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25t)). So, after 18 seconds Michael’s location will be M(18) = (0.9004, 0.4350). !!! CAUTION !!! Interpreting the coordinates of the point P = (cos(θ), sin(θ)) in Figure 17.11 only works if the angle θ is viewed in central standard position. You must do some additional work if the angle is placed in a different position; see the next Example. y-axis r = 1 kilometer 0.025 rad sec x-axis Michael starts here Angela starts here 0.03 rad sec (a) Angela and Michael on the same test track. y-axis M(t) Angela starts here 0.03 rad sec β(t) θ(t) α(t) A(t) 0.025 rad sec x-axis Michael starts here (b) Modeling the motion of Angela and Michael. Example 17.4.3. Both Angela and Michael are test driving vehicles counterclockwise around a desert test track which is circular of radius 1 kilometer. They start at the locations shown in Figure 17.14(a). Michael is traveling 0.025 rad/sec and Angela is traveling 0.03 rad/sec. Impose coordinates as pictured. Where are the drivers located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds. Likewise, let A(t) be the point on the circle of motion representing Angela’s location after t seconds. Let θ(t) be the angle swept out the by Michael and α(t) the angle swept out by Angela after t seconds. Since we are given the angular speeds, we get θ(t) = 0.025t radians, and α(t) = 0.03t radians. From the previous Example 17.4.2, Figure 17.14: Visualizing motion on a circular track. M(t) = (cos(0.025t), sin(0.025t)), and M(18) = (0.9004, 0.4350). Angela’s angle α(t) is NOT in central standard position, so we must observe that α(t) + π = β(t), where β(t) is in central standard position: See Figure 17.14(b). We conclude that A(t) = (cos(β(t)), sin(β(t))) = (cos(π + 0.03t), sin(π + 0.03t)). (−0.8577, −0.5141). So, after 18 seconds Angela’s location will be A(18) = 17.4. CIRCULAR FUNCTIONS 229 17.4.2 Relating circular functions and right triangles If the point P on the unit circle is located in the first quadrant, then we can compute cos(θ) and sin(θ) using trigonometric ratios. In general, it’s useful to relate right triangles, the unit circle and the circular functions. To describe this connection, given θ we place it in central standard position in the unit circle, where ∠ROP = θ. Draw a line through P perpendicular to the x-axis, obtaining an inscribed right triangle. Such a right triangle has hypotenuse of length 1, vertical side of length labeled b and horizontal side of length labeled a. There are four cases: See Figure 17.16. unit circle (radius = 1) P θ sin(θ) O R cos(θ) Figure 17.15: The point P in the first quadrant CASE I CASE II CASE III CASE IV Figure 17.16: Possible positions of θ on the unit circle. Case I has already been discussed, arriving at cos(θ) = a and sin(θ) = b. In Case II , we can interpret cos(θ) = −a, sin(θ) = b. We can reason similarly in the other Cases III and IV, using Figure 17.16, and we arrive at this conclusion: Important Facts 17.4.4 (Circular functions and triangles). View θ as in Figure 17.16 and form the pictured inscribed right triangles. Then we can interpret cos(θ) and sin(θ) in terms of these right triangles as follows: sin(θ) = b Case I : cos(θ) = a, Case II : cos(θ) = −a, sin(θ) = b Case III : cos(θ) = −a, sin(θ) = −b sin(θ) = −b Case IV : cos(θ) = a, 230 CHAPTER 17. THE CIRCULAR FUNCTIONS 17.5 What About Other Circles? T P θ O R S Cr unit circle Figure 17.17: Points on other circles. What happens if we begin with a circle Cr with radius r (possibly different than 1) and want to compute the coordinates of points on this circle? The circular functions can be used to answer this more general question. Picture our circle Cr centered at the origin in the same picture with unit circle C1 and the angle θ in standard central position for each circle. As pictured, we can view θ = ∠ROP = ∠SOT . If P = (x,y) is our point on the unit circle corresponding to the angle θ, then the calculation below shows how to compute coordinates on general circles: P = (x,y) = (cos(θ), sin(θ)) C1 ∈ x2 + y2 = 1 r2x2 + r2y2 = r2 (rx)2 + (ry)2 = r2 T = (rx, ry) = (r cos(θ), r sin(θ)) Cr. ∈ ⇔ ⇔ ⇔ ⇔ Important Fact 17.5.1. Let Cr be a circle of radius r centered at the origin and θ = ∠SOT an angle in standard central position for this circle, as in Figure 17.17. Then the coordinates of T = (r cos(θ), r sin(θ)). y-axis β = π − α = 2.9416 -axis circle radius = 1 circle radius = 2 circle radius = 3 α B Figure 17.18: Coordinates of points on circles. Examples 17.5.2. Consider the picture below, with θ = 0.8 radians and α = 0.2 radians. What are the coordinates of the labeled points? Solution. The angle θ is in standard central position; α is a central angle, but it is not in standard position. Notice, β = π − α = 2.9416 is an angle in standard central position which locates the same points U, T, S as the angle α. Applying Definition 17.4.1 on page 227: = (0.6967, 0.7174) P = (cos(0.8), sin(0.8)) = (1.3934, 1.4347) Q = (2 cos(0.8), 2 sin(0.8)) = (2.0901, 2.1521) R = (3 cos(0.8), 3 sin(0.8)) S = (cos(2.9416), sin(2.9416)) = (−0.9801, 0.1987) T = (2 cos(2.9416), 2 sin(2.9416)) = (−1.9602, 0.3973) U = (3 cos(2.9416), 3 sin(2.9416)) = (−2.9403, 0.5961). 17.6. OTHER BASIC CIRCULAR FUNCTION 231 Example 17.5.3. Suppose Cosmo begins at the position R in the figure, walking around the circle of radius 20 feet with an angular speed of 4 5 RPM counterclockwise. After 3 minutes have elapsed, describe Cosmo’s precise location. S 5 = 12 Solution. Cosmo has traveled 3 4 5 revolutions. If θ is = the angle traveled after 3 minutes, θ = 24π 5 radians = 15.08 radians. By (15.5.1), we have x = = 20 cos 11.76 feet. Conclude that Cosmo is located at the point S = (−16.18, 11.76). Using (15.1), θ = 864◦ = 2(360◦) + 144◦; this means that Cosmo walks counterclockwise around the circle two complete revolutions, plus 144◦. = −16.18 feet and y = 20 sin 2π radians 24π 5 rad 24π 5 rad 12 5 rev rev P 20 feet R Figure 17.19: Where is Cosmo after 3 minutes? 17.6 Other Basic Circular Function Given any angle θ, our constructions offer a concrete link between the cosine and sine functions and right triangles inscribed inside the unit circle: See Figure 17.20. P P θ O R θ O R CASE I CASE II θ O P CASE III R O θ R P CASE IV Figure 17.20: Computing the slope of a line using the function tan(θ). The slope of the hypotenuse of these inscribed triangles is just the slope of the line through OP. Since P = (cos(θ), sin(θ)) and O = (0, 0): Slope = ∆y ∆x = sin(θ) cos(θ) ; this would be valid as long as cos(θ) new circular function called the tangent of θ by the rule = 0. This calculation motivates a tan(θ) = sin(θ) cos(θ) , provided cos(θ) = 0. 6 6 232 CHAPTER 17. THE CIRCULAR FUNCTIONS The only time cos(θ) = 0 is when the corresponding point P on the unit circle has x-coordinate 0. But, this only happens at the positions (0, 1) and (0, −1) on the unit circle, corresponding to angles of the form θ = . These are the cases when the inscribed right triangle would “degenerate” to having zero width and the line segment OP becomes vertical. In summary, we then have this general idea to keep in mind: 5π 2 , 3π 2 , · · · π 2 , ± ± ± Important Fact 17.6.1. The slope of a line = tan(θ), where θ is the angle the line makes with the x-axis (or any other horizontal line) Three other commonly used circular functions come up from time to time. The cotangent function y = cot(θ), the secant function y = sec(θ) and the cosecant function y = csc(θ) are defined by the formulas: sec(θ) def= 1 cos(θ) , csc(θ) def= 1 sin(θ) , cot(θ) def= 1 tan(θ) . Just as with the tangent function, one needs to worry about the values of θ for which these functions are undefined (due to division by zero). We will not need these functions in this text. North Alaska Northwest West SeaTac 0 115 South 0 50 East 0 20 Delta (a) The flight paths of three airplanes. Alaska x = 30 Q N Northwest P y = 20 E R Delta W x = −50 S (b) Modeling the paths of each flight. Figure 17.21: Visualizing and modeling departing airplanes. Example 17.6.2. Three airplanes depart SeaTac Airport. A NorthWest flight is heading in a direction 50◦ counterclockwise from East, an Alaska flight is heading 115◦ counterclockwise from East and a Delta flight is heading 20◦ clockwise from East. Find the location of the Northwest flight when it is 20 miles North of SeaTac. Find the location of the Alaska flight when it is 50 miles West of SeaTac. Find the location of the Delta flight when it is 30 miles East of SeaTac. impose a coordinate Solution. We system in Figure 17.21(a), where “East” (resp. “North”) points along the positive x-axis (resp. positive y-axis). To solve the problem, we will find the equation of the three lines representing the flight paths, then determine where they intersect the appropriate horizontal or vertical line. The Northwest and Alaska directions of flight are angles in standard central position; the Delta flight direction will be −20◦. We can imagine right triangles with their hypotenuses along the directions of flight, then using the tangent function, we have these three immediate conclusions: slope NW line = tan(50◦) = 1.19, slope Alaska line = tan(115◦) = −2.14, and slope Delta line = tan(−20◦) = −0.364. 17.6. OTHER BASIC CIRCULAR FUNCTION 233 All three flight paths pass through the origin (0,0) of our coordinate system, so the equations of the lines through the flight paths will be: NW flight : y = 1.19x, Alaska flight : y = −2.14x, Delta flight : y = −0.364x. The Northwest flight is 20 miles North of SeaTac when y = 20; plugging into the equation of the line of flight gives 20 = 1.19x, so x = 16.81 and the plane location will be P = (16.81, 20). Similarly, the Alaska flight is 50 miles West of SeaTac when x = −50; plugging into the equation of the line of flight gives y = (−2.14)(−50) = 107 and the plane location will be Q = (−50, 107). Finally, check that the Delta flight is at R = (30, −10.92) when it is 30 miles Eas
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t of SeaTac. 234 CHAPTER 17. THE CIRCULAR FUNCTIONS 17.7 Exercises Problem 17.1. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the world’s greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. Recall, we worked on this in Exercise 16.5. 12 RPM θ P 4 feet (a) Impose a coordinate system and find the coordinates T (t) = (x(t),y(t)) of Tiff at time t seconds after she starts the ride. (b) Tiff becomes a human missile after 6 seconds on the ride. Find Tiff’s coordinates the instant she becomes a human missile. (c) Find the equation of the tangential line along which Tiff travels the instant she becomes a human missile. Sketch a picture indicating this line and her initial direction of motion along it when the seat detaches. Problem 17.2. (a) Find the equation of a line passing through the point (-1,2) and making an angle of 13o with the x-axis. (Note: There are two answers; find them both.) (b) Find the equation of a line making an angle of 8o with the y-axis and passing through the point (1,1). (Note: There are two answers; find them both.) Problem 17.3. The crew of a helicopter needs to land temporarily in a forest and spot a flat horizontal piece of ground (a clearing in the forest) as a potential landing site, but are uncertain whether it is wide enough. They make two measurements from A (see picture) finding α = 25o and β = 54o. They rise vertically 100 feet to B and measure γ = 47o. Determine the width of the clearing to the nearest foot. B 100 feet γ A β α C E clearing D Problem 17.4. Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12 seconds to reach the northermost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive y-axis. (a) Give Marla’s coordinates at her starting point. (b) Give Marla’s coordinates when she has been running for 10 seconds. (c) Give Marla’s coordinates when she has been running for 901.3 seconds. Problem 17.5. A merry-go-round is rotating at the constant angular speed of 3 RPM counterclockwise. The platform of this ride is a circular disc of radius 24 feet. You jump onto the ride at the location pictured below. rotating 3 RPM jump on here θ 17.7. EXERCISES 235 (a) If θ = 34o, then what are your xy- coordinates after 4 minutes? (b) If θ = 20o, then what are your xy- coordinates after 45 minutes? (c) If θ = −14o, then what are your xycoordinates after 6 seconds? Draw an accurate picture of the situation. (d) If θ = −2.1 rad, then what are your xy-coordinates after 2 hours and 7 seconds? Draw an accurate picture of the situation. (e) If θ = 2.1 rad, then what are your xycoordinates after 5 seconds? Draw an accurate picture of the situation. Problem 17.6. Shirley is on a ferris wheel which spins at the rate of 3.2 revolutions per minute. The wheel has a radius of 45 feet, and the center of the wheel is 59 feet above the ground. After the wheel starts moving, Shirley takes 16 seconds to reach the top of the wheel. How high above the ground is she when the wheel has been moving for 9 minutes? Problem 17.7. The top of the Boulder Dam has an angle of elevation of 1.2 radians from a point on the Colorado River. Measuring the angle of elevation to the top of the dam from a point 155 feet farther down river is 0.9 radians; assume the two angle measurements are taken at the same elevation above sea level. How high is the dam? downriver dam 0.9 1.2 155 ft a Problem 17.8. A radio station obtains a permit to increase the height of their radio tower on Queen Anne Hill by no more than 100 feet. You are the head of the Queen Anne Community Group and one of your members asks you to make sure that the radio station does not exceed the limits of the permit. After finding a relatively flat area nearby the tower (not necessarily the same altitude as the bottom of the tower), and standing some unknown distance away from the tower, you make three measurements all at the same height above sea level. You observe that the top of the old tower makes an angle of 39◦ above level. You move 110 feet away from the original measurement and observe that the old top of the tower now makes an angle of 34◦ above level. Finally, after the new construction is complete, you observe that the new top of the tower, from the same point as the second measurement was made, makes an angle of 40◦ above the horizontal. All three measurements are made at the same height above sea level and are in line with the tower. Find the height of the addition to the tower, to the nearest foot. Problem 17.9. Charlie and Alexandra are running around a circular track with radius 60 meters. Charlie started at the westernmost point of the track, and, at the same time, Alexandra started at the northernmost point. They both run counterclockwise. Alexandra runs at 4 meters per second, and will take exactly 2 minutes to catch up to Charlie. Impose a coordinate system, and give the x- and y-coordinates of Charlie after one minute of running. Problem 17.10. George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at the easternmost point. The illustration below shows their starting positions and running directions. They start running toward each other at constant speeds. George runs at 9 feet per second. Paula takes 50 seconds to run a lap of the track. George and Paula pass each other after 11 seconds. N George Paula After running for 3 minutes, how far east of his starting point is George? 