If just one variable is squared, the graph is a parabola. Put the equation in the form of Equation 7.2 (if x is squared) or Equation 7.3 (if y is squared). If both variables are squared, look at the coeﬃcients of x2 and y2, A and B. If A = B, the graph is a circle. Put the equation in the form of Equation 7.1. If A = B but A and B have the same sign, the graph is an ellipse. Put the equation in the form of Equation 7.4. If A and B have the diﬀerent signs, the graph is a hyperbola. Put the equation in the form of either Equation 7.6 or Equation 7.7. aThat is, a parabola, circle, ellipse, or hyperbola – see Section 7.1. 4First solve each hyperbola for y, and choose the correct equation (branch) before proceeding. 5See Section 11.6 to see why we skip B. 6Examples 7.2.3, 7.3.4, 7.4.3, and 7.5.3, in particular. 7We formalize this in Exercise 34. 7.5 Hyperbolas 7.5.1 Exercises 541 In Exercises 1 - 8, graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 1. 3. 5. 7. − x2 y2 16 9 (x − 2)2 4 (x + 4)2 16 (y + 2)2 16 = 1 − − − (y + 3)2 9 (y − 4)2 1 (x − 5)2 20 = 1 = 1 = 1 2. 4. 6. 8. − x2 y2 9 16 (y − 3)2 11 (x + 1)2 9 (x − 4)2 8 = 1 − − − (x − 1
)2 10 (y − 3)2 4 (y − 2)2 18 = 1 = 1 = 1 In Exercises 9 - 12, put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 9. 12x2 − 3y2 + 30y − 111 = 0 10. 18y2 − 5x2 + 72y + 30x − 63 = 0 11. 9x2 − 25y2 − 54x − 50y − 169 = 0 12. −6x2 + 5y2 − 24x + 40y + 26 = 0 In Exercises 13 - 18, ﬁnd the standard form of the equation of the hyperbola which has the given properties. 13. Center (3, 7), Vertex (3, 3), Focus (3, 2) 14. Vertex (0, 1), Vertex (8, 1), Focus (−3, 1) 15. Foci (0, ±8), Vertices (0, ±5). 16. Foci (±5, 0), length of the Conjugate Axis 6 17. Vertices (3, 2), (13, 2); Endpoints of the Conjugate Axis (8, 4), (8, 0) 18. Vertex (−10, 5), Asymptotes y = ± 1 2 (x − 6) + 5 In Exercises 19 - 28, ﬁnd the standard form of the equation using the guidelines on page 540 and then graph the conic section. 19. x2 − 2x − 4y − 11 = 0 20. x2 + y2 − 8x + 4y + 11 = 0 21. 9x2 + 4y2 − 36x + 24y + 36 = 0 22. 9x2 − 4y2 − 36x − 24y − 36 = 0 542 Hooked on Conics 23. y2 + 8y − 4x + 16 = 0 24. 4x2 + y2 − 8x + 4 = 0 25. 4x2 + 9y2 − 8x + 54y + 49 = 0 26. x2 + y2 − 6x + 4y + 14 = 0 27. 2x2 + 4y2 + 12x − 8y + 25 = 0 28. 4x2 − 5
y2 − 40x − 20y + 160 = 0 29. The graph of a vertical or horizontal hyperbola clearly fails the Vertical Line Test, Theorem 1.1, so the equation of a vertical of horizontal hyperbola does not deﬁne y as a function of x.8 However, much like with circles, horizontal parabolas and ellipses, we can split a hyperbola into pieces, each of which would indeed represent y as a function of x. With the help of your classmates, use your calculator to graph the hyperbolas given in Exercises 1 - 8 above. How many pieces do you need for a vertical hyperbola? How many for a horizontal hyperbola? 30. The location of an earthquake’s epicenter − the point on the surface of the Earth directly above where the earthquake actually occurred − can be determined by a process similar to how we located Sasquatch in Example 7.5.5. (As we said back in Exercise 75 in Section 6.1, earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a course in Geology or the U.S. Geological Survey’s Earthquake Hazards Program found here.) Our technique works only for relatively small distances because we need to assume that the Earth is ﬂat in order to use hyperbolas in the plane.9 The P-waves (“P” stands for Primary) of an earthquake in Sasquatchia travel at 6 kilometers per second.10 Station A records the waves ﬁrst. Then Station B, which is 100 kilometers due north of Station A, records the waves 2 seconds later. Station C, which is 150 kilometers due west of Station A records the waves 3 seconds after that (a total of 5 seconds after Station A). Where is the epicenter? 31. The notion of eccentricity introduced for ellipses in Deﬁnition 7.5 in Section 7.4 is the same for hyperbolas in that we can deﬁne the eccentricity e of a hyperbola as e = distance from the center to a focus distance from the center to a vertex (a) With the help of your classmates, explain why e > 1 for any hyperbola. (b) Find the equation of the hyperbola with vertices (±3, 0) and eccentricity
e = 2. (c) With the help of your classmates, ﬁnd the eccentricity of each of the hyperbolas in Exercises 1 - 8. What role does eccentricity play in the shape of the graphs? 32. On page 510 in Section 7.3, we discussed paraboloids of revolution when studying the design of satellite dishes and parabolic mirrors. In much the same way, ‘natural draft’ cooling towers are often shaped as hyperboloids of revolution. Each vertical cross section of these towers 8We will see later in the text that the graphs of certain rotated hyperbolas pass the Vertical Line Test. 9Back in the Exercises in Section 1.1 you were asked to research people who believe the world is ﬂat. What did you discover? 10Depending on the composition of the crust at a speciﬁc location, P-waves can travel between 5 kps and 8 kps. 7.5 Hyperbolas 543 is a hyperbola. Suppose the a natural draft cooling tower has the cross section below. Suppose the tower is 450 feet wide at the base, 275 feet wide at the top, and 220 feet at its narrowest point (which occurs 330 feet above the ground.) Determine the height of the tower to the nearest foot. 275 ft 220 ft 450 ft 330 ft 33. With the help of your classmates, research the Cassegrain Telescope. It uses the reﬂective property of the hyperbola as well as that of the parabola to make an ingenious telescope. 34. With the help of your classmates show that if Ax2 + Cy2 + Dx + Ey + F = 0 determines a non-degenerate conic11 then AC < 0 means that the graph is a hyperbola AC = 0 means that the graph is a parabola AC > 0 means that the graph is an ellipse or circle NOTE: This result will be generalized in Theorem 11.11 in Section 11.6.1. 11Recall that this means its graph is either a circle, parabola, ellipse or hyperbola. 544 7.5.2 Answers Hooked on Conics 1. 2. − = 1 y2 x2 9 16 Center (0, 0) Transverse axis on y = 0 Conjugate axis on x = 0 Vertices (4, 0), (−4, 0) Foci (5, 0),
(−5, 0) Asymptotes y = ± 3 4 x − = 1 y2 x2 16 9 Center (0, 0) Transverse axis on x = 0 Conjugate axis on y = 0 Vertices (0, 3), (0, −3) Foci (0, 5), (0, −5) Asymptotes y = ± 3 4 x 3. (x − 2)2 4 − (y + 3)2 9 = 1 Center (2, −3) Transverse axis on y = −3 Conjugate axis on x = 2 Vertices (0, −3), (4, −3) √ 13, −3), (2 − Foci (2 + Asymptotes y = ± 3 √ 13, −3) 2 (x − 26 −5 −4 −3 −2 −1 −1 −2 1 2 3 4 5 6 x −3 −4 −5 −6 y 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 y 4 3 2 1 −3 −2 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 7.5 Hyperbolas 545 4. (y − 3)2 11 − (x − 1)2 10 = 1 Center (1, 3) Transverse axis on x = 1 Conjugate axis on y = 3 Vertices (1, 3 + Foci (1, 3 + Asymptotes y = ± √ √ 11), (1, 3 − √ 21), (1, 3 − 21) √ 110 10 (x − 1) + 3 √ 11) 5. (x + 4)2 16 − (y − 4)2 1 = 1 Center (−4, 4) Transverse axis on y = 4 Conjugate axis on x = −4 Vertices (−8, 4), (0, 4) Foci (−4 + Asymptotes y = ± 1 17, 4), (−4 − √ √ 17, 4) 4 (x + 4) + 4 6. (x + 1)2 9 − (y − 3)2 4 = 1 Center (−1, 3) Transverse axis on y = 3 Conjugate axis on x = −1 Vertices (2, 3),
(−4, 3) Foci −1 + Asymptotes y = ± 2 13, 3, −1 − √ 3 (x + 1) + 3 √ 13, 3 7. (y + 2)2 16 − (x − 5)2 20 = 1 Center (5, −2) Transverse axis on x = 5 Conjugate axis on y = −2 Vertices (5, 2), (5, −6) Foci (5, 4), (5, −8) Asymptotes y = ± 2 √ 5 5 (x − 55 −4 −3 −2 −1 −1 −2 −11−10 −9 −8 −7 −6 −5 −4 −3 −2 −7 −6 −5 −4 −3 −2 − 10 11 x 4 3 2 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 546 Hooked on Conics 8. (x − 4)2 8 − (y − 2)2 18 = 1 Center (4, 2) Transverse axis on y = 2 Conjugate axis on x = 4 Vertices 4 + 2 Foci 4 + Asymptotes y = ± 3 26, 2, 4 − √ √ 2, 2, 4 − 2 2, 2 √ √ 26, 2 2 (x − 4) + 2 9. − = 1 (y − 5)2 12 x2 3 Center (0, 5) Transverse axis on y = 5 Conjugate axis on x = 0 Vertices ( 3, 5) 15, 5), (− Foci ( 15, 5) Asymptotes y = ±2x + 5 3, 5), (− √ √ √ √ 10 x −2−1 −1 −2 10. (y + 2)2 5 − (x − 3)2 18 = 1 Center (3, −2) Transverse axis on x = 3 Conjugate axis on y = −2 Vertices (3, −2 + Foci (3, −2 + Asymptotes y = ± √ √ 5), (3, −2 − √ 23), (3, −2 − 23) √ 10 6 (x − 3) − 2 √ 5) 11. (x − 3)2 25 − (y + 1)2 9 = 1 12. Center (3
, −1) Transverse axis on y = −1 Conjugate axis on x = 3 Vertices (8, −1), (−2, −1) Foci 3 + 34, −1, 3 − Asymptotes y = ± 3 √ √ 34, −1 5 (x − 3) − 1 − = 1 (y + 4)2 6 (x + 2)2 5 Center (−2, −4) Transverse axis on x = −2 Conjugate axis on y = −4 Vertices −2, −4 + √ Foci −2, −4 + Asymptotes y = ± √ 6, −2, −4 − √ 11, −2, −4 − √ 30 5 (x + 2) − 4 √ 6 11 13. (y − 7)2 16 − (x − 3)2 9 = 1 14. (x − 4)2 16 − (y − 1)2 33 = 1 15. y2 25 − x2 39 = 1 16. x2 16 − y2 9 = 1 17. (x − 8)2 25 − (y − 2)2 4 = 1 18. (x − 6)2 256 − (y − 5)2 64 = 1 7.5 Hyperbolas 547 19. (x − 1)2 = 4(y + 3) 20. (x − 4)2 + (y + 2)2 = 9 y y −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 21. (x − 2)2 4 + (y + 3)1 −2 −3 −4 −5 −6 1 4 7 x 1 −2 −5 22. (x − 2)2 4 − (y + 3)3 −2 −1 −1 −3 −4 −5 −6 −7 −8 −9 −10 (x − 1)2 1 y2 4 + = 0 The graph is the point (1, 0) only. 23. (y + 4)2 = 4x y −1 1 2 3 4 x 24. −1 −2 −3 −4 −5 −6 −7 −8 548 Hooked on Conics 25. (x − 1)2 9 + (y + 3)2 4 = 1 y −2 −1 1 2 3 4 x 26. (x − 3)2 + (y
+ 2)2 = −1 There is no graph. −1 −2 −3 −4 −5 27. (x + 3)2 2 + There is no graph. (y − 1)2 1 = − 3 4 28. (y + 2)2 16 − (x − 5)2 20 = 10 11 x 4 3 2 1 −1 −1 −2 −3 −4 −5 −6 −7 −8 30. By placing Station A at (0, −50) and Station B at (0, 50), the two second time diﬀerence yields the hyperbola y2 2464 = 1 with foci A and B and center (0, 0). Placing Station C at (−150, −50) and using foci A and C gives us a center of (−75, −50) and the hyperbola 225 − (y+50)2 (x+75)2 5400 = 1. The point of intersection of these two hyperbolas which is closer to A 36 − x2 than B and closer to A than C is (−57.8444, −9.21336) so that is the epicenter. 31. (b) x2 9 − y2 27 = 1. 