236 CHAPTER 17. THE CIRCULAR FUNCTIONS Problem 17.11. A kite is attached to 300 feet of string, which makes a 42 degree angle with the level ground. The kite pilot is holding the string 4 feet above the ground. kite o 42 4 feet ground level (a) How high above the ground is the kite? (b) Suppose that power lines are located Is 250 feet in front of the kite flyer. any portion of the kite or string over the power lines? Problem 17.12. In the pictures below, a bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle of motion. In each of the cases, answer these questions: (a) Find an angle θ0 in standard central position that gives the bugs initial location. (In some cases, this is the angle given in the picture; in other cases, you will need to do something.) (b) The location angle of the bug at time t is given by the formula θ(t) = θ0 + ωt. Plug in the values for θ0 and ω to explicitly obtain a formula for θ(t). (c) Find the coordinates of the bug at time t. (d) What are the coordinates of the bug after 1 second? After 0 seconds? After 3 seconds? After 22 seconds? ω=4π/9rad/sec bug lands here 1.2 rad 2 in bug lands here ω=4π/9 rad/sec ω= 4π/9rad/sec bug lands here 0.5 rad 2 in 2 in Chapter 18 Trigonometric Functions Our definitions of the circular functions are based upon the unit circle. This makes it easy to visualize many of their properties. 18.1 Easy Properties of Circular Functions How can we determine the range of function values for cos(θ) and sin(θ)? To begin with, recall the abstract definition for the range of a function f(θ): Range of f = {f(θ) : θ is in the domain}. Using the unit circle constructions of the basic circular functions, it is easy to visualize the range of cos(θ) and sin(θ). Beginning at the position (1, 0), imagine a ball If we moving counterclockwise around the unit circle. “freeze” the motion at any point in time, we will have swept out an angle θ and the corresponding position P(θ) on the circle will have coordinates P(θ) = (cos(θ), sin(θ)). ball moves from 0 to π 2 radians around unit circle light source x-axis y-axis y-axis light source ball moves from 0 π 2 radians around unit circle to x-axis (a) What do you see on the y-axis? (b) What do you see on the x-axis? Figure 18.2: Projecting the coordinates of points onto the y-axis and the x-axis. By studying the coordinates of the ball as it moves in the first quadπ/2 radians. rant, we will be studying cos(θ) and sin(θ), for 0 θ 237 ≤ ≤ ball moves counterclockwise y-axis (0,1) P(θ) sin(θ) 1 (−1,0) θ (1,0) cos(θ) x-axis UNIT CIRCLE (0, − 1) Figure 18.1: Visualizing the range of sin(θ) and cos(θ). 238 CHAPTER 18. TRIGONOMETRIC FUNCTIONS We can visualize this very concretely. Imagine a light source as in Figure 18.2(a); then a shadow projects onto the vertical y-axis. The shadow locations you would see on the y-axis are precisely the values sin(θ), for π/2 radians. Similarly, imagine a light source as in Figure 18.2(b); 0 then a shadow projects onto the horizontal x-axis. The shadow locations you would see on the x-axis are precisely the values cos(θ), for 0 π/2 radians. ≤ ≤ ≤ ≤ θ θ There are two visual conclusions: First, the function values of sin(θ) vary from 0 to 1 as θ varies from 0 to π/2. Secondly, the function values of cos(θ) vary from 1 to 0 as θ varies from 0 to π/2. Of course, we can go ahead and continue analyzing the motion as the ball moves into the second, third and fourth quadrant, ending up back at the starting position (1, 0). See Figure 18.3. y-axis #2 #3 #1 #4 ball moves from 0 to 2π radians around unit circle x-axis light source light source #2 #1 #3 #4 ball moves from 0 to 2π radians around nit circle x-axis (a) What do you see on the y-axis? (b) What do you see on the x-axis? Figure 18.3: Analyzing the values of the sine and cosine functions. The conclusion is that after one complete counterclock
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wise rotation, the values of sin(θ) and cos(θ) range over the interval [−1, 1]. As the ball moves through the four quadrants, we have indicated the “order” in which these function values are assumed by labeling arrows #1 — #4: For example, for the sine function, look at Figure 18.3(a). The values of the sine function vary from 0 up to 1 while the ball moves through the first quadrant (arrow labeled #1), then from 1 down to 0 (arrow labeled #2), then from 0 down to −1 (arrow labeled #3), then from −1 up to 0 (arrow labeled #4). What about the tangent function? We have seen that the tangent function computes the slope of the hypotenuse of an inscribed triangle. This means we can determine the range of values of tan(θ) by investigating the possible slopes for these inscribed triangles. We will maintain the above model of a ball moving around the unit circle. We look at two cases, each starting at (1, 0). In the first quadrant, the ball moves counterclockwise and in the fourth quadrant it moves clockwise: In the first quadrant, we notice that these hypotenuse slopes are always non-negative, beginning with slope 0 (the degenerate right triangle when θ = 0) then increasing. In fact, as the angle θ approaches π/2 radians, the ball is getting closer to the position (0, 1) and the hypotenuse 18.1. EASY PROPERTIES OF CIRCULAR FUNCTIONS 239 y-axis 8 7 6 5 4 3 2 1 x-axis y-axis x-axis 1 2 3 4 5 6 7 (a) What happens to the slopes of these triangles? (b) What happens to the slopes of these triangles? Figure 18.4: Analyzing the values of the tangent function. θ < π/2 will be 0 is approaching a vertical line. This tells us that as θ varies from 0 to π/2 (but not equal to π/2), these slopes attain all possible non-negative In other words, the range of values for tan(θ) on the domain values. 0 . Similar reasoning shows that as the ball moves in the fourth quadrant, the slopes of the hypotenuses of the triangles are always non-positive, varying from 0 to ANY negative value. In other words, the range of values for tan(θ) on the domain −π/2 < θ 0 will be − z < < z ∞ 0. ≤ ≤ ≤ On your calculator, you can verify the visual conclusions we just established by studying the values of tan(θ) for θ close (but not equal) to π 2 radians = 90◦: ∞ ≤ tan(89◦) = 57.29 tan(89.9◦) = 572.96 tan(89.99◦) = 5729.58 ... tan(−89◦) = −57.29 tan(−89.9◦) = −572.96 tan(−89.99◦) = −5729.58 ... The fact that the values of the tangent function become arbitrarily large π/2 radians means the function output values are as we get close to unbounded. ± Important Fact 18.1.1 (Circular function values). For any angle θ, we cos(θ) always have −1 2π, the range of both cos(θ) and sin(θ) is −1 In contrast, on the domain of all θ values for which tangent is defined, the range of tan(θ) is all real numbers. 1. On domain 0 1 and −1 sin(θ) ≤ z 1. ≤ ≤ ≤ ≤ ≤ ≤ ≤ θ For the sine and cosine functions, if the domain is not 0 2π, then we need to consider the “periodic qualities” of the circular functions to determine the range. This is discussed below. ≤ ≤ θ 240 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 18.2 Identities There are dozens of formulas that relate the values of two or more circular functions; these are usually lumped under the heading of Trigonometric Identities. In this course, we only need a couple frequently used identities. If we take the point P = (cos(θ), sin(θ)) on the unit circle, corresponding to the standard central position angle θ, then recall the equation of the unit circle tells us x2 +y2 = 1. But, since the x coordinate is cos(θ) and the y coordinate is sin(θ), we have (cos(θ))2 + (sin(θ))2 = 1. It is common notational practice to write (cos(θ))2 = cos2(θ) and (sin(θ))2 = sin2(θ). This leads to the most important of all trigonometric identities: Important Fact 18.2.1 (Trigonometric identity). For any angle θ, we have the identity cos2(θ) + sin2(θ) = 1. Adding any multiple of 2π radians (or 360◦) to an angle will not change the values of the circular functions. If we focus on radians for a moment, this says that knowing the values of cos(θ) and sin(θ) on the domain 0 θ There is something very general going on here, so let’s pause a mo- 2π determines the values for any other possible angle. ≤ ≤ ment to make a definition and then an observation. Definition 18.2.2 (Periodic function). For c > 0, a function f(θ) is called c-periodic if two things are true: (i) f(θ + c) = f(θ) holds for all θ; (ii) There is no smaller d, 0 < d < c, such that f(θ + d) = f(θ) holds for all θ. We usually call c the period of the function. Using this new terminology, we conclude that the sine and cosine circular functions are 2π-periodic. In the case of the tangent circular function, it is also true that tan(θ) = tan(θ + 2πn), for every integer n. However, referring back to the unit circle definitions of the circular functions, we have tan(θ) = tan(θ + nπ), for all integers n. If you take n = 1, then this tells us that the tangent circular function is π-periodic. We summarize this information below. Important Fact 18.2.3 (Periodicity identity). For any angle θ and any integer n = 0, 3, . . . , we have cos(θ) = cos(θ+2πn), sin(θ) = sin(θ+ 1, 2πn), and tan(θ) = tan(θ + nπ). 2, ± ± ± 18.2. IDENTITIES Next, we draw an angle θ and its negative in the same unit circle picture in standard central position. We have indicated the points Pθ and P−θ used to define the circular functions. It is clear from the picture in Figure 18.5 that Pθ and P−θ have the same x-coordinate, but the ycoordinates are negatives of one another. This gives the next identity: 241 (cos(θ), sin(θ)) = Pθ unit circle R θ O −θ (cos(−θ), sin(−θ)) = P−θ Figure 18.5: Visualizing a trigonometric identity. Important Fact 18.2.4 (Even/Odd identity). For any angle θ, sin(−θ) = − sin(θ), and cos(−θ) = cos(θ). We can use the terminology of even and odd functions here. In this language, this result says that the cosine function is an even function and the sine function is an odd function. Next, draw the angles θ and θ + π in the same unit circle picture in standard central position. We have indicated the corresponding points Pθ and Pθ+π on the unit circle and their coordinates in terms of the circular functions: From the picture in Figure 18.6, the x-coordinate of Pθ must be the “negative” of the x-coordinate of Pθ+π and similarly, the y-coordinate of Pθ must be the “negative” of the y-coordinate of Pθ+π. This gives us the next identity: Pθ = (cos(θ), sin(θ)) θ θ + π R unit circle Pθ+π = (cos(θ + π), (sin(θ + π)) Figure 18.6: Visualizing Fact 18.2.5. Important Fact 18.2.5 (Plus π identity). For any angle θ, we have sin(θ+ π) = − sin(θ), and cos(θ + π) = − cos(θ). Important Fact 18.2.6. For any angle θ, we have sin(π − θ) = sin(θ) and cos(π − θ) = − cos(θ). 5π 6 For example, we have sin = sin 2. This calculation leads to a computational observation: Combining Table 17.1 with the previous two identities we can compute the EXACT value of cos(θ), sin(θ), and tan(θ) at an angle θ which is a multiple of 30◦ = π 4 radians. Here are some sample calculations together with a reference as to “why” each equality is valid: 6 radians or 45◦ = π = 1 π 6 242 CHAPTER 18. TRIGONOMETRIC FUNCTIONS Example 18.2.7. (i) cos(−45◦) = cos(45◦) Fact 18.2.4 on page 241 = √2 2 Table 17.1 on page 223 (ii) sin(225◦) = sin(45◦ + 180◦) Fact 18.2.3 on page 240 Fact 18.2.5 on page 241 = − sin(45◦) √2 2 = − Table 17.1 on page 223 (iii) cos 2π 3 = cos − = − cos π 3 − π 3 = − cos = − 1 2 + π Fact 18.2.3 on page 240 π 3 Fact 18.2.5 on page 241 Fact 18.2.4 on page 241 Table 17.1 on page 223 18.3 Graphs of Circular Functions We have introduced three new functions of the variable θ and it is important to understand and interpret the pictures of their graphs. To do this, we need to settle on a coordinate system in which to work. The horizontal axis will correspond to the independent variable, so this should be the θaxis. We will label the vertical axis, which corresponds to the dependent variable, the z-axis. With these conventions, beginning with any of the circular functions z = sin(θ), z = cos(θ), or z = tan(θ), the graph will be a subset of the θz-coordinate system. Precisely, given a circular function z = f(θ), the graph consists of all pairs (θ, f(θ)), where θ varies over a domain of allowed values. We will record and discuss these graphs below; a graphing device will painlessly produce these for us! There is a point of possible confusion that needs attention. We purposely did not use the letter “y” for the dependent variable of the circular functions. This is to avoid possible confusion with our construction of the sine and cosine functions using the unit circle. Since we viewed the unit circle inside the xy-coordinate system, the x-coordinates (resp. ycoordinates) of points on the unit circle are computed by cos(θ) (resp. sin(θ)). 18.3. GRAPHS OF CIRCULAR FUNCTIONS 243 z y P = (cos(θ), sin(θ)) θ θ x Coordinate system used to GRAPH the circular functions. Coordinate system used to DEFINE the circular functions. Figure 18.7: The zθ versus xy coordinate systems. 18.3.1 A matter of scaling The first issue concerns scaling of the axes used in graphing the circular functions. As we know, the definition of radian measure is directly tied to the lengths of arcs subtended by angles in the unit circle: Important Fact 18.3.1. An angle of measure 1 radian inside the unit circle will subtend an arc of length 1. Since length is a good intuitive scaling quantity, it is natural to scale the θ-axis so that the length of 1 radian on the θ-axis (horizontal axis) is the same length as 1 unit on the vertical axis. For this reason, we will work primarily with radian measure when sketching the graphs of circular functions. If we need to work explicitly with degree measure for angles, then we can always convert radians to degrees using the fact: 360◦ = 2π radians. 18.3.2 The sine and cosine graphs Using Fact 18.1.1, we know that −1 1. Pictorially, this tells us that the graphs of z = sin(θ) and z = cos