32. The tower may be modeled (approximately)12 by x2 34203 = 1. To ﬁnd the height, we plug in x = 137.5 which yields y ≈ 191 or y ≈ 469. Since the top of the tower is above the narrowest point, we get the tower is approximately 469 feet tall. 12100 − (y−330)2 12The exact value underneath (y − 330)2 is 52707600 1541 in case you need more precision. Chapter 8 Systems of Equations and Matrices 8.1 Systems of Linear Equations: Gaussian Elimination Up until now, when we concerned ourselves with solving diﬀerent types of equations there was only one equation to solve at a time. Given an equation f (x) = g(x), we could check our solutions geometrically by ﬁnding where the graphs of y = f (x) and y = g(x) intersect. The x-coordinates of these intersection points correspond to the solutions to the equation f (x) = g(x), and the ycoordinates were largely ignored. If we modify the problem and ask for the intersection points of the
graphs of y = f (x) and y = g(x), where both the solution to x and y are of interest, we have what is known as a system of equations, usually written as y = f (x) y = g(x) The ‘curly bracket’ notation means we are to ﬁnd all pairs of points (x, y) which satisfy both equations. We begin our study of systems of equations by reviewing some basic notions from Intermediate Algebra. Deﬁnition 8.1. A linear equation in two variables is an equation of the form a1x + a2y = c where a1, a2 and c are real numbers and at least one of a1 and a2 is nonzero. For reasons which will become clear later in the section, we are using subscripts in Deﬁnition 8.1 to indicate diﬀerent, but ﬁxed, real numbers and those subscripts have no mathematical meaning beyond that. For example, 3x − y 2 = 0.1 is a linear equation in two variables with a1 = 3, a2 = − 1 2 and c = 0.1. We can also consider x = 5 to be a linear equation in two variables1 by identifying a1 = 1, a2 = 0, and c = 5. If a1 and a2 are both 0, then depending on c, we get either an equation which is always true, called an identity, or an equation which is never true, called a If c = 0, then we’d have contradiction. (If c = 0, then we get 0 = 0, which is always true. 0 = 0, which is never true.) Even though identities and contradictions have a large role to play 1Critics may argue that x = 5 is clearly an equation in one variable. It can also be considered an equation in 117 variables with the coeﬃcients of 116 variables set to 0. As with many conventions in Mathematics, the context will clarify the situation. 550 Systems of Equations and Matrices in the upcoming sections, we do not consider them linear equations. The key to identifying linear equations is to note that the variables involved are to the ﬁrst power and that the coeﬃcients of the variables are numbers. Some examples of equations which are non-linear are x2 + y = 1, xy = 5 and e2
x + ln(y) = 1. We leave it to the reader to explain why these do not satisfy Deﬁnition 8.1. From what we know from Sections 1.2 and 2.1, the graphs of linear equations are lines. If we couple two or more linear equations together, in eﬀect to ﬁnd the points of intersection of two or more lines, we obtain a system of linear equations in two variables. Our ﬁrst example reviews some of the basic techniques ﬁrst learned in Intermediate Algebra. Example 8.1.1. Solve the following systems of equations. Check your answer algebraically and graphically. 2x − y = 1 y = 3 1. 3x + 4y = −2 −3x − y = 5 2. 2x − 4y = 6 3x − 6y = 9 4. 6x + 3y = 9 4x + 2y = 12 5. 3. 6. Solution. x 3 − 4y 9 + y 5 = 7 3 = 1 2x 2x + y = −2 1. Our ﬁrst system is nearly solved for us. The second equation tells us that y = 3. To ﬁnd the corresponding value of x, we substitute this value for y into the the ﬁrst equation to obtain 2x − 3 = 1, so that x = 2. Our solution to the system is (2, 3). To check this algebraically, we substitute x = 2 and y = 3 into each equation and see that they are satisﬁed. We see 2(2) − 3 = 1, and 3 = 3, as required. To check our answer graphically, we graph the lines 2x − y = 1 and y = 3 and verify that they intersect at (2, 3). 2. To solve the second system, we use the addition method to eliminate the variable x. We take the two equations as given and ‘add equals to equals’ to obtain 3x + 4y = −2 + (−3x − y = 5) 3 3y = This gives us y = 1. We now substitute y = 1 into either of the two equations, say −3x−y = 5, to get −3x − 1 = 5 so that x = −2. Our solution is (−2, 1). Substituting x = −2 and
y = 1 into the ﬁrst equation gives 3(−2) + 4(1) = −2, which is true, and, likewise, when we check (−2, 1) in the second equation, we get −3(−2) − 1 = 5, which is also true. Geometrically, the lines 3x + 4y = −2 and −3x − y = 5 intersect at (−2, 1). 8.1 Systems of Linear Equations: Gaussian Elimination 551 y 4 2 1 (2, 3) −1 1 2 3 4 x 2x − y = 1 y = 3 y 2 1 (−2, 1) −4 −3 −2 −1 x −1 −2 3x + 4y = −2 −3x − y = 5 3. The equations in the third system are more approachable if we clear denominators. We multiply both sides of the ﬁrst equation by 15 and both sides of the second equation by 18 to obtain the kinder, gentler system 5x − 12y = 21 4x + 6y = 9 Adding these two equations directly fails to eliminate either of the variables, but we note that if we multiply the ﬁrst equation by 4 and the second by −5, we will be in a position to eliminate the x term 20x − 48y = 84 + (−20x − 30y = −45) 39 −78y = From this we get y = − 1 2. We can temporarily avoid too much unpleasantness by choosing to substitute y = − 1 2 into one of the equivalent equations we found by clearing denominators, say into 5x − 12y = 21. We get 5x + 6 = 21 which gives x = 3. Our answer is 3, − 1. 2 At this point, we have no choice − in order to check an answer algebraically, we must see if the answer satisﬁes both of the original equations, so we substitute x = 3 and y = − 1 2 into both x 2. We leave it to the reader to verify that the solution is correct. Graphing both of the lines involved with considerable care yields an intersection point of 3, − 1 2 5 and 2x 3 − 4y. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly integer coeﬃcients belie something more sinister? We note that if we multiply both
sides of the ﬁrst equation by 3 and both sides of the second equation by −2, we are ready to eliminate the x 552 Systems of Equations and Matrices 6x − 12y = 18 + (−6x + 12y = −18) 0 0 = 2 x − 3 2. For each value of x, the formula y = 1 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This means that these two diﬀerent linear equations are, in fact, equivalent. In other words, if an ordered pair (x, y) satisﬁes the equation 2x − 4y = 6, it automatically satisﬁes the equation 3x − 6y = 9. One way to describe the solution set to this system is to use the roster method2 and write {(x, y) \| 2x − 4y = 6}. While this is correct (and corresponds exactly to what’s happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric solution to a system. Our ﬁrst step is to solve 2x − 4y = 6 for one of the variables, say y = 1 2 x − 3 2 determines the corresponding y-value of a solution. Since we have no restriction on x, it is called a free variable. We let x = t, a so-called ‘parameter’, and get y = 1 2. Our set of solutions can then be described as t, 1 \| − ∞ < t < ∞.3 For speciﬁc values of t, we can generate solutions. For example, t = 0 gives us the solution 0, − 3 ; t = 117 2 gives us (117, 57), and while we can readily check each of these particular solutions satisfy both equations, the question is how do we check our general answer algebraically? Same as always. We claim that for any real number t, the pair t, 1 satisﬁes both equations. = 6. Simplifying, Substituting x = t and we get 2t − 2t + 6 = 6, which is always true. Similarly, when we make these substitutions in the equation 3x − 6y = 9, we get 3t − 6 1 = 9 which reduces to 3t − 3t + 9 = 9,
so it 2 t − 3 2 checks out, too. Geometrically, 2x − 4y = 6 and 3x − 6y = 9 are the same line, which means that they intersect at every point on their graphs. The reader is encouraged to think about how our parametric solution says exactly that. 2 t − 3 2 into 2x − 4y = 6 gives 2t −, − 1 2 1 −1 −1 −2 −3 −4 x 3 − 4y 2x y 2 1 −1 1 2 3 4 x 2x − 4y = 6 3x − 6y = 9 (Same line.) 2See Section 1.2 for a review of this. 3Note that we could have just as easily chosen to solve 2x − 4y = 6 for x to obtain x = 2y + 3. Letting y be the parameter t, we have that for any value of t, x = 2t + 3, which gives {(2t + 3, t) \| − ∞ < t < ∞}. There is no one correct way to parameterize the solution set, which is why it is always best to check your answer. 8.1 Systems of Linear Equations: Gaussian Elimination 553 5. Multiplying both sides of the ﬁrst equation by 2 and the both sides of the second equation by −3, we set the stage to eliminate x 12x + 6y = 18 + (−12x − 6y = −36) 0 = −18 As in the previous example, both x and y dropped out of the equation, but we are left with an irrevocable contradiction, 0 = −18. This tells us that it is impossible to ﬁnd a pair (x, y) which satisﬁes both equations; in other words, the system has no solution. Graphically, the lines 6x + 3y = 9 and 4x + 2y = 12 are distinct and parallel, so they do not intersect. 6. We can begin to solve our last system by adding the ﬁrst two equations x − y = 0 + (x + y = 2) 2x = 2 which gives x = 1. Substituting this into the ﬁrst equation gives 1 − y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the ﬁr
st two equations, when we substitute x = 1 and y = 1 into the third equation, we get −2(1)+(1) = −2 which simpliﬁes to the contradiction −1 = −2. Graphing the lines x−y = 0, x + y = 2, and −2x + y = −2, we see that the ﬁrst two lines do, in fact, intersect at (1, 1), however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the system is to be found. y 6 5 4 3 2 1 −1 −2 −3 1 2 x 6x + 3y = 9 4x + 2y = 12 y 1 −2x + y = −2 A few remarks about Example 8.1.1 are in order. It is clear that some systems of equations have solutions, and some do not. Those which have solutions are called consistent, those with no solution are called inconsistent. We also distinguish the two diﬀerent types of behavior among 554 Systems of Equations and Matrices consistent systems. Those which admit free variables are called dependent; those