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(θ) lie between the horizontal lines z = 1 and z = −1; i.e. the graphs lie inside the darkened band pictured in Figure 18.8. 1 and −1 cos(θ) sin(θ) ≤ ≤ ≤ ≤ By Fact 18.2.3, we know that the values of the sine and cosine repeat themselves every 2π radians. Consequently, if we know the graphs of the θ sine and cosine on the domain 0 2π, then the picture will repeat for ≤ 0, etc. θ the interval 2π 4π, −2π θ ≤ ≤ ≤ ≤ ≤ 244 CHAPTER 18. TRIGONOMETRIC FUNCTIONS z-axis z = 1 θ-axis z = −1 Figure 18.8: Visualizing the range of sin(θ) and cos(θ). y = 1 z-axis Repeat Repeat Repeat Repeat θ-axis −4π −2π 0 2π 4π 6π y = −1 picture in here repeats each 2π units Figure 18.9: On what intervals will the graph repeat? Sketching the graph of z = sin (θ) for 0 achieved by plotting points. For example, θ ≤ 6 , 1 π , 2 , ≤ 2π can be roughly 3 , √3 4 , √2 π , and 4 , − √2 6 , − 1 11π 2 , , π 2 2 lie on the graph, as do 3 , − √3 π 2 , 1 , and 2 (2π, 0), etc. If we return to our analysis of the range of values for the sine function in Figure 18.2, it is easy to see where sin(θ) is positive or negative; combined with Chapter 4, this tells us where the graph is above and below the horizontal axis (see Figure 18.10). 3π 2 , −1 5π 7π , 2 z-axis −2π −π π 2π 3π θ-axis Figure 18.10: Where is the graph positive or negative? We now include a software plot of the graph of sine function, observing the three qualitative features just isolated: bounding, periodicity and sign properties (see Figure 18.11). 18.3. GRAPHS OF CIRCULAR FUNCTIONS 245 z-axis 1 −2π −π π 2π 3π θ-axis −1 one period Figure 18.11: The graph of z = sin(θ). We could repeat this analysis to arrive at the graph of the cosine. Instead, we will utilize an identity. Given an angle θ, place it in central standard position in the unit circle, as one of the four cases of Figure 17.16. For example, we have pictured Case I in this figure. Since the sum of the angles in a triangle is 180◦ = π radians, we know that θ, π/2, and π 2 − θ are the three angles of the inscribed right triangle. From the picture in Figure 18.12, it then follows that π 2 − θ θ π 2 unit circle Figure 18.12: Visualizing the conversion identity. cos(θ) = = side adjacent to θ hypotenuse side opposite to hypotenuse − θ . = sin π 2 π 2 − θ Using the same reasoning this identity is valid for all θ. This gives us another useful identity: Important Fact 18.3.2 (Conversion identity). For any angle θ, cos(θ) = sin , and sin(θ) = cos . π 2 − θ π 2 − θ This identity can be used to sketch the graph of the cosine function. First, we do a calculation using our new identity: cos(θ) = cos(−θ) Fact 18.2.4 on page 241 = sin = sin π 2 θ − − (−θ) π 2 − Fact 18.3.2 (above) Since: (a + b) = (a − (−b)) = (b − (−a)) By the horizontal shifting principle in Fact 13.3.1 on page 170, the graph of z = cos(θ) is obtained by horizontally shifting the graph of z = sin(θ) by π 2 units to the left. Here is a plot of the graph of the cosine function: See Figure 18.13. 246 CHAPTER 18. TRIGONOMETRIC FUNCTIONS z-axis 1 −2π −π π 2π 3π θ-axis −1 one period Figure 18.13: The graph of z = cos(θ). 18.3.3 The tangent graph θ = − π 2 graph heads this way, getting close to vertical line as θ gets close to π 2 − π 2 π 2 θ-axis graph heads this way, getting close to vertical line as θ gets close to − π 2 z-axis θ = π 2 Figure 18.14: The behavior of tan(θ) as θ approaches asymptotes. As we have already seen, unlike the sine and cosine circular functions, the tangent function is NOT defined for all values of θ. Since tan(θ) = sin(θ) cos(θ) , here are some properties we can immediately deduce: • • • • The function z = tan(θ) is undefined if and only if θ = π 1, 2 + kπ, where k = 0, 2, 3, . ± ± ± · · · The function z = tan(θ) = 0 if and only if θ = kπ, where k = 0, . 1, 2, 3, ± ± ± · · · By Fact 18.2.3, the tangent function is π-periodic, so the picture of the graph will repeat itself every πunits and it is enough to understand the graph when − π 2 < θ < π 2 . On the domain 0 − π 0, tan(θ) 2 < θ ≤ θ < π 0. ≤ ≤ 2 , tan(θ) 0; on the domain ≥ ± ± 1, 2, · · · In the θz-coordinate system, the vertical lines θ = π 2 + kπ, where k = 0, will be vertical asymptotes for the graph of the tangent function. Using our slope interpretation in Figure 18.4, what becomes clear is this: As the values of θ get close to π 2 , the graph is getting close to the vertical line θ = π 2 AND becoming farther and farther away from the horizontal axis: To understand this numerically, first suppose θ is π slightly smaller than π 2 − 0.1 . Then the calculation of tan(θ) involves dividing a number very close to 1 by a very small positive number: 2 , say θ = π 2 − 0.001 π 2 − 0.01 , and , = 9.9666, = 99.9967, and − 0.1 π 2 π − 0.01 2 π − 0.001 2 tan tan tan = 1000. 18.4. TRIGONOMETRIC FUNCTIONS Conclude that as θ “approaches π 2 from below”, the values of tan(θ) are becoming larger and larger. This says that the function values become “unbounded”. Likewise, imagine the case when θ is slightly bigger than − π 2 , say θ = . Then the cal, and culation of tan(θ) involves dividing a number very close to −1 by a very small positive number: 2 + 0.001 2 + 0.01 2 + 0.1 − π − π − π , 247 θ = − π 2 z-axis one period etc. θ = 5π 2 etc. θ-axis −2π −π π 2π 3π θ = − 3π 2 2 θ = 3π θ = π Figure 18.15: The graph of z = tan(θ). 2 − − + 0.1 π 2 π + 0.01 2 π + 0.001 2 tan tan tan − = −9.9666, = −99.9967, and = −1000. Conclude that as θ “approaches − π becoming negative numbers of increasingly larger magnitude: 2 from above”, the values of tan(θ) are Again, this tells us the function values are becoming “unbounded”. The graph of z = tan(θ) for − π 2 can be roughly achieved by combining the calculations as in Example 18.2.7 and the qualitative features highlighted. Figure 18.15 shows a software plot. 2 < θ < π 18.4 Trigonometric Functions To become successful mathematical modelers, we must have wide variety of functions in our toolkit. As an illustration, the graph below might represent the height of the tide above some reference level over the course of several days. The curve drawn is clearly illustrating that the height of the tide is “periodic” as a function of time t; in other words, the behavior of the tide repeats itself as time goes by. However, if we try to model this periodic behavior, the only weapon at our disposal would be the circular functions and these require an angle variable, not a time variable such as t; we are stuck! Modeling the tide graph requires the trigonometricfunctions, which lie at the heart of studying all kinds of periodic behavior. We have no desire to “reinvent the wheel”, so let’s use our previous work on the circular functions to define the trigonometric functions. 18.4.1 A Transition Given a real number t, is there a sensible way to define cos(t) and sin(t)? The answer is yes and depends on the ideas surrounding radian measure of angles. Given the positive real number t, we can certainly imagine an angle of measure t radians inside the unit circle (in standard position) and we know the arc subtended by this angle has length t (this is why we use the unit circle). 248 CHAPTER 18. TRIGONOMETRIC FUNCTIONS feet +20 +15 +10 +5 0 −5 −10 −15 −20 t (time) Figure 18.16: A periodic function with input variable t. P(t) = (x,y) arc length = t t rads (1,0) unit circle Figure 18.17: The circular functions with input variable t. We already know that cos(t radians) and sin(t radians) compute the x and y coordinates of the point P(t). In effect, we are just using the measure of the angle t to help us locate the point P(t). An alternate way to locate P(t) is to move along the circumference counterclockwise, beginning at (1, 0), until we have an arc of length t; that again puts us at the point P(t). In the case of an angle of measure −t radians, the point P(−t) can be located by moving along the circumference clockwise, beginning at (1, 0), until we have an arc of length t. Definition 18.4.1 (Trigonometric functions). Let t be a real number. We DEFINE the sine function y = sin(t), the cosine function y = cos(t) and the tangent function y = tan(t) by the rules sin(t) def= y-coordinate of P(t) = sin(t radians) cos(t) def= x-coordinate of P(t) = cos(t radians) tan(t) def= sin(t) cos(t) = tan(t radians) We refer to these as the basic trigonometric functions. If we are working with radian measure and t is a real number, then there is no difference between evaluating a trigonometric function at the real number t and evaluating the corresponding circular function at the angle of measure t radians. Example 18.4.2. Assume that the number of hours of daylight in Seattle + 12, where during 1994 is given by the function d(t) = 3.7sin 2π 366 (t − 80.5) 18.4. TRIGONOMETRIC FUNCTIONS 249 t represents the day of the year and t = 0 corresponds to January 1. How many hours of daylight will there be on May 11? Solution. To solve the problem, you need to consult a calendar, finding every month has 31 days, except: February has 28 days and April, June, September and November have 30 days. May 11 is the 31+28+31+30+11 = 131st day of the year. So, there will be d(131) = 3.7 sin(2(50.5)π/366) + 12 = 14.82 hours of daylight on May 11. 18.4.2 Graphs of trigonometric functions The graphs of the trigonometric functions y = sin(t), y = cos(t), and y = tan(t) will look just like Figures 18.11, 18.13, and 18.15, except that the horizontal axis becomes the t-axis and the vertical axis becomes the y-axis. 18.4.3 Notation for trigonometric functions In many texts, you will find the sine function written as y = sin t; i.e. the parenthesis around the “t” are omitted. A similar comment applies to all of the trigonometric functions. We will never do this and the reasoning is simply this: Maintaining the parenthesis, as in y = sin(t), emphasizes the fact that we are dealing with a function and the “input values” are located between the parenthesis. For example, if we write the function y = sin(t2 +2t+1), it is crystal clear that the sine function is applie
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d to the expression “t2 + 2t + 1”; using the alternate notation yields the expression y = sin t2 + 2t + 1, which is interpreted to mean y = (sin t2) + (2t + 1). ), As a rule, whenever you see an expression involving sin( or tan( ” is in units of RADIANS, unless otherwise noted. When computing values on your calculator, MAKE SURE YOU ARE USING RADIAN MODE! ), we assume “ ), cos( · · · · · · · · · · · · !!! CAUTION !!! 250 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 18.5 Exercises Problem 18.1. Work the following problems without using ANY calculators. (b) If sin(θ) = −0.8 and θ is in the third quadrant of the xy plane, what is cos(θ)? (a) Sketch y = sin(x). (b) Sketch y = sin2(x). (c) Sketch y = 1 1 + sin2(x) . Problem 18.2. Sketch the graphs of these functions: (a) f(t) = | sin(t)|. (b) f(t) = | cos(t)|. (c) f(t) = | tan(t)|. Problem 18.3. Solve the following: (a) If cos(θ) = 24 25 , what are the two possible values of sin(θ)? (c) If sin(θ) = 3 7 , what is sin( π 2 − θ)? Problem 18.4. These graphs represent periodic functions. Describe the period in each case. Problem 18.5. Start with the equation sin(θ) = cos(θ). Use the unit circle interpretation of the circular functions to find the solutions of this equation; make sure to describe your reasoning. Chapter 19 Sinusoidal Functions A migrating salmon is heading up a portion of the Columbia River. It’s depth d(t) (in feet) below the water surface is measured and plotted over a 30 minute period, as a function of time t (minutes). What is the formula for d(t)? In order to answer the question, we need to introduce an important new family of functions called the sinusoidal functions. These functions will play a central role in modeling any kind of periodic phenomena. The amazing fact is that almost any function you will encounter can be approximated by a sum of sinusoidal functions; a result that has far-reaching implications in all of our lives. 5 10 20 25 30 15 time Figure 19.1: The depth of a salmon as a function of time. 19.1 A special class of functions Beginning with the trigonometric function y = sin(x), what is the most general function we can build using the graphical techniques of shifting and stretching? horizontally shift y-axis 1 y = sin(x) horizontally dilate −π − π 2 π 2 π −1 x-axis vertically dilate vertically shift Figure 19.2: Visualizing the geometric operations available for curve sketching. The graph of y = sin(x) can be manipulated in four basic ways: horizontally shift, vertically shift, horizontally dilate or vertically dilate. Each of these “geometric operations” corresponds to a simple change in the “symbolic formula” for the function, as discussed in Chapter 13. 251 252 CHAPTER 19. SINUSOIDAL FUNCTIONS If we vertically shift the graph by D units upward, the resulting curve would be the graph of the function y = sin(x)+D; see Facts 13.3.1. Recall, the effect of the sign of D: If D is negative, the effect of shifting D units upward is the same as shifting |D| units downward. Notice, the function y = sin(x) + D is still a periodic function, having the same period 2π as y = sin(x). Notice, whereas the graph of the function y = sin(x) oscillates between the horizontal lines y = 1, the graph of y = sin(x) + D oscillates between y = D 1. For this reason, we sometimes refer to the constant D as the mean of the function y = sin(x) + D. In Figure 19.3, notice that the graph of y = sin(x) +D is symmetrically split by the horizontal “mean” line y = D. ± ± y-axis y-axis y = sin(x) + D y = sin(x) y = D D x-axis shift D units x-axis Figure 19.3: Interpreting the mean. Next, consider the effect of horizontally shifting the graph of y = sin(x) by C units to the right. By Facts 13.3.1, the new curve is the graph of the function y = sin(x − C). Also, recall the effect of the sign of C: If C is negative, the effect of shifting C units right is the same as shifting |C| units left. If the domain of sin(x) is 0 2π, then the domain of sin(x − C) is 0 x − C 2π, again by Facts 13.3.1. Rewriting this, the ≤ domain of sin(x − C) is C 2π + C and the graph will go through precisely one period on this domain. In other words, the new function sin(x − C) is still 2π-periodic. The constant C is usually called the phase shift of y = sin(x − C). Looking at Figure 19.4, it is possible to interpret C graphically: C will be a point where the graph crosses the horizontal axis on its way up from a minimum to a maximum. ≤ ≤ ≤ ≤ ≤ x x y-axis y-axis C y = sin(x) y = sin(x − C) x-axis x-axis shift C units Figure 19.4: Interpreting the phase shift. Vertically dilating the graph, either by vertical expansion or compression, leads to a new curve. The graph of this vertically dilated curve is y = A sin(x), for some positive constant A. Furthermore, if A > 1, 19.1. A SPECIAL CLASS OF FUNCTIONS 253 the graph of y = A sin(x) is a vertically expanded version of y = sin(x), whereas, if 0 < A < 1, then the graph of y = A sin(x) is a vertically compressed version of y = sin(x). Notice, the function y = A sin(x) is still 2πperiodic. What has changed is the band of oscillation: whereas the graph of the function y = sin(x) stays between the horizontal lines y = 1, the ± graph of y = A sin(x) oscillates between the horizontal lines y = A. We ± usually refer to A as the amplitude of the function y = A sin(x). y-axis y = sin(x) y-axis y = A sin(x) A x-axis x-axis stretch A units Figure 19.5: Interpreting the amplitude. Finally, horizontally dilating the graph, either by horizontal expansion or compression, leads to a new curve. The equation of this horizontally dilated curve is y = sin(cx), for some constant c > 0. We know that y = sin(x) is a 2π-periodic function and observe that horizontally dilation still results in a periodic function, but the period will typically NOT be 2π. For future purposes, it is useful to rewrite the equation for the horizontally stretched curve in a way more directly highlighting the period. To begin with, once the horizontal stretching factor c is known, we could rewrite c = 2π B , for some B = 0. y-axis y = sin(x) x-axis y-axis y = sin(( 2π B )x) stretch x-axis Figure 19.6: Interpreting the period. Here is the point of this yoga with the horizontal dilating constant: If we let the values of x range over the interval [0, B], then 2π B x will range over 2π the interval [0,2π]. In other words, the function y = sin is B-periodic B x and we can read off the period of y = sin by viewing the constant in this mysterious way. The four constructions outlined lead to a new family of functions. 2π B x 6 254 CHAPTER 19. SINUSOIDAL FUNCTIONS Definition 19.1.1 (The Sinusoidal Function). Let A, B, C and D be fixed constants, where A and B are both positive. Then we can form the new function y = A sin 2π B (x − C) + D, which is called a sinusoidal function. The four constants can be interpreted graphically as indicated: y-axis y = sin(x) x-axis all four operations B D y-axis C A x-axis y = A sin(( 2π B )(x − c)) + D Figure 19.7: Putting it all together for the sinusoidal function. 19.1.1 How to roughly sketch a sinusoidal graph Important Procedure 19.1.2. Given a sinusoidal function in the standard form y = A sin 2π B (x − C) + D, once the constants A, B, C, and D are specified, any graphing device can produce an accurate graph. However, it is pretty straightforward to sketch a rough graph by hand and the process will help reinforce the graphical meaning of the constants A, B, C, and D. Here is a “five step procedure” one can follow, assuming we are given A, B, C, and D. It is a good idea to follow Example 19.1.3 as you read this procedure; that way it will seem a lot less abstract. 1. Draw the horizontal line given by the equation y = D; this line will + D into symmetrical upper 2π B (x − C) “split” the graph of y = A sin and lower halves. 2. Draw the two horizontal lines given by the equations y = D A. These two lines determine a horizontal strip inside which the graph of the ± 19.1. A SPECIAL CLASS OF FUNCTIONS 255 sinusoidal function will oscillate. Notice, the points where the sinusoidal function has a maximum value lie on the line y = D + A. Likewise, the points where the sinusoidal function has a minimum value lie on the line y = D − A. Of course, we do not yet have a prescription that tells us where these maxima (peaks) and minima (valleys) are located; that will come out of the next steps. 3. Since we are given the period B, we know these important facts: (1) The period B is the horizontal distance between two successive maxima (peaks) in the graph. Likewise, the period B is the horizontal (2) distance between two successive minima (valleys) in the graph. The horizontal distance between a maxima (peak) and the successive minima (valley) is 1 2B. 4. Plot the point (C, D). This will be a place where the graph of the sinusoidal function will cross the mean line y = D on its way up from a minima to a maxima. This is not the only place where the graph crosses the mean line; it will also cross at the points obtained from (C, D) by horizontally shifting by any integer multiple of 1 2B. For example, here are three places the graph crosses the mean line: (C, D), (C + 1 2B, D), (C + B, D) 5. Finally, midway between (C, D) and (C+ 1 2B, D) there will be a maxima (peak); i.e. at the point (C + 1 4 B, D + A). Likewise, midway between (C + 1 2B, D) and (C + B, D) there will be a minima (valley); i.e. at the point (C + 3 4B,D − A). It is now possible to roughly sketch the graph on C + B by connecting the points described. Once the domain C this portion of the graph is known, the fact that the function is periodic C + 2B, tells us to simply repeat the picture in the intervals C + B C − B C, etc. ≤ ≤ ≤ ≤ x x x ≤ ≤ To make sense of this procedure, let’s do an explicit example to see how these five steps produce a rough sketch. 256 eplacements CHAPTER 19. SINUSOIDAL FUNCTIONS Example 19.1.3. The temperature (in ◦C) of Adri-N’s dorm room varies during the day according to the sinusoidal function d(t) = 6 sin