with no free variables are called independent.4 Using this new vocabulary, we classify numbers 1, 2 and 3 in Example 8.1.1 as consistent independent systems, number 4 is consistent dependent, and numbers 5 and 6 are inconsistent.5 The system in 6 above is called overdetermined, since we have more equations than variables.6 Not surprisingly, a system with more variables than equations is called underdetermined. While the system in number 6 above is overdetermined and inconsistent, there exist overdetermined consistent systems (both dependent and independent) and we leave it to the reader to think about what is happening algebraically and geometrically in these cases. Likewise, there are both consistent and inconsistent underdetermined systems,7 but a consistent underdetermined system of linear equations is necessarily dependent.8 In order to move this section beyond a review of Intermediate Algebra, we now deﬁne what is meant by a linear equation in n variables. Deﬁnition 8.2. A linear equation in n variables, x1, x2,..., xn, is an equation of the form a1x1 + a2x2 +... + anxn = c where a1, a2,... an and c are
real numbers and at least one of a1, a2,..., an is nonzero. Instead of using more familiar variables like x, y, and even z and/or w in Deﬁnition 8.2, we use subscripts to distinguish the diﬀerent variables. We have no idea how many variables may be involved, so we use numbers to distinguish them instead of letters. (There is an endless supply of distinct numbers.) As an example, the linear equation 3x1 −x2 = 4 represents the same relationship between the variables x1 and x2 as the equation 3x − y = 4 does between the variables x and y. In addition, just as we cannot combine the terms in the expression 3x − y, we cannot combine the terms in the expression 3x1 − x2. Coupling more than one linear equation in n variables results in a system of linear equations in n variables. When solving these systems, it becomes increasingly important to keep track of what operations are performed to which equations and to develop a strategy based on the kind of manipulations we’ve already employed. To this end, we ﬁrst remind ourselves of the maneuvers which can be applied to a system of linear equations that result in an equivalent system.9 4In the case of systems of linear equations, regardless of the number of equations or variables, consistent independent systems have exactly one solution. The reader is encouraged to think about why this is the case for linear equations in two variables. Hint: think geometrically. 5The adjectives ‘dependent’ and ‘independent’ apply only to consistent systems – they describe the type of solu- tions. Is there a free variable (dependent) or not (independent)? 6If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown quantities, we need two pieces of information - i.e., two equations. Having more than two equations suggests we have more information than necessary to determine the values of the unknowns. While this is not necessarily the case, it does explain the choice of terminology ‘overdetermined’. 7We need more than two variables to give an example of the latter. 8Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra. 9That is, a system with the same solution set. 8.1 Systems of Linear Equations: Gaussian Elimination
555 Theorem 8.1. Given a system of equations, the following moves will result in an equivalent system of equations. Interchange the position of any two equations. Replace an equation with a nonzero multiple of itself.a Replace an equation with itself plus a nonzero multiple of another equation. aThat is, an equation which results from multiplying both sides of the equation by the same nonzero number. We have seen plenty of instances of the second and third moves in Theorem 8.1 when we solved the systems in Example 8.1.1. The ﬁrst move, while it obviously admits an equivalent system, seems silly. Our perception will change as we consider more equations and more variables in this, and later sections. Consider the system of equations    1 3 7 2 3, 7 2 and z = −1 into the ﬁrst equation to get x − 1 2 (−1) = 4 to obtain y = 7 Clearly z = −1, and we substitute this into the second equation y − 1 2. + 1 Finally, we substitute y = 7 2 (−1) = 1, so that x = 8 3. The reader can verify that these values of x, y and z satisfy all three original equations. It is tempting for us to write the solution to this system by extending the usual (x, y) 2, −1. The question quickly becomes what does notation to (x, y, z) and list our solution as 8 3, 7 an ‘ordered triple’ like 8 2, −1 represent? Just as ordered pairs are used to locate points on the two-dimensional plane, ordered triples can be used to locate points in space.10 Moreover, just as equations involving the variables x and y describe graphs of one-dimensional lines and curves in the two-dimensional plane, equations involving variables x, y, and z describe objects called surfaces in three-dimensional space. Each of the equations in the above system can be visualized as a plane situated in three-space. Geometrically, the system is trying to ﬁnd the intersection, or common point, of all three planes. If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line,
11 so our intersection point is where all three of these lines meet. 10You were asked to think about this in Exercise 40 in Section 1.1. 11In fact, these lines are described by the parametric solutions to the systems formed by taking any two of these equations by themselves. 556 Systems of Equations and Matrices Since the geometry for equations involving more than two variables is complicated, we will focus our eﬀorts on the algebra. Returning to the system    1 we note the reason it was so easy to solve is that the third equation is solved for z, the second equation involves only y and z, and since the coeﬃcient of y is 1, it makes it easy to solve for y using our known value for z. Lastly, the coeﬃcient of x in the ﬁrst equation is 1 making it easy to substitute the known values of y and z and then solve for x. We formalize this pattern below for the most general systems of linear equations. Again, we use subscripted variables to describe the general case. The variable with the smallest subscript in a given equation is typically called the leading variable of that equation. Deﬁnition 8.3. A system of linear equations with variables x1, x2,... xn is said to be in triangular form provided all of the following conditions hold: 1. The subscripts of the variables in each equation are always increasing from left to right. 2. The leading variable in each equation has coeﬃcient 1. 3. The subscript on the leading variable in a given equation is greater than the subscript on the leading variable in the equation above it. 4. Any equation without variablesa cannot be placed above an equation with variables. anecessarily an identity or contradiction 8.1 Systems of Linear Equations: Gaussian Elimination 557 In our previous system, if we make the obvious choices x = x1, y = x2, and z = x3, we see that the system is in triangular form.12 An example of a more complicated system in triangular form is    x1 − 4x3 + x4 − x6 = 6 x2 + 2x3 = 1 x4 + 3x5 − x6 = 8 x5 + 9x
6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. Example 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent.    1. 3x − y + z = 3 2x − 4y + 3z = 16. 2x + 3y − z = 1 10x − z = 2 4x − 9y + 2z = 5    3. 3x1 + x2 + x4 = 6 2x1 + x2 − x3 = 4 x2 − 3x3 − 2x4 = 0 Solution. 1. For deﬁnitiveness, we label the topmost equation in the system E1, the equation beneath that E2, and so forth. We now attempt to put the system in triangular form using an algorithm known as Gaussian Elimination. What this means is that, starting with x, we transform the system so that conditions 2 and 3 in Deﬁnition 8.3 are satisﬁed. Then we move on to the next variable, in this case y, and repeat. Since the variables in all of the equations have a consistent ordering from left to right, our ﬁrst move is to get an x in E1’s spot with a coeﬃcient of 1. While there are many ways to do this, the easiest is to apply the ﬁrst move listed in Theorem 8.1 and interchange E1 and E3.    3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→    x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) To satisfy Deﬁnition 8.3, we need to eliminate the x’s from E2 and E3. We accomplish this by replacing each of
them with a sum of themselves and a multiple of E1. To eliminate the x from E2, we need to multiply E1 by −2 then add; to eliminate the x from E3, we need to multiply E1 by −3 then add. Applying the third move listed in Theorem 8.1 twice, we get    x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) Replace E2 with −2E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −3E1 + E3    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 12If letters are used instead of subscripted variables, Deﬁnition 8.3 can be suitably modiﬁed using alphabetical order of the variables instead of numerical order on the subscripts of the variables. 558 Systems of Equations and Matrices Now we enforce the conditions stated in Deﬁnition 8.3 for the variable y. To that end we need to get the coeﬃcient of y in E2 equal to 1. We apply the second move listed in Theorem 8.1 and replace E2 with itself times − 1 2.    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 Replace E2 with − 1 2 E2 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 To eliminate the y in E3, we add −2E2 to it.    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 Replace E3 with −2E2 + E3 −−−−