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+ 19, where t represents hours after midnight. Roughly sketch the graph of d(t) over a 24 hour period.. What is the temperature of the room at 2:00 pm? What is the maximum and minimum temperature of the room? π 12(t − 11) Solution. We begin with the rough sketch. Start by taking an inventory of the constants in this sinusoidal function: d(t) = 6 sin π 12 (t − 11) + 19 = A sin 2π B (t − C) + D. Conclude that A = 6, B = 24, C = 11, D = 19. Following the first four steps of the procedure outlined, we can sketch the lines y = D = 19, y = D 6 and three points where the graph crosses the mean line (see Figure19.8). A = 19 ± ± d(t) graph will oscillate inside this strip 30 25 20 15 10 (11, 19) (23, 19) (35, 19) y = 25 y = 19 y = 13 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.8: Sketching the mean D and amplitude A. According to the fifth step in the sketching procedure, we can plot the 4B, D − A) = (29, 13). maxima (C + 1 4B, D + A) = (17, 25) and the minima (C + 3 We then “connect the dots” to get a rough sketch on the domain 11 35. ≤ ≤ t d(t) graph will oscillate inside this strip 30 25 20 15 10 maxima (17, 25) (11, 19) (35, 19) (23, 19) (29, 13) minima y = 25 y = 19 y = 13 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.9: Visualizing the maximum and minimum over one period. 19.1. A SPECIAL CLASS OF FUNCTIONS 257 Finally, we can use the fact the function has period 24 to sketch the graph to the right and left by simply repeating the picture every 24 horizontal units. maxima (−7,25) y-axis 30 25 20 (−13,19) (−1,19) 15 maxima (17,25) maxima (41,25) (11,19) (23,19) (35,19) (47,19) (59,19) −12 −8 −14 −10 10 (5,13) minima (29,13) minima (53,13) minima −4 −6 −2 2 6 10 14 18 20 16 12 8 22 24 26 28 30 32 34 36 38 4 Figure 19.10: Repeat sketch for every full period. y = 25 y = 19 y = 13 t-axis We restrict the picture to the domain 0 24 and obtain the computer generated graph pictured in Figure 19.11; as you can see, our rough graph is very accurate. The temperature at 2:00 p.m. is just d(14) = 23.24◦ C. From the graph, the maximum value of the function will be D + A = 25◦ C and the minimum value will be D − A = 13◦ C 25 20 15 10 5 10 15 20 t (hours) Figure 19.11: The computer generated solution. 19.1.2 Functions not in standard sinusoidal form Any time we are given a trigonometric function written in the standard form y = A sin 2π B (x − C) + D, for constants A, B, C, and D (with A and B positive), the summary in Definition 19.1.1 tells us everything we could possibly want to know 258 CHAPTER 19. SINUSOIDAL FUNCTIONS about the graph. But, there are two ways in which we might encounter a trigonometric type function that is not in this standard form: • • The constants A or B might be negative. For example, y = −2 sin(2x− + 4 are examples that fail to be in 7) − 3 and y = 3 sin standard form. 2 x + 1 − 1 We might use the cosine function in place of the sine function. For example, something like y = 2 cos(3x + 1) − 2 fails to be in standard sinusoidal form. Now what do we do? Does this mean we need to repeat the analysis that led to Definition 19.1.1? It turns out that if we use our trig identities just right, then we can move any such equation into standard form and read off the amplitude, period, phase shift and mean. In other words, equations that fail to be in standard sinusoidal form for either of these two reasons will still define sinusoidal functions. We illustrate how this is done by way of some examples: Examples 19.1.4. (i) Start with y = −2 sin(2x−7) −3, then here are the steps with reference to the required identities to put the equation in standard form: y = −2 sin(2x − 7) − 3 = 2 (− sin(2x − 7)) − 3 = 2 sin(2x − 7 + π) + (−3) = 2 sin 2π π x − 7 − π 2 + (−3). Fact 18.2.5 on page 241 This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 7−π 2 = 1.93, mean D = −3, amplitude A = 2 and period B = π. (ii) Start with y = 3 sin(− 1 2x+1) +4, then here are the steps with reference to the required identities to put the equation in standard form: y = 3 sin − sin − − sin = 3 = 3 sin = 3 sin 1 2 2π 4π x − 1 + π (x − [2 − 2π]) + Fact 18.2.4 on page 241 + 4 Fact 18.2.5 on page 241 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 259 This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 2 − 2π, mean D = 4, amplitude A = 3 and period B = 4π. (iii) Start with y = 2 cos(3x+1) −2, then here are the steps to put the equation in standard form. A key simplifying step is to use the identity: cos(t) = sin( π 2 + t). y = 2 cos(3x + 1) − 2 = 2 sin = 2 sin = 2 sin + 3x + 1 π 2 3x − 2π 2π 3 h − 2 π −1 − + (−2) π 2 i! + (−2) This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 1 2 ], mean D = −2, amplitude A = 2 and period B = 2π 3 . 3[−1 − π 19.2 Examples of sinusoidal behavior Problems involving sinusoidal behavior come in two basic flavors. On the one hand, we could be handed an explicit sinusoidal function y = A sin 2π B (x − C) + D and asked various questions. The answers typically require either direct calculation or interpretation of the constants. Example 19.1.3 is typical of this kind of problem. On the other hand, we might be told a particular situation is described by a sinusoidal function and provided some data or a graph. In order to further analyze the problem, we need a “formula”, which means finding the constants A, B, C, and D. This is a typical scenario in a “mathematical modeling problem”: the process of observing data, THEN obtaining a mathematical formula. To find A, take half the difference between the largest and smallest values of f(x). The period B is most easily found by measuring the distance between two successive maxima (peaks) or minima (valleys) in the graph. The mean D is the average of the largest and smallest values of f(x). The phase shift C (which is usually the most tricky quantity to get your hands on) is found by locating a “reference point”. This “reference point” is a location where the graph crosses the mean line y = D on its way up from a minimum to a maximum. The funny thing is that the phase shift C is NOT unique; there are an infinite number of correct choices. One choice that will work 260 CHAPTER 19. SINUSOIDAL FUNCTIONS is C = (x-coordinate of a maximum) − B from this one by a multiple of the period B. 4 . Any other choice of C will differ A = max value − min value 2 B = distance between two successive peaks (or valleys) C = x-coordinate of a maximum − D = max value + min value 2 . B 4 Example 19.2.1. Assume that the number of hours of daylight in Seattle is given by a sinusoidal function d(t) of time. During 1994, assume the longest day of the year is June 21 with 15.7 hours of daylight and the shortest day is December 21 with 8.3 hours of daylight. Find a formula d(t) for the number of hours of daylight on the tth day of the year. 15 12.5 10 7. 50 100 150 200 250 300 350 t-axis (days) Solution. Because the function d(t) is assumed to be sinusoidal, it has the form y = A sin + D, for constants A, B, C, and D. We simply need to use the given information to find these constants. The largest value of the function is 15.7 and the smallest value is 8.3. Knowing this, from the above discussion we can read off : 2π B (t − C) Figure 19.12: Hours of daylight in Seattle in 1994. D = 15.7 + 8.3 2 = 12 A = 15.7 − 8.3 2 = 3.7. To find the period, we need to compute the time between two successive maximum values of d(t). To find this, we can simply double the time length of one-half period, which would be the length of time between successive maximum and minimum values of d(t). This gives us the equation B = 2(days between June 21 and December 21) = 2(183) = 366. Locating the final constant C requires the most thought. Recall, the longest day of the year is June 21, which is day 172 of the year, so C = (day with max daylight) − B 4 = 172 − 366 4 = 80.5. In summary, this shows that d(t) = 3.7 sin 2π 366 (t − 80.5) + 12. A rough sketch, following the procedure outlined above, gives this graph 366; we have included the mean line y = 12 for on the domain 0 reference. ≤ ≤ t 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 261 We close with the example that started this section. Example 19.2.2. The depth of a migrating salmon below the water surface changes according to a sinusoidal function of time. The fish varies It takes the fish between 1 and 5 feet below the surface of the water. 1.571 minutes to move from its minimum depth to its successive maximum depth. It is located at a maximum depth when t = 4.285 minutes. What is the formula for the function d(t) that predicts the depth of the fish after t minutes? What was the depth of the salmon when it was first spotted? During the first 10 minutes, how many times will the salmon be exactly 4 feet below the surface of the water? Solution. We know that d(t) = A sin( 2π B (x − C)) + D, for appropriate constants A, B, C, and D. We need to use the given information to extract these four constants. The amplitude and mean are easily found using the above formulas = max depth − min depth 2 max depth + min depth . 2 4 6 time 8 10 Figure 19.13: Depth of a migrating salmon. The period can be found by noting that the information about the time between a successive minimum and maximum depth will be half of a period (look at the picture in Figure 19.13): B = 2(1.571) = 3.142 Finally, to find C we C = (time of maximum depth) − B 4 = 4.285 − 3.142 4 = 3.50. The formula is now d(t) = 2 sin 2π 3.142 (t − 3.5) + 3 = 2 sin(2t − 7) + 3 The depth of the salmon when it was first spotted is just d(0) = 2 sin(−7) + 3 = 1.686 feet. Finally, graphically, the last question amounts to determining how many times the graph of d(t) crosses the line y = 4 on the domain [0,10]. This can be done using Figure 19.13. A simultaneous picture of the two graphs is given, from which we can see the salmon is exactly 4 feet below t
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he surface of the water six times during the first 10 minutes. 262 CHAPTER 19. SINUSOIDAL FUNCTIONS 19.3 Summary A sinusoidal function is one of the form • f(t) = A sin (t − C) + D where A, B, C, and D are constants. 2π B – A is the amplitude of the function; this is half the vertical dis- tance between a high point and a low point on its graph. – B is the period of the function; this is the horizontal distance between two consecutive high points (or low points) on its graph. – C is the phase shift of the function; it is multi-valued, but one choice for C is a value of t at which the function is increasing and equal to D. – D is the mean value of the function; it is the y-value of the horizontal line about which the graph of the function is balanced. • The graph of a sinusoidal function is a shifted, scaled version of the graph of y = sin t. 19.4. EXERCISES 19.4 Exercises Problem 19.1. Find the amplitude, period, a phase shift and the mean of the following sinusoidal functions. (a) y = sin(2x − π) + 1 (b) y = 6 sin(πx) − 1 (c) y = 3 sin(x + 2.7) + 5.2 (d) y = 5.6 sin (e) y = 2.1 sin − 12.1 − 9.8 2 3 x − 7 x π + 44.3 (f) y = 3.9 (sin(22.34(x + 18)) − 11) (g) y = 11.2 sin 5 π (x − 9.2) + 8.3 Problem 19.2. A weight is attached to a spring suspended from a beam. At time t = 0, it is pulled down to a point 10 cm above the ground and released. After that, it bounces up and down between its minimum height of 10 cm and a maximum height of 26 cm, and its height h(t) is a sinusoidal function of time t. It first reaches a maximum height 0.6 seconds after starting. (a) Follow the procedure outlined in this section to sketch a rough graph of h(t). Draw at least two complete cycles of the oscillation, indicating where the maxima and minima occur. (b) What are the mean, amplitude, phase shift and period for this function? (c) Give four different possible values for the phase shift. (d) Write down a formula for the function h(t) in standard sinusoidal form; i.e. as in 19.1.1 on Page 254. (e) What is the height of the weight after 0.18 seconds? (f) During the first 10 seconds, how many times will the weight be exactly 22 cm above the floor? (Note: This problem does not require inverse trigonometry.) Problem 19.3. A respiratory ailment called “Cheyne-Stokes Respiration” causes the volume per breath to increase and decrease in a sinusoidal manner, as a function of time. For one particular patient with this condition, a 263 machine begins recording a plot of volume per breath versus time (in seconds). Let b(t) be a function of time t that tells us the volume (in liters) of a breath that starts at time t. During the test, the smallest volume per breath is 0.6 liters and this first occurs for a breath that starts 5 seconds into the test. The largest volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into the test. (a) Find a formula for the function b(t) whose graph will model the test data for this patient. (b) If the patient begins a breath every 5 seconds, what are the breath volumes during the first minute of the test? Problem 19.4. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which time the water is 10 feet above the height of low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally. (a) Find a formula for the function y = h(t) that computes the height of the tide above low tide at time t. (In other words, y = 0 corresponds to low tide.) (b) What is the tide height at 11:00 a.m.? Problem 19.5. Your seat on a Ferris Wheel is at the indicated position at time t = 0. Start 53 feet Let t be the number of seconds elapsed after the wheel begins rotating counterclockwise. You find it takes 3 seconds to reach the top, which is 53 feet above the ground. The wheel is rotating 12 RPM and the diameter of the wheel is 50 feet. Let d(t) be your height above the ground at time t. 264 CHAPTER 19. SINUSOIDAL FUNCTIONS (a) Argue that d(t) is a sinusoidal function, describing the amplitude, phase shift, period and mean. (b) When are the first and second times you are exactly 28 feet above the ground? (c) After 29 seconds, how many times will you have been exactly 28 feet above the ground? Problem 19.6. In Exercise 17.12, we studied the situation below: A bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle of motion. In each of the cases, the earlier exercise found the coordinates P(t) of the bug at time t. For each of the scenarios below, answer these two questions: (a) Both coordinates of P(t) = (x(t),y(t)) are sinusoidal functions in the variable t. Sketch a rough graph of the functions x(t) and y(t) on the domain 0 9. t ≤ ≤ ω=4π/9rad/sec bug lands here 1.2 rad 2 in bug lands here ω=4π/9 rad/sec ω= 4π/9rad/sec bug lands here 0.5 rad 2 in 2 in Problem 19.7. The voltage output(in volts) of an electrical circuit at time t seconds is given by the function V(t) = 23 sin(5πt−3π)+1. (a) What is the initial voltage output of the circuit? (b) Is the voltage output of the circuit ever equal to zero? Explain. (c) The function V(t) = 2p(t), where p(t) = 3 sin(5πt − 3π) + 1. Put the sinusoidal function p(t) in standard form and sketch the graph for 0 1. Label the coordinates of the extrema on the graph. ≤ ≤ t (b) Use the graph sketches to help you find the the amplitude, mean, period and phase shift for each function. Write x(t) and y(t) in standard sinusoidal form. (d) Calculate the maximum and minimum voltage output of the circuit. (e) During the first second, determine when the voltage output of the circuit is 10 volts. 19.4. EXERCISES 265 (f) A picture of the graph of y = V(t) on the 1 is given; label the coor- domain 0 dinates of the extrema on the graph. ≤ ≤ t volts 15 12.5 10 7.5 5 2.5 free to move back and forth along the x-axis. The point A is at (2,0) at time t = 0, and the wheel rotates counterclockwise at 3 rev/sec. 2 -2 time t=0 time t > 0 (a) As the point A makes one complete revolution, indicate in the picture the direction and range of motion of the point B. 0.2 0.4 0.6 0.8 1 t axis (b) Find the coordinates of the point A as a t ≤ ≤ (g) Restrict the function V(t) to the domain 0.3; explain why this function 0.1 has an inverse and find the formula for the inverse rule. Restrict the function V(t) to the domain 0.3 0.5; explain why this function has an inverse and find the formula for the inverse rule. ≤ ≤ t Problem 19.8. A six foot long rod is attached at one end A to a point on a wheel of radius 2 feet, centered at the origin. The other end B is function of time t. (c) Find the coordinates of the point B as a function of time t. (d) What is the x-coordinate of the point B when t = 1? You should be able to find this two ways: with your function from part (c), and using some common sense (where is point A after one second?). (e) Is the function you found in (c) a sinu- soidal function? Explain. 266 CHAPTER 19. SINUSOIDAL FUNCTIONS Chapter 20 Inverse Circular Functions An aircraft is flying at an altitude 10 miles above the elevation of an airport. If the airplane begins a steady descent 100 miles from the airport, what is the angle θ of descent? The only natural circular function we can use is z = tan(θ), leading to the equation: s e l i m 0 1 θ 100 miles Figure 20.1: An aircraft decending toward an airport. tan(θ) = 10 100 = 1 10 . The problem is that this equation does not tell us the value of θ. Moreover, none of the equation solving techniques at our disposal (which all amount to algebraic manipulations) will help us solve the equation for θ. What we need is an inverse function θ = f−1(z); then we could use the fact that tan−1(tan(θ)) = θ and obtain: θ = tan−1(tan(θ)) = tan−1 1 10 . Computationally, without even thinking about what is going on, any scientific calculator will allow us to compute values of an inverse circular function and leads to a solution of our problem. In this example, you will find θ = tan−1 = 5.71◦. Punch this into your calculator and verify it! 1 10 20.1 Solving Three Equations Example 20.1.1. Find all values of θ (an angle) that make this equation true: sin(θ) = 1 2. Solution. We begin with a graphical reinterpretation: the solutions correspond to the places where the graphs of z = sin(θ) and z = 1 2 intersect in the θz-coordinate system. Recalling Figure 18.11, we can picture these two graphs simultaneously as below: 267 268 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS cross cross z-axis 1 A B cross cross Graph of z = 1 2 θ-axis −2π −π π 2π 3π Graph of z = sin(θ) −1 one period Figure 20.2: Where does sin(θ) cross z = 1 2 ? The first thing to notice is that these two graphs will cross an infinite number of times, so there are infinitely many solutions to Example 20.1.1! However, notice there is a predictable spacing of the crossing points, which is just a manifestation of the periodicity of the sine function. In fact, if we can find the two crossing points labeled “A” and “B”, then all other crossing points are obtained by adding multiples of 2π to either “A” or “B”. By Table 17.1, θ = π 6 radians is a special angle where we π = 1 computed sin 2, which tells us that the crossing point labeled “A” 6 is the point ( π 6 , 1 2). Using the identities in Facts 18.2.4 and 18.2.5, notice that sin 5π 6 = sin − = − sin π 6 − − sin π 6 = − = sin = 1 2 . + π π 6 π 6 So, θ = 5π is the only other angle θ between 0 and 2π such that Ex6 ample 20.1.1 holds. This corresponds to the crossing point labeled “B”, 6 , 1 . In view of the remarks above, the crossing which has coordinates points come in two flavors: 5π 2 π 6 5π 6 + 2kπ, + 2kπ, 1 2 1 2 , k = 0, 1, ± 2, ± 3, ± · · · , and , k = 0, 1, ± 2, ± 3, ± · · · . Taking this example as a model, we can tackle the more general problem: For a fixed real number c, describe the solu
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tion(s) of the equation 20.1. SOLVING THREE EQUATIONS 269 c = f(θ) for each of the circular functions z = f(θ). Studying solutions of these equations will force us to come to grips with three important issues: • • • For what values of c does f(θ) = c have a solution? For a given value of c, how many solutions does f(θ) = c have? Can we restrict the domain so that the resulting function is one-toone? All of these questions must be answered before we can come to grips with any understanding of the inverse functions. Using the graphs of the circular functions, it is an easy matter to arrive at the following qualitative conclusions. Important Fact 20.1.2. None of the circular functions is one-to-one on the domain of all θ values. The equations c = sin(θ) and c = cos(θ) have a solution if and only if −1 1; if c is in this range, there are infinitely many solutions. The equation c = tan(θ) has a solution for any value of c and there are infinitely many solutions. ≤ ≤ c Example 20.1.3. If two sides of a right triangle have lengths 1 and √3 as pictured below, what are the acute angles α and β? Solution. By the Pythagorean Theorem the remaining side has length 1 + √3 2 = 2. r Since tan(α) = √3, we need to solve this equation for α. Graphically, we need to determine where z = √3 crosses the graph of the tangent function: β √3 α 1 Figure 20.3: What are the values for α and β? From Fact 20.1.2, there will be infinitely many solutions to our equa2 , π tion, but notice that there is exactly one solution in the interval and we can find it using Table 17.1: − π 2 tan = tan(60◦) = π 3 sin(60◦) cos(60◦) = = √3. √3 2 1 2 So, α = π 3 radians = 60◦ is the only acute angle solution and β = 180◦ − 60◦ − 90◦ = 30◦. 270 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS θ = − π 2 z-axis one period etc. only place graphs cross in this period is ( π 3 ,√3) θ = 5π 2 etc. θ-axis −2π −π π 2π 3π θ = − 3π 2 θ = π 2 θ = 3π 2 Figure 20.4: Where does the line z = √3 cross z = tan θ? 20.2 Inverse Circular Functions Except for specially chosen angles, we have not addressed the serious problem of FINDING values of the inverse rules attached to the circular function equations. (Our previous examples were “rigged”, so that we could use Table 17.1.) To proceed computationally, we need to obtain the inverse circular functions. If we were to proceed in a sloppy manner, then a first attempt at defining the inverse circular functions would be to write 8 sin−1(z) = solutions θ of the equation z = sin(θ). 8 cos−1(z) = solutions θ of the equation z = cos(θ). 8 tan−1(z) = solutions θ of the equation z = tan(θ). (20.1) z ≤ ≤ There are two main problems with these rules as they stand. First, to have a solution θ in the case of sin−1(z) and cos−1(z), we need to restrict z so that −1 1. Secondly, having made this restriction on z in the first two cases, there is no unique solution; rather, there are an infinite number of solutions. This means that the rules sin−1, cos−1, and tan−1 as they now stand do not define functions. Given what we have reviewed about inverse functions, the only way to proceed is to restrict each circular function to a domain of θ values on which it becomes one-to-one, then we can appeal to Fact 9.3.1 and conclude the inverse function makes sense. 20.2. INVERSE CIRCULAR FUNCTIONS 271 At this stage a lot of choice (flexibility) enters into determining the domain on which we should try to invert each circular function. In effect, there are an infinite number of possible choices. If z = f(θ) denotes one of the three circular functions, there are three natural criteria we use to guide the choice of a restricted domain, which we will call a principal domain: • • • The domain of f(θ) should include the angles between 0 and π these are the possible acute angles in a right triangle. 2 , since On the restricted domain, the function f(θ) should take on all possible values in the range of f(θ). In addition, the function should be one-to-one on this restricted domain. The function f(θ) should be “continuous” on this restricted domain; i.e. the graph on this domain could be traced with a pencil, without lifting it off the paper. In the case of z = sin(θ), the principal domain − π π 2 satisfies our criteria and the picture is given below. Notice, we would not want to take π, since z = sin(θ) doesn’t achieve negative values on the interval 0 this domain; in addition, it’s not one-to-one there. 2 ≤ ≤ ≤ ≤ θ θ z-axis 1 −π − π 2 π 2 π θ-axis −1 sine restricted to principal domain y-axis principal domain on the unit circle θ x-axis Figure 20.5: Principal domain for sin(θ). θ θ ≤ 2 ≤ In the case of z = cos(θ), the principal domain 0 π satisfies our criteria and the picture is given below. Notice, we would not want to take the interval − π π 2 , since z = cos(θ) doesn’t achieve negative values on this domain; in addition, it’s not one-to-one there. In the case of z = tan(θ), the principal domain − π 2 satisfies our criteria and the picture is given below. Notice, we would not want to π, since z = tan(θ) does not have a continuous take the interval 0 graph on this interval; in other words, we do not include the endpoints since tan(θ) is undefined there Important Facts 20.2.1 (Inverse circular functions). Restricting each circular function to its principal domain, its inverse rule f−1(z) = θ will define a function. 272 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS z-axis 1 y-axis principal domain on the unit circle −π − π 2 π 2 π 3π 2 θ-axis θ x-axis −1 cosine restricted to principal domain Figure 20.6: Principal domain for cos(θ). (i) If −1 z domain − π ≤ ≤ 2 ≤ 1, then sin−1(z) is the unique angle θ in the principal θ π 2 with the property that sin(θ) = z. ≤ (ii) If −1 z domain 0 ≤ 1, then cos−1(z) is the unique angle θ in the principal π with the property that cos(θ) = z. ≤ ≤ θ ≤ (iii) For any real number z, tan−1(z) is the unique angle θ in the principal domain − π 2 < θ < π 2 with the property that tan(θ) = z. We refer to the functions defined above as the inverse circular functions. These are sometimes referred to as the “arcsine”, “arccosine” and “arctangent” functions, though we will not use that terminology. The inverse circular functions give us one solution for each of these equations: c = cos(θ) c = sin(θ) c = tan(θ); !!! CAUTION !!! these are called the principal solutions. We also can refer to these as the principal values of the inverse circular function rules θ = f−1(z). As usual, be careful with “radian mode” and “degree mode” when making calculations. For example, if your calculator is in “degree” mode and you type in “tan−1(18)”, the answer given is “86.82”. This means that an angle of measure θ = 86.82◦ has tan(86.82) = 18. If your calculator is in “radian” mode and you type in “sin−1(0.9)”, the answer given is “1.12”. This means that an angle of measure θ = 1.12 radians has sin(1.12) = 0.9. There is a key property of the inverse circular functions which is useful in equation solving; it is just a direct translation of Fact 9.3.2 into our current context: Important Facts 20.2.2 (Composition identities). We have the following equalities involving compositions of circular functions and their inverses: 20.3. APPLICATIONS 273 z-axis 1 −1 −π − π 2 π 2 π θ-axis tangent function restricted to principal domain y-axis principal domain on the unit circle θ x-axis Figure 20.7: Principal domain for tan(θ). (a) If − π θ 2 ≤ θ π 2 , then sin−1(sin(θ)) = θ. ≤ π, then cos−1(cos(θ)) = θ. (b) If 0 ≤ ≤ (c) If − π 2 < θ < π 2 , then tan−1(tan(θ)) = θ. We have been very explicit about the allowed θ values for the equations in Fact 20.2.2. This is important and an Exercise will touch on this issue. 20.3 Applications As a simple application of Fact 20.2.2, we can return to the beginning of this section and justify the reasoning used to find the angle of descent of the aircraft: θ = tan−1((tan(θ)) = tan−1 1 10 = 0.09967 rad = 5.71◦. Let’s look at some other applications. Example 20.3.1. Find two acute angles θ so that the following equation is satisfied: 9 4 cos2(θ) = 252 16 1 − cos2(θ) . Solution. Begin by multiplying each side of the equation by cos2(θ) and rearranging terms: 9 4 = 0 = 252 16 252 16 1 − cos2(θ) cos2(θ) cos4(θ) − 252 16 cos2(θ) + 9 4 . 274 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS To solve this equation for θ, we use what is called the technique of substitution. The central idea is to bring the quadratic formula into the picture by making the substitution z = cos2(θ): 0 = 252 16 z2 − 252 16 z + 9 4 . Applying the quadratic formula, we obtain 252 z = 16 ± r 2 − 4 252 16 252 16 9 4 2(252) 16 = 0.9386 or 0.06136. We now use the fact that z = cos2(θ) and note the cosine of an acute angle is non-negative to conclude that cos2(θ) = 0.9386 or cos2(θ) = 0.06136 cos(θ) = 0.2477. cos(θ) = 0.9688 or Finally, we use the inverse cosine function to arrive at our two acute angle solutions: cos(θ) = 0.9688 cos(θ) = 0.2477 θ = cos−1(0.9688) = 14.35◦ θ = cos−1(0.2477) = 75.66◦ ⇒ ⇒ h α 32 d Example 20.3.2. A 32 ft ladder leans against a building (as shown below) making an angle α with the wall. OSHA (Occupational Safety and Health Administration) specifies a “safety range” for the angle α to be 15◦ 44◦. If the base of the ladder is d = 10 feet from the house, is this a safe placement? Find the highest and lowest points safely accessible. Solution. If d = 10, then sin(α) = 10 lution is α = sin−1 ≤ ≤ α 32, so the principal so10 = 18.21◦; this lies within the safety 32 zone. From the picture, it is clear that the highest point safely reached will occur precisely when α = 15◦ and as this angle increases, the height h decreases until we reach the lowest safe height when α = 44◦. We need to solve two right triangles. If α = 15◦, then h = 32 cos(15◦) = 30.9 ft. If α = 44◦, then h = 32 cos(44◦) = 23.02 ft. Figure 20.8: A ladder problem. Example 20.3.3. A Coast Guard jet pilot makes contact with a small unidentified propeller plane 15 miles away at the same altitude in a direction 0.5 radians counterclockwise from East.
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The prop plane flies in the direction 1.0 radians counterclockwise from East. The jet has been instructed to allow the prop plane to fly 10 miles before intercepting. In what direction should the jet fly to intercept the prop plane? If the prop plane is flying 200 mph, how fast should the jet be flying to intercept? 20.3. APPLICATIONS 275 North Intercept point Solution. A picture of the situation is shown in Figure 20.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in the Figure 20.9(b), by imposing a coordinate system and labeling the various sides of our triangles. We will work in radian units and label θ to be the required intercept heading. We will first determine the sides x + y and u + w of the large right triangle. To do this, we have x = 15 cos(0.5) = 13.164 miles, y = 10 cos(1.0) = 5.403 miles, w = 15 sin(0.5) = 7.191 miles, and u = 10 sin(1.0) = 8.415 miles. We now have tan(θ) = w+u x+y = 0.8405, so the principal solution is θ = 0.699 radians, which is about 40.05◦. This is the only acute angle solution, so we have found the required intercept heading. find speed to intercept Jet West 0.5 rad South 10 miles 1 rad East prop spotted here plane 15 miles East θ = intercept heading (a) The physical layout. y-axis 1 rad Intercept point u 10 w u + w prop plane spotted here x-axis d 15 θ 0.5 rad Jet x y x + y (b) Modeling the problem. Figure 20.9: Visualizing the Coast Guard problem. the hypotenuse of To find the intercept speed, first compute your distance to the intercept point, which is the length d = of (18.567)2 + (15.606)2 = 24.254 miles. You need to travel this distance in the same amount of time T it takes the p prop plane to travel 10 miles at 200 mph; i.e. T = 10 the intercept speed s is the big right triangle: 200 = 0.05 hours. Thus, s = distance traveled time T elapsed = 24.254 0.05 = 485 mph. Later you will use an alternative approach to this problem using ve- locity vectors. In certain applications, knowledge of the principal solutions for the equations c = cos(θ), c = sin(θ), and c = tan(θ) is not sufficient. Here is a typical example of this, illustrating the reasoning required. Example 20.3.4. A rigid 14 ft pole is used to vault. The vaulter leaves and returns to the ground when the tip is 6 feet high, as indicated. What are the angles of the pole with the ground on takeoff and landing? 276 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 14 ft α 6 ft 6 ft β leaving ground airborne returning to ground Figure 20.10: Various angles of a vaulter’s pole. Q y-axis -axis Figure 20.11: Modeling the problem with a unit circle. Solution. From the obvious right triangles in the picture, we are interested in finding angles θ where sin(θ) = 6 14 = 3 7. The idea is to proceed in three steps: Find the principal solution of the equation sin(θ) = 3 7; Find all solutions of the equation sin(θ) = 3 7; Use the constraints of the problem to find α and β among the set of all solutions. • • • Solving the equation sin(θ) = 3 on the unit circle with y-coordinate equal to 3 7 involves finding the points 7 . From the picture, we see there are two such points, labeled P and Q. The coordinates of these points will be P = cos(β), 3 7 Notice, α is the principal solution of our equation sin(θ) = 3 7, since 0 In general, the solutions come in two basic flavors: 90◦; so α = sin−1 cos(α), 3 7 = 25.38◦. and + 2k(180◦) = 25.38◦ + 2k(180◦), or θ = β + 2k(180◦), where k = 0, 1, properties of the circular functions: 2, ± ± ± 3, . . . . To find the angle β, we can use basic sin(β) = sin(α) = − sin(−α) = sin(180◦ − α) = sin(154.62◦). This tells us β = 154.62◦. 20.4 How to solve trigonometric equations So far, our serious use of the inverse trigonometric functions has focused on situations that ultimately involve triangles. However, many trigonometric modeling problems have nothing to do with triangles and so we 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 277 need to free ourselves from the necessity of relying on such a geometric picture. There are two general strategies for finding solutions to the equations c = sin(θ), c = cos(θ), and c = tan(θ): • • The first strategy is summarized in Procedure 20.4.1. This method has the advantage of offering a “prescription” for solving the equations; the disadvantage is you can lose intuition toward interpreting your answers. The second strategy is graphical in nature and is illustrated in Example 20.4.2 below. This method usually clarifies interpretation of the answers, but it does require more work since an essential step is to roughly sketch the graph of the trigonometric function (following the procedure of Chapter 19 or using a graphing device). Each approach has its merits as you will see in the exercises. Important Procedure 20.4.1. To find ALL solutions to the equations c = sin(θ), c = cos(θ), and c = tan(θ), we can lay out a foolproof strategy. Step Sine case Cosine case Tangent case 1. Find principal solution 2. Find symmetry solution θ = sin−1(c) θ = cos−1(c) θ = tan−1(c) θ = − sin−1(c) + π θ = − cos−1(c) not applicable 3. Write out multiples of period k = 0, 2, 1, ± ± · · · 2kπ 2kπ kπ 4. Obtain general principal solutions 5. Obtain general symmetry solutions θ = sin−1(c) + 2kπ θ = cos−1(c) + 2kπ θ = tan−1(c) + kπ θ = − sin−1(c) + π + 2kπ θ = − cos−1(c) + 2kπ not applicable Example 20.4.2. Assume that the number of hours of daylight in your hometown during 1994 is given by the function d(t) = 3.7 sin + 12, where t represents the day of the year. Find the days of the year during which there will be approximately 14 hours of daylight?1 2π 366 (t − 80.5) t ≤ ≤ Solution. To begin, we want to roughly sketch the graph of y = d(t) on the domain 0 366. If you apply the graphing procedure discussed in Chapter 19, you obtain the sinusoidal graph below on the larger domain −366 732. (The reason we use a larger domain is so that the “pattern” that will arise in the strategy described below is more evident. Ultimately, we will restrict our attention to the smaller domain 0 366.) To determine when there will be 14 hours of daylight, we need to solve the equation 14 = d(t). Graphically, this amounts to finding ≤ ≤ ≤ ≤ t t 278 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS y-axis 15 12.5 10 7.5 t-axis −200 200 400 600 Figure 20.12: Where does the sinusoidal function d(t) cross the line y = 14. the places where the line y = 14 intersects the graph of d(t). As can be seen in Figure 20.12, there are several such intersection points. We now outline a “three step strategy” to find all of these intersection points (which amounts to solving the equation 14 = d(t)): 1. Principal Solution. We will find one solution by using the inverse sine function. If we start with the function y = sin(t) on its principal domain −π π 2 , then we can compute the domain of d(t) = + 12: 3.7 sin 2 ≤ t ≤ 2π 366(t − 80.5) π − 2 ≤ −11 ≤ 2π 366 t ≤ (t − 80.5) π 2 ≤ 172. Now, using the inverse sine function we can find the principal solution to the equation 14 = d(t): 14 = 3.7 sin (t − 80.5) + 12 (20.2) 2π 366 2π 366 0.54054 = sin (t − 80.5) 0.57108 = sin−1(0.54054) = t = 113.8 π 183 (t − 80.5) Notice, this answer is in the domain −11 In effect, we have found THE ONLY SOLUTION on this domain. Conclude that there will be about 14 hours of daylight on the 114th day of the year. 172. ≤ ≤ t 1You can get the actual data from the naval observatory at this world wide web address: <http://tycho.usno.navy.mil/time.html> 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 279 2. Symmetry Solution. To find another solution to the equation 14 = d(t), we will use symmetry properties of the graph of y = d(t). This is where having the graph of y = d(t) is most useful. We know the a maxima on the graph occurs at the point M = (172, 15.7); review Example 19.2.1 for a discussion of why this is the case. From the graph, we can see there are two symmetrically located intersection points on either side of M. The principal solution gives the intersection point (113.8, 14). This point is 58.2 horizontal units to the left of M; see the picture below. So a symmetrically positioned intersection point will be (172 + 58.2, 14) = (230.2, 14). y-axis M 15 12.5 10 7.5 −200 200 400 600 t-axis Figure 20.13: Finding the symetry solution. In other words, t = 230.2 is a second solution to the equation 14 = d(t). We call this the symmetry solution. 3. Other Solutions. To find all other solutions of 14 = d(t), we add integer multiples of the period B = 366 to the t-coordinates of the principal and symmetric intersection points. On the domain −366 t the graph: ≤ 732 we get these six intersection points; refer to the picture of ≤ (−252.2, 14), (113.8, 14), (479.8, 14), (−135.8, 14), (230.2, 14), (596.2, 14). So, on the domain −366 equation 14 = d(t): t ≤ ≤ 732 we have these six solutions to the t = −252.2, −135.8, 113.8, 230.2, 479.8, 596.2. To conclude the problem, we only are interested in solutions in the 366, so the answers are t = 113.8, 230.2; i.e. on days 114 domain 0 and 230 there will be about 14 hours of daylight. ≤ ≤ t 280 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 20.5 Summary • • • The inverse sine function, sin−1 x, is defined as the inverse of the sine function, sin x, restricted to the domain −π/2 π/2. x ≤ ≤ Inverse cosine and inverse tangent are defined similarly. To find the solutions to an equation of the form f(t) = A sin 2π B (t − C) + D = k, proceed as follows: 1. Sketch the graph of the function f(t), including a few periods and t = C. 2. Use algebra and the inverse sine function (sin−1) to find one solution. This is the principal solution; it is the solution nearest to t = C. Call this solution P. 3. Use the principal solution and your knowledge of the graph of the function to find the symmetry solution. Call this solution S. 4. All solutions are then of the form P, P ± B, P ± 2B, P ± 3B, . . . or S, S ± B, S ± 2B, S ± 3B, . . . . 20.6. EXERCISES 20.6 Exercises 281 Problem 20.1. Let’s make sure we can handle the symbolic and mechanical aspects
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of working with the inverse trigonometric functions: (a) Set your calculator to “radian mode” and compute to four decimal places: Problem 20.4. Hugo bakes world famous scones. The key to his success is a special oven whose temperature varies according to a sinusoidal function; assume the temperature (in degrees Fahrenheit) of the oven t minutes after inserting the scones is given by (a1) sin−1(x), −3 11 ,2. (a2) cos−1(x), −3 11 ,2. (a3) tan−1(x), −3 11 ,2. for x = 0,1,−1, √3 2 ,0.657, for x = 0,1,−1, √3 2 ,0.657, for x = 0,1,−1, √3 2 ,0.657, (b) Redo part (a) with your calculator set in “degree mode”. (c) Find four values of x that satisfy the equation 5 sin(2x2 + x − 1) = 2. (d) Find four solutions to the equation 5 tan(2x2 + x − 1) = 2 Problem 20.2. For each part of the problem below: • • • Sketch the graphs of f(x) and g(x) on the same set of axes. Set f(x) = g(x) and find the principal and symmetry solutions. Calculate at least two other solutions to f(x) = g(x). Indicate them on your graph. (a) f(x) = sin (b) f(x) = sin (x) = 1 3 . , g(x) = −1. (c) f(x) = sin (2x − 1), g(x) = 1 4 . (d) f(x) = 10 cos (2x + 1) − 5, g(x) = −1. Problem 20.3. Assume that the number of hours of daylight in New Orleans in 1994 is given by the function D(x) = 7 + 35 3 , where x represents the number of days after March 21. 2π 365 x 3 sin (a) Find the number of hours of daylight on January 1, May 18 and October 5. (b) On what days of the year will there be approximately 10 hours of daylight? y = s(t) = 15 sin t − + 415 π 5 3π 2 (a) Find the amplitude, phase shift, period and mean for s(t), then sketch the graph t on the domain 0 ≤ (b) What is the maximum temperature of the oven? Give all times when the oven achieves this maximum temperature during the first 20 minutes. 20 minutes. ≤ (c) What is the minimum temperature of the oven? Give all times when the oven achieves this minimum temperature during the first 20 minutes. (d) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at least 410oF. (e) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at most 425oF. (f) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is between 410oF and 425oF. Problem 20.5. The temperature in Gavin’s oven is a sinusoidal function of time. Gavin sets his oven so that it has a maximum temperature of 300◦F and a minimum temperature of 240◦. Once the temperature hits 300◦, it takes 20 minutes before it is 300◦ again. Gavin’s cake needs to be in the oven for 30 minutes at temperatures at or above 280 ◦. He puts the cake into the oven when it is at 270◦ and rising. How long will Gavin need to leave the cake in the oven? Problem 20.6. Maria started observing Elasticman’s height at midnight. At 3 AM, he was at his shortest: only 5 feet tall. At 9 AM, he was at his tallest: 11 feet tall. Elasticman’s height is a sinusoidal function of time. 282 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS In the 24 hours after Maria began observing Elasticman, how much of the time will Elasticman be less than 6 feet tall? Problem 20.7. Suppose T (t) = 23 sin 2π 24 (t − 7) + 66 (i) Find where Tiffany passes Michael the first time. (j) Find when Tiffany passes Michael the second time. (k) Find where Tiffany passes Michael the second time. is the temperature (in degrees Fahrenheit) at time t, where t is measured in hours after midnight on Sunday. You paint the exterior door to your house at 5 p.m. on Monday. The paint information states that 48 hours of 75◦ F drying time is required; i.e., you can only count time periods when the temperature is at least 75◦ F. When will the door be dry? Problem 20.9. A communications satellite orbits the earth t miles above the surface. Assume the radius of the earth is 3,960 miles. The satellite can only “see” a portion of the earth’s surface, bounded by what is called a horizon circle. This leads to a two-dimensional cross-sectional picture we can use to study the size of the horizon slice: Problem 20.8. Tiffany and Michael begin running around a circular track of radius 100 yards. They start at the locations pictured. is running 0.025 rad/sec counterMichael clockwise and Tiffany is running 0.03 rad/sec counterclockwise. Impose coordinates as pictured. r= 100 yards 0.025 rad/sec Michael starts here Tiff starts here 0.03 rad/sec (a) Where is each runner located (in xy-coordinates) after 8 seconds? (b) How far has each runner traveled after 8 seconds? (c) Find the angle swept out by Michael after t seconds. (d) Find the angle swept out by Tiffany after t seconds. (e) Find the xy-coordinates of Michael and Tiffany after t seconds. (f) Find the first time when Michael’s x-coordinate is -50. (g) Find the first time when Tiffany’s x-coordinate is -50. Earth horizon circle center of Earth Earth satellite satellite α α t CROSS−SECTION (a) Find a formula for α in terms of t. (b) If t = 30,000 miles, what is alpha? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? (c) If t = 1,000 miles, what is alpha? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? (d) Suppose you wish to place a satellite into orbit so that 20% of the circumference is covered by the satellite. What is the required distance t? (h) Find when Tiffany passes Michael the first time. Problem 20.10. Answer the following questions: 20.6. EXERCISES 283 (a) If y = sin(x) on the domain − π π 2 , what is the domain D and range R of y = 2 sin(3x − 1) + 3? How many solutions does the equation 4 = 2 sin(3x−1)+3 have on the domain D and what are they? 2 ≤ ≤ x (b) If y = sin(t) on the domain − π π 2 , what is the domain D and range R of y = 8 sin( 2π 1.2 (t−0.3))+18. How many solutions does the equation 22 = 8 sin( 2π 1.2 (t − 0.3))+18 have on the domain D and what are they? 2 ≤ ≤ t (c) If y = sin(t) on the domain − π t 2 ≤ ≤ π 2 , what is the domain D and range R of y = 27 sin( 2π 366 (t − 80.5)) + 45. How many solutions does the equation 40 = 27 sin( 2π 366 (t − 80.5)) + 45 have on the domain D and what are they? (d) If y = cos(x) on the domain 0 π, what is the domain D and range R of y = 4 cos(2x + 1)−3? How many solutions does the equation −1 = 4 cos(2x + 1) − 3 have on the domain D and what are they? ≤ ≤ x (e) If y = tan(x) on the domain − π 2 < x < π 2 , what is the domain D and range R of y = 2 tan(−x + 5) + 13? How many solutions does the equation 100 = 2 tan(−x + 5) + 13 have on the domain D and what are they? Problem 20.11. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket filled with laughing gas. Ac- cording to her design, if the atmospheric pressure exerted on the rocket is less than 10 pounds/sq.in., the laughing gas chamber inside the rocket will explode. Tiff worked from a formula p = (14.7)e−h/10 pounds/sq.in. for the atmospheric pressure h miles above sea level. Assume that the rocket is launched at an angle of α above level ground at sea level with an initial speed of 1400 feet/sec. Also, assume the height (in feet) of the rocket at time t seconds is given by the equation y(t) = −16t2 + 1400 sin(α)t. (a) At what altitude will the rocket explode? (b) If the angle of launch is α = 12o, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? (c) If the angle of launch is α = 82o, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? (d) Find the largest launch angle α so that the rocket will not explode. Problem 20.12. Let’s make sure we can handle the symbolic and mechanical aspects of working with the inverse trigonometric functions: (a) Find four solutions of tan(2x2 + x −1) = 5. (b) Solve for x: tan−1(2x2 + x − 1) = 0.5 284 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix 285 286 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix A Useful Formulas Abbreviations inch = in feet = ft yard = yd mile = mi millimeter = mm centimeter = cm meter = m kilometer = km second = sec = s minute = min hour = hr year = yr ounce = oz pound = lb gram = g kilogram = kg quart = qt gallon = gal milliliter = ml liter = L Joule = J calorie = cal atmosphere = atm Coulomb = C radian = rad degree = deg = ◦ miles per hour = mi/hr = mph feet per second = ft/sec = ft/s meters per second = m/sec = m/s revolutions per minute = rev/min = RPM 287 288 APPENDIX A. USEFUL FORMULAS Conversion Factors Length Energy 1 in = 2.54 cm 1 ft = 0.3048 m 1 mi = 1.609344 km Volume 1 gal = 3.7854 L 1 qt = 0.946353 L 1 J = 1 kg m2/s2 1 cal = 4.184 J Mass 1 oz = 28.3495 g 1 lb = 0.453592 kg Formulas from Plane and Solid Geometry Rectangle Perimeter = 2ℓ + 2w Area = ℓw • • ℓ Triangle Perimeter = a + b + c Area = 1 2bh • • r w Rectanglular prism Surface Area = 2(ℓw + ℓh + wh) Volume = ℓhw • • h w ℓ a c h Right circular cylinder b Circle Perimeter = 2πr Area = πr2 • • Surface Area = 2πr2 + 2πrh Volume = πr2h • • r h Sphere Right circular cone 289 Surface Area = 4πr2 Volume = 4 3πr3 • • r Constants Surface Area = πr2 + πrs Volume = 1 3 πr2h • • r Avogadro’s number = N = 6.022142 1023 speed of light = c = 2.99792 density of water = 1 g/cm3 × × 108 m/s2 mass of earth = 5.9736 1024 kg × earth’s equatorial radius = 3,960 mi = 6.38 acceleration of gravity at earth’s surface = 32 ft/sec2 = 9.8 m/s2 × 106 m Algebra axay = ax+y (ax)y = axy a0 = 1 x = ax bx = ax−y a b ax ay n√a = a1/n (a + b)(c + d) = ac + ad + bc + bd (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 Quadratic Formula: If ax2 + bx + c = 0, then x = −b ± √b2 − 4ac 2a • • • • • • • • • • 290 • Completing the Square: ax2 + bx + c = a x + 2 − b2 4a b 2a + c APPENDIX A. USEFUL FORMULAS Trigonometry sin(−x) = − sin x cos(−x) = cos x sin cos sin cos = cos x − x = sin x +
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x = cos x + x = − sin x sin(π − x) = sin x • • • • • • • • • • • • • • • cos(π − x) = − cos x sin(π + x) = − sin x cos(π + x) = − cos x sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y sin2 x + cos2 x = 1 sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x Appendix B Answers Answer 1.1 (b) 150 ft/sec. (c) Gina. (d) 6300 hours. Answer 1.14 (a) N = 12 + 0.45(60 − x). (b) Reduced competition for resources (eg. water, nutrients, etc.). Answer 1.2 68.4444 km. Answer 1.3 r = 10.172 cm for lead; r = 16.433 cm for aluminum. Answer 1.4 The mass of the air is 9.740 million kg, more than the mass of the Tower. Answer 1.5 (a) 5.5 min/mi = 5:30 pace. (b) 14 2 Adrienne. 3 ft/sec. (c) × ≤ × ≤ ≥ (d) 1500 $56000 and (taxes) = 0.15 (c) (John’s Salary) (John’s Salary). ≤ 1800. (e) (cost red Porsche) > 3 Answer 1.6 (a) (John’s Salary) = $56000 and (taxes) = $0. (b) (John’s Salary) (John’s $56000 and (John’s taxes) Salary). > 0.28 (number of 120 students each year) (cost F-150 pickup). (f) 2 hours (weekly study time per credit hour) (number of happy math students) ≤ × > 5 (number of happy chemistry students). (happy math students) + (happy chemistry students) < 1 (number of cheerful biology students). (h) (Cady’s high score) - (Cady’s (i) low score) = 10%. 1 (0.9999) 2 × (total votes cast). (Cady’s final exam score) = 97%. (1.0001) (Tush votes) 1 2 (total votes cast) 3 hours (g) 2 2 × ≤ ≤ × ≤ × × Answer 1.7 Go for the 15 inch pie. Answer 1.8 (a) m90 = 151kg,m99 = 468kg,m99.9 = 1476kg. (b) v = 2.974 108 m sec . × Answer 1.9 1080 pizzas sold in 4 hours. Profit reaches $1000 at about 8 : 22 pm. Answer 1.10 Lee has 2.265 in2 more pie. Answer 1.15 Formula simplifies to r(1+x) 1+2x . Rates are about 0.67r, 0.6r, 0.54r and 0.51r at the indicated times. Rates decrease over time, but will never be less than 0.5r. Answer 1.16 (a) Initial time = 7am, Initial temperature = 44◦F, Final time = 10am, Final temperature = 50◦F, rate of change = 2◦F/hr. (b) 58◦F. (c) Initial time = 4.5pm, Initial temperature = 54◦F, Final time = 6.25pm, Final temperature = 26◦F, rate of change = -16◦F/hr. Answer 1.17 (a) t = 9. (b) a = 1/8. (c) x = −3a/4. (d) t > 3. (e) x+2 x(x+1) . Answer 2.1 (a) d = √2, ∆x = 1 = ∆y = 1. ∆x = −1, ∆y = −2. d = √10t2 + 2t + 1, ∆x = 3t, ∆y = 1 + t. (c) d = √34, ∆x = 5, ∆y = −3. (b) d = √5, (d) Answer 2.2 (a-c) True. (e) ∆x = a − s, ∆y = b − t. (f) ∆x = 0 means the points line on the same vertical line; ∆y = 0 means the points line on the same horizontal line. (d) ∆x = s − a, ∆y = t − b. Answer 2.3 Just after 12:29 PM that afternoon. Answer 2.4 (a) Erik= 6.818 mph, Ferry = 17.6 ft/sec. (b) Impose coordinates with Kingston the origin and units of miles on each axis; then Edmonds is located at (6,0) and Erik’s sailboat is at (3,2). The table rows have these entries: (0,0), (0.1,0), (1.4,0), (12t,0). (3,2), (3,1.9432), (3,1.2045), (3,2 − 6.818t) 3.606, 3.491, 2.003, (12t − 3)2 + (2 − 6.818t)2 (c) Use coordinates as in (b), then when the ferry reaches (3,0), Erik is at (3,0.296). (d) CG vessel does not catch the ferry before Edmonds. p Answer 1.11 About 5 × 106 times around the equator. Answer 2.5 (a) d(t) = (65.3)t (b) 227 minutes, 168.4 miles (c) t = 80.86 seconds. Answer 1.12 (a) Radii are about r = 0.2416, 2.317, 4.184 and 9.167 cm at the indicated times. (b) No. Answer 1.13 (a) 140 million gallons per week; about 7300 million gallons per year. (b) 100 yds 20 yds. 50 yds × × Answer 2.6 (a) Allyson’s coordinate position: (0 ft,20 ft). (b) After 2 Adrienne’s coordinate position: seconds there will still be slack in the bungee cord. (c) t 2.34. Use this time to find where Allyson and Adrienne are located. (d) t = 5.5 seconds. (−16 ft,0 ft). ≈ 291 292 APPENDIX B. ANSWERS Answer 2.7 Impose a coordinate system with origin at A and the shore the horizontal x axis; let x be the location on the shore where Brooke beaches. Equation 1 (6 − x) gives time to reach Kono’s. T (0) = 4 2 hr, T (6) = 3.9 hr. Neither time will be the minimum time. √25 + x2 + 1 4 Answer 2.8 (b) t = 2; t = 5 ), 2 . ant=( 41 ). (e) 1.5 feet. (f) Spider reaches (9,6) when t = 4; ant reaches (9,6) when t = 3. (g) spider speed is √5 ft/sec; ant speed is √8 ft/sec. (c) (3,3) (d) spider=( 7 3 , 4 3 , 8 3 3 Answer 2.9 49.92 mph (exactly). Answer 2.10 141.46 miles. They are 300 miles apart at time 0.826 hr = 49.6 minutes. Answer 2.11 (a) Final answer is correct, but second (b) Final answer should be 4xy; key equality is wrong. fact is that (x + y)2 = x2 + 2xy + y2, etc. (c) Answer and steps correct. Answer 2.12 (a) x = 1+β2 α2 ±q (b) x = αβ α+β (c) x = β αβ+1 Answer 2.13 (a) 5t2 + 6t + 5 (b) 2t2 + 4t (c) √5t2 + 4t + 4 2 t2 −1 (d) Answer 3.1 (a) (x+3)2+(y−4)2 = 9 (b) (x−3)2+ 1 16 (c) Draw a vertical and horizontal line through (1, 1). On the vertical line, a circle of radius 2 will have a center at either (1, 3) or (1, −1). Likewise, the horizontal line will have circles at either (−1, 1) or (3, 1). y + 11 3 = 2 Eqn 1. 