−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Finally, we apply the second move from Theorem 8.1 one last time and multiply E3 by −1 to satisfy the conditions of Deﬁnition 8.3 for the variable z.    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Replace E3 with −1E3 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 6 z = y − 1 Now we proceed to substitute. Plugging in z = 6 into E2 gives y − 3 = −3 so that y = 0. With y = 0 and z = 6, E1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6). We leave it to the reader to check that substituting the respective values for x, y, and z into the original system results in three identities. Since we have found a solution, the system is consistent; since there are no free variables, it is independent. 2. Proceeding as we did in 1, our ﬁrst step is to get an equation with x in the E1 position with 1 as its coeﬃcient. Since there is no easy ﬁx, we multiply E1 by 1 2.    2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E1 with 1 2 E1 −−−−−−−−−−−−−→    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Now it’s time to take care of the x’s in E2 and E3E1) 2 10x − z =
2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3    2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 3 2 z = 8.1 Systems of Linear Equations: Gaussian Elimination 559 Our next step is to get the coeﬃcient of y in E2 equal to 1. To that end, we have    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) Replace E2 with − 1 15 E2 −−−−−−−−−−−−−−−→    Finally, we rid E3 of y. (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3)    (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3) Replace E3 with 15E2 + E3 −−−−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) y − 1 5 2 z = −3 6 0 = (E3) The last equation, 0 = 6, is a contradiction so the system has no solution. According to Theorem 8.1, since this system has no solutions, neither does the original, thus we have an inconsistent system. 3. For our last system, we begin by multiplying E1 by 1 3 to get a coeﬃcient of 1 on x1.    3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2
− 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Next we eliminate x1 from E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E2 −−−−−−−−−−→ with −2E1 + E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 We switch E2 and E3 to get a coeﬃcient of 1 for x2.    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 Switch E2 and E3 −−−−−−−−−−−→    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Finally, we eliminate x2 in E3. 560 Systems of Equations and Matrices    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Replace
E3 −−−−−−−−−−→ with − 1 3 E2 + E3    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 0 = 0 (E3) Equation E3 reduces to 0 = 0,which is always true. Since we have no equations with x3 or x4 as leading variables, they are both free, which means we have a consistent dependent system. We parametrize the solution set by letting x3 = s and x4 = t and obtain from E2 that x2 = 3s + 2t. Substituting this and x4 = t into E1, we have x1 + 1 3 t = 2 which gives x1 = 2 − s − t. Our solution is the set {(2 − s − t, 2s + 3t, s, t) \| − ∞ < s, t < ∞}.13 We leave it to the reader to verify that the substitutions x1 = 2 − s − t, x2 = 3s + 2t, x3 = s and x4 = t satisfy the equations in the original system. 3 (3s + 2t) + 1 Like all algorithms, Gaussian Elimination has the advantage of always producing what we need, but it can also be ineﬃcient at times. For example, when solving 2 above, it is clear after we eliminated the x’s in the second step to get the system    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) that equations E2 and E3 when taken together form a contradiction since we have identical left hand sides and diﬀerent right hand sides. The algorithm takes two more steps to reach this contradiction. We also note that substitution in Gaussian Elimination is delayed until all the elimination is done, thus it gets called back-substitution. This may also be ineﬃcient in many cases. Rest assured, the technique of substitution as you may have learned it in Intermediate Algebra will once again take center stage in Section 8.7. Lastly, we note that the system in 3 above is under
determined, and as it is consistent, we have free variables in our answer. We close this section with a standard ‘mixture’ type application of systems of linear equations. Example 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has stock solutions of 30% and 90% acid as well as all of the distilled water he wants. Set-up and solve a system of linear equations which determines all of the possible combinations of the stock solutions and water which would produce the required solution. Solution. We are after three unknowns, the amount (in mL) of the 30% stock solution (which we’ll call x), the amount (in mL) of the 90% stock solution (which we’ll call y) and the amount (in mL) of water (which we’ll call w). We now need to determine some relationships between these variables. Our goal is to produce 500 milliliters of a 40% acid solution. This product has two deﬁning characteristics. First, it must be 500 mL; second, it must be 40% acid. We take each 13Here, any choice of s and t will determine a solution which is a point in 4-dimensional space. Yeah, we have trouble visualizing that, too. 8.1 Systems of Linear Equations: Gaussian Elimination 561 of these qualities in turn. First, the total volume of 500 mL must be the sum of the contributed volumes of the two stock solutions and the water. That is amount of 30% stock solution + amount of 90% stock solution + amount of water = 500 mL Using our deﬁned variables, this reduces to x + y + w = 500. Next, we need to make sure the ﬁnal solution is 40% acid. Since water contains no acid, the acid will come from the stock solutions only. We ﬁnd 40% of 500 mL to be 200 mL which means the ﬁnal solution must contain 200 mL of acid. We have amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution is 0.90y. We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes We
ﬁrst eliminate the x from the second equation x + y + w = 500 10 y = 200 10 x + 9 3 (E1) x + y + w = 500 10 x + 9 10 y = 200 (E2) 3 Replace E2 with − 3 10 E1 + E2 −−−−−−−−−−−−−−−−−−→ (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Next, we get a coeﬃcient of 1 on the leading variable in E2 (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Replace E2 with 5 3 E2 −−−−−−−−−−−−−→ (E1) x + y + w = 500 y − 1 2 w = 250 (E2) 3 3 3, 1 2 t + 250 2 t + 250 2 t + 250 3. Substituting into E1 gives x + 1 + t = 500 so that x = − 3 2 t + 1250 Notice that we have no equation to determine w, and as such, w is free. We set w = t and from E2 get y = 1 2 t + 1250 3. This system is consistent, dependent and its solution set is {− 3 3, t \| − ∞ < t < ∞}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The 3 ≥ 0 or t ≥ − 500 constraint x ≥ 0 gives us − 3 3 ≥ 0, or t ≤ 2500 3. The condition z ≥ 0 yields t ≥ 0, and we see that when we take the set theoretic intersection of 3, t \| 0 ≤ t ≤ 2500 9. Our ﬁnal answer is {− 3 these intervals, we get 0 ≤ t ≤ 2500 9 }. Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining and it is due to expire. Can we use all of it to make our required solution? We would have y = 100 so that 1 3. This means the amount of 30% solution required is x = − 3 3 mL. The reader is invited to check
that mixing these three amounts of our constituent solutions produces the required 40% acid mix. 3 = 100, and we get t = 100 + 1250 3 = − 3 3 mL, and for the water, w = t = 100 9. From y ≥ 0, we get 1 2 t + 250 2 t + 1250 3 = 1100 2 t + 1250 2 t + 1250 2 t + 250 2 t + 250 3, 1 100 3 2 14We do this only because we believe students can use all of the practice with fractions they can get! 562 Systems of Equations and Matrices 8.1.1 Exercises (Review Exercises) In Exercises 1 - 8, take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically. 1. 3. 5. 7. x + 2y = 5 x = 6 x+2y 4 3x−y 2 = −5 = 1 1 2 x − 1 2y − 3x = 3 y = −1 6 2 x = − 15 3y −. 4. 6. 8. 2y − 3x = 1 y = − + 4y = 6 3 y = 1 12 − 20 3 y = − 7 3 3 y = 10 − 10 In Exercises 9 - 26, put each system of linear equations into triangular form and solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent. −5x + y = 17 x + y = 5 9. 11. 13. 15. 17. 19.    4x − y + z = 5 2y + 6z = 30 17 y − 3z = 0          3x − 2y + z = −5 x + 3y − z = 12 0 x + y + 2z = x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9        �
�           10. 12. 14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 4x − y + z = 5 2y + 6z = 30 x + z = 6 x − 2y + 3z = 7 −3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 8.1 Systems of Linear Equations: Gaussian Elimination 563 21. 23. 25.          x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1 4x − 4y + 4z = −13 6x − 5y + 7z = −25 x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 22. 24. 26.          x − y + z = 8 3x + 3y − 9z = −6 7x − 2y + 5z = 39 2x1 + x2 − 12x3 − x4 = 16 −x1 + x2 + 12x3 − 4x4 = −5 3x1 + 2x2 − 16x3 − 3x4 = 25 x1 + 2x2 − 5x4 = 11 x1 − x2 − 5x3 + 3x4 =
−1 x1 + x2 + 5x3 − 3x4 = 0 x2 + 5x3 − 3x4 = 1 x1 − 2x2 − 10x3 + 6x4 = −1 27. Find two other forms of the parametric solution to Exercise 11 above by reorganizing the equations so that x or y can be the free variable. 28. A local buﬀet charges $7.50 per person for the basic buﬀet and $9.25 for the deluxe buﬀet (which includes crab legs.) If 27 diners went out to eat and the total bill was $227.00 before taxes, how many chose the basic buﬀet and how many chose the deluxe buﬀet? 29. At The Old Home Fill’er Up and Keep on a-Truckin’ Cafe, Mavis mixes two diﬀerent types of coﬀee beans to produce a house blend. The ﬁrst type costs $3 per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound? 30. Skippy has a total of $10,000 to split between two investments. One account oﬀers 3% simple interest, and the other account oﬀers 8% simple interest. For tax reasons, he can only earn $500 in interest the entire year. How much money should Skippy invest in each account to earn $500 in interest for the year? 31. A 10% salt solution is to be mixed with pure water to produce 75 gallons of a 3% salt solution. How much of each are needed? 32. At The Crispy Critter’s Head Shop and Patchouli Emporium along with their dried up weeds, sunﬂower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile? 33. Discuss with your classmates how

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