2. 3. 4. C(h,k) (3, 1) (1, 3) (−1, 1) (1, −1) (x − h)2 + (y − k)2 = r2 (x − 3)2 + (y − 1)2 = 4 (x − 1)2 + (y − 3)2 = 4 (x + 1)2 + (y − 1)2 = 4 (x − 1)2 + (y + 1)2 = 4 (d) (1,1), (1, − 3), (1 + √3,0), (0, − 1 − √3) Answer 3.2 (a) center (3,-1), radius 2√3 (b) center (-2,-3), 203/12 (d) center radius 2 (c) center (-1/6,5/3), radius (3/4, -1/2), radius √3 p Answer 3.3 (a) Wet 25.4154041 minutes. in 24.0 minutes. (b) Wet in Answer 3.4 (a) Imposing with x-axis along ground and yaxis along tower, wheel modeled by x2 + (y − 62)2 = 602. (b) 46.43 ft. to right of tower. (c) (−24,7) and (−24,117). Answer 3.5 (a) Impose a coordinate system so that the tractor is at the origin at t = 0 seconds. With this coordinate system, the south edge of the sidewalk is modeled by ys = 100; the north edge is modeled by, yn = 110. (b) t = 32 minutes. (c) t = 52 minutes. (d) 20 minutes. Answer 3.6 (a)] The equation for eastward travel from Kingston is y = 8. Southward travel along x = −1. (b) Boundary: (x+2)2+(y−10)2 = 9; Interior:(x+2)2+(y−10)2 < 9; Exterior:(x + 2)2 + (y − 10)2 > 9. (d) Plug x = −1 into circle equation to find exit point. Use this to find when the ferry exits the radar zone. Be careful to reference all times relative to when the ferry departed Kingston. (e) 20.32 minutes. (c) 3.82 minutes. Answer 3.7 (a) 6.92 seconds. (b) 7.67 seconds. (c) 38.16 seconds. (d) 7332 ft2 < area < 7632 ft2. Answer 3.8 (a) x = 11/5. (b) no solutions. (c) (−2, − 2) and (−2,4). (d) (−1 2/5,3). ± p Answer 4.1 (a) y = − 5 3 (c) y = −2x − 2 (d) y = 11 (e) m = 3 y = 40x − 14 (g) y = − 3 (x − 1) − 1 (b) y = 40(x + 1) − 2 5 , y = 3 (x − 1) + 1 (f) 5 4 (h Answer 4.2 (a)( 2 1+2α , 1+4α 1+2α ). (b) α = −2/5. (c) α = −1/4. Answer 4.3 (a) area = 25 13 (b) area = − b 2m (c) m = − 1 4 2 Answer 4.4 In some cases, answers in this problem are unique. In other cases, the answers are not unique. Here is a possible solution set: Eqn Slope y-int Pt on line Pt on line y = 2x + 1 2 y = − 11 4 x + 17 4 − 11 4 y = −2x + 1 −,000 y = 0 1 2 0 0 1 17 4 1 1 (0, 1) (− 1 2 , 0) (3, − 4) (−1,7) (0, 1) ( 1 2 , 0) (0, 1) (−2, 0) 1,000 (0, 1,000) (0, 1,000) 0 (0, 0) (1, 0) x = 3 Undef None (3, 3) (3, −2) y = x − 14 1 −14 (5, − 9) (0, −14) Answer 4.5 (a) y = 6,850(x − 1970) + 38,000 (b) y = 8,000(x − 1970) + 8,400 (d) Tabularize your answer: Year 1983 1998 Seattle $127,050 $229,800 Port Townsend $112,400 $232,400 (e) 1995.74, $214,313 (f) 1982.70, Sea = $124,965 (g) 2008.78, Sea = $303,661 (h) 1972.32, Sea = $53,892 (i) No. Answer 4.6 For these answers, we take the hole to be the origin of our coordinate system. Then, the ball’s line of travel is y = (0.667)(x − 35). (a) (−13.46, −32.31) (b) 3.19 seconds (c) 5.82 seconds (d) (10.78, −16.17) at about t = 6.1 sec. (a) Allyson at (0,70); Adrian at (−44,0). Answer 4.7 Bungee is 83.1401 ft. long. (b) Occurs at time t = 7.7546 seconds and Allyson is at (0,77.546). Allyson’s final location is 77.546 ft. from her starting point. Answer 4.8 The lines of tangency have equations y = The non-visible portion of the y-axis is (x − 12). 2 √5 ± − 24√5 5 ≤ y ≤ 24√5 5 . Answer 4.9 C = 5 9 temperature is −9.4◦F. (F − 32) and F = 9 5 C + 32. In Oslo, the Answer 4.10 At closest 26.22471828 miles from Paris. the point, she will be Answer 4.11 Impose coordinates with Angela’s initial location as the origin. Angela is closest to Mary at (18.8356,41.4383); this takes approximately 3.8 seconds. Answer 4.12 (a) Impose coordinates with sprinkler initial location as the origin (0,0). Line equation becomes y = − 1 (b) Sprinkler is located at (0,82.5) at time t = 33 minutes. Circular boundary of watered zone hits southern edge of sidewalk at the points (−6.708,101.3416) and (13.4386,97.3123). (c) y = − 1 5 x + 100. 5 x + 110.198. Answer 4.13 (a) With the origin set at the statue, x = 30 − 3.123475t, y = 2.49878t. (b) The distance to the statue is √16t2 − 187.4085t + 900 feet. (c) Margot will be 28 feet from the statue after 0.655672 seconds, and after 11.057360 seconds (assuming she continues past the point due north of the statue). Answer 4.14 18.923076923 seconds Answer 4.15 (a) (−1 −0.463325, 0.863325; or the exact answer would be (1 √11)/5. (c) t = 0.2. (d) No real number solutions. 7/3)/2 = (−3 √21)/6. p ± ± (b) t = 293 Answer 5.8 (a) − 3 , 2 ,− 5 − 3)( ). ( ♥ ). (c) Always 4π2. (b) 0, 20, 6x + 2x2, Answer 5.9 Use the vertical line test. For example, (a) is not a function, by the vertical line test; you can split it into two function graphs by slicing the ellipse symmetrically into upper and lower halves. On the other hand, (o) is a function, by the vertical line test; etc. Answer 5.10 (a) x = 5. 242−√56400 2 (b) x = 4, 8. (c) x = 36. (d) Answer 6.1 (a) 0, 2, 3. (b) x = Intersect at (4,4) and (− 4 3 , 4 3 ± ). Area is 16 3 . 4, x = 0, no solution. (c) Answer 6.2 (a) | − 0.5x − 1| = −0.5x − 1 0.5x + 1 −2 if x ≤ if x > −2 y 3 2.5 2 1.5 1 0.5 -4 -2 2 x 4 ± Answer 6.3 (a) x = 4 and x = − 22 equation has no solutions. 3 (b) x = 0.75 (c) the Answer 4.16 (a) There are four answers: x = 2 − √2 (b) y
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= 6 + 2√5. and x = ±p 2 + √2 ±p Answer 5.1 (a) −2 + h + 2x. (b) 2. (c) h + 2x. (d) −h − 2x. (e) −π(h + 2x) (f) . 1 √h+x−1+√x−1 Answer 6.4 (a) x = −7 and x = 3 (b) a = 3 (c) x = 8/3 Answer 6.5 For 0 x ≤ ≤ 6, the area is 1 6 x(x + 18). Answer 6.6 (a) The rule is Answer 5.2 g(x) = 9 v(5) = 25, minimum v(x) = 20, maximum v(x) = 40. 5 x + 24, f(x) = 4 5 x + 4, v(x) = x + 20, y = 2x −x + 30 if 0 if 10 x ≤ x ≤ ≤ ≤ 10 30 Answer 5.3 For example, in (a), suppose Dave has constant speed v ft/min. Then the function s = d(t) = vt will compute the distance Dave travels in t minutes. The graph would be a line with s-intercept 0 and slope v; the domain would be 0 2400 t v ; etc. ≤ ≤ Answer 5.4 Answer 5.5 Several possible answers for each one. Answer 5.6 (a) x-intercepts= 1 6 (b) ( 1 (3 − √93),5) and ( 1 6 6 No, Yes. (e) ( √33). y-intercept = −2. (3 + √93),5). (c) None. (d) Yes, No, 1 + √2 + 3(1 + √2)). 1 + √2, − 2 − 3 (3 ± p p (4 − x)2 + 1. (c) The Answer 5.7 (b) f(x) = 30000x + 50000 (0., 206155), (0.5, 197003), table of values (x,f(x)) will be: (1, 188114), (1.5, 179629), (2,171803), (2.5, 165139), (3., 160711), (3.5, 160902), (4,170000). Minimal cost occurs for some 2.5 < x < 3.5; the exact answer is x = 13 4 , but we cannot solve this in our class since it requires the tools of Calculus. p and the range is 0 y 20. ≤ (b) The rule for the area function a(x) is ≤ a(x) = x2 2 x2 + 30x − 150 − 1 (c) x = 12.6795 inches. Answer 6.7 (a) if 0 if 10 x ≤ x ≤ ≤ ≤ 10 30 j(t) = 0 62.22(t + 1) if t < −1 if −1 t ≤ 3.5 ≤ s(t) = 280 280 − 70t if t < 0 t if 0 ≤ 4 ≤ (b) (c) 280 280 − 62.22(t + 1) 217.78 − 132.22t 132.22t − 217.78 70t d(t) = if t < −1 if −1 if 0 ≤ if 1.6471 if 3.5 ≤ t < 0 ≤ t < 1.6471 t < 3.5 ≤ t < 4 294 APPENDIX B. ANSWERS Answer 6.8 The distance to his starting point t seconds after he starts is Answer 7.2 (a) y = − 2 (c) y = −0.5x2 + 3.5x − 4. (d) No solution. 3 x2 + 5 3 x. (b) y = 1.125x2 −1.5x−1.625. 10t 2502 + (12(t − 25))2 4002 + (250 − 9(t − 175 3 ))2 d(t) = p q Answer 6.9 25, t if 0 if 25 if 175 ≤ ≤ 3 ≤ ≤ t ≤ t ≤ 175 3 , 205 3 . y = d(t) = 18t 902 + 182(t − 5)2 902 + (90 − 18(t − 10))2 90 − 18(t − 15) p p 5 10 if 0 if 5 if 10 if 15 x ≤ x ≤ x ≤ x ≤ ≤ ≤ ≤ ≤ 15 20 Here is the graph of d(t) ft 140 120 100 80 60 40 20 5 10 15 20 t sec Answer 6.10 (a) v(x) = 2x3 − 70x2 + 500x; degree 3. (b) a(x) = 1000 − 50x − 2x2, degree 2. To get 600 sq. in. dimensions are: 6.3746 18.6254. 7.2508 × × (b) Impose coordinates with x Answer 6.11 (a) 2 hours. axis the bottom of the ditch and the y axis the pictured centerline. 20 100 − (x + 40)2 100 − (x + 20)2 0 100 − (x − 20)2 100 − (x − 40)2 20 p p p p y = 10 + 10 − 10 − 10 + if x ≤ if −40 if −30 if −20 if 20 if 30 if 40 −40 30 −20 20 ≤ ≤ ≤ 30 40 (c) 2(40 − √91) feet. (d) 42 ft. wide: 0.3008 minutes. 50 ft. wide: 8.038 minutes. 73 ft. wide: 116.205 minutes. Answer 6.12 (a) 0 x and 2 ≤ 4. Decreasing: −2 ≤ 6. (b) Increasing: −6 x 2 and 4 y x ≤ ≤ ≤ ≤ (c) ≤ −2 x 6. ≤ ≤ ≤ y = 4 + 4 − 6. (e) 2 (d) 2 y ≤ ≤ 2x + 12 4 − (x + 2)2 4 − (x − 2)2 −2x + 12 p p y ≤ ≤ 4. if −6 if −4 if 0 if 4 ≤ ≤ ≤ ≤ x x − Answer 6.13 (a) 0. (b) a = 13 3. Answer 7.1 (a) 2(x − 4)2 + 9, Vertex: (4, 9), Axis: x = 4. (b) 3(x − 5/2)2 − 383/4, Vertex: (5/2, -383/4), Axis: x = 5/2. (c) (x − 3/14)2 + 2539/196, Vertex: (3/14, 2539/196), Axis: x = 3/14. (e) (1/100)(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. (d)2(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. Answer 7.3 (a) maximum value is 22; minimum value is 1.75 (b) maximum value is 32; minimum value is 2 (c) maximum value is -6; minimum value is -46 Answer 7.4 In the first case, d = 0 or d = 12. In the second case, d = . 2 √13 ± Answer 7.5 (a) Multipart function: s(t) = − 1 114.031, s(t) = 0 for t for 0 Sell at time t = 50 days; $2500. 160 t2 + 5 8 t + 10 114.031.; $23,125. (b) ≥ ≤ ≤ t Answer 7.6 The parabola has x intercepts at −1, 3 and y intercept at -3. The vertex of the parabola is (1, − 4). |x2 − 2x − 3| = x2 − 2x − 3 −x2 + 2x + 3 x2 − 2x − 3 ≤ −1 if x if −1 < x < 3 3 if x ≥ Answer 7.7 (a) 100 ft. (d) When x = 54.81 ft. or x = 570.19 ft.; i.e. at (54.81,39.04) and (570.19, − 64.04). (c) (625, − 125). (b) 156.25 ft. Answer 7.8 (a) No, since f(1) = 1 (b) The points (1,1 + 2b) and (−(1 + b),1 + 2b). (c) Only the point (a,1 − a2). (d) The points (−2.2701, − 7.1168) and (2.9368,1.5612). = 2. Answer 7.9 She should have 225 trees in the orchard. Answer 7.10 She should charge $7.72 to make the most money. Answer 7.11 The radius of the circular part should be 24 4 + π ≈ 3.36059492 feet. The short side of the rectangular part should also be equal to this. Answer 7.12 The enclosure should be 50 meters by 75 meters. Answer 7.13 The maximum possible area is 4285.71 m2. Answer 7.14 To achieve the minimal combined area, cut the wire so that the pieces have lengths 26.394 in. and 33.606 in.; bend the 26.394 inch piece into a circle. Answer 7.15 (a) y = 1 3.8681948. 2 x − 4 (b) y = 4 + 1 6 x (c) t = 1350/349 ≈ Answer 7.16 (a) The distance between Sven and Rudyard 41t2 − 2400t + 36900 feet, where t is the number of is seconds after they start moving. (b) They will be closest together after they have been moving for 29.268292 seconds, and will be 42.166916 feet apart at that time. p Answer 7.17 Michael M(t) = (5.547t,8.32t), Tina T (t) = (400 − 8.944t,50 + 4.472t). Study the distance SQUARED from M(t) to T (t). Michael and Tina will be closest when t = 26.641 sec; they are 54.3 ft at that instant. 6 Answer 7.18 −2α2 x = ± √4α4 − 4α 2α α = −x2 ± √x4 − 8x 4x Answer 7.19 (d) There are two possible values for α: If α = 8 + 2√17, then the unique solution of the equation will be x = −4 − √17. If α = 8 − 2√17, then the unique solution of the equation will be x = −4 + √17. 295 Answer 8.5 (a) f(s) = s 60 ; C(f(s)) = 70s2 10(602) + s2 C(f(s)) computes mph when you input seconds s. g(h) = 60h; (b) C(g(h)) = 252000h2 10 + 3600h2 C(g(h)) computes mph when you input hours h. (c) v(s) = 22 15 s; Answer 7.20 (a) t = 1 2 (2 ± √2√s − 1). (b) x = −1 √y − 2. ± v(C(m)) = 308m2 30 + 3m2 Answer 8.1 (a) h(f(t)) = −t + 1 t − 1 if t if 1 1 t ≤ ≤ h(g(t)) = −t − 1 t + 1 if t ≤ if −1 −1 t ≤ f(h(t)) = −t − 1 t − 1 g(h(t)) = t − 1 −t − 1 if t if 0 0 t ≤ ≤ if t if 0 0 t ≤ ≤ (b) (c) v(C(m)) computes ft/sec when you input minutes m. Answer 8.6 (a) f(g(x)) = (x+3)2, f(f(x)) = x4, g(f(x)) = x2 + 3. (b) f(g(x)) = 1 , f(f(x)) = x, g(f(x)) = 1 . (c) f(g(x)) = x, √x √x (d) f(g(x)) = 6(x − 4)2 + 5, f(f(x)) = 81x + 20, g(f(x)) = x. f(f(x)) = 6(6x2 +5)2 +5, g(f(x)) = 6x2 +1. (e) f(g(x)) = 8x+21, f(f(x)) = 4(4x3 − 3)3 − 3, g(f(x)) = 2x. (f) f(g(x)) = 2x3 + 1, f(f(x)) = 4x+3, g(f(x)) = (2x+1)3. (g) f(g(x)) = 3, f(f(x)) = 3, g(f(x)) = 43. (h) f(g(x)) = −4, f(f(x)) = −4, g(f(x)) = 0. Answer 8.7 −5/2 x ≤ ≤ −1/2 Answer 8.8 (a) 7 (d) −9 y x 7. (e) B = 5, C = 39 2 ≤ ≤ 6. (b) −3 y 5. (cf) A = 1 8 . ≤ ≤ 6. x ≤ ≤ Answer 8.9 (a) −1 (x−1)(x+h−1) , set h = 0 to get 4 + 4h + 8x, set h = 0 to get 4 + 8x. (c) set h = 0 to get −x √25−x2 . −1 (x−1)2 (b) √25−x2+√25−(x+h)2 , −h−2x if t ≤ if −1 if 0 if 1 ≤ ≤ −1 t ≤ 1 ≤ ≤ t t h(h(t) − 1) = −t − 1 t + 1 −t + 1 t − 1 y 3 2.5 2 1.5 1 0.5 0 Answer 9.1 (a) domain of f ={x|x (b) f−1(y) = 2+4y 3y . = 4 3 }; range=f= {y|y = 0}. Answer 9.2 Answer 9.3 (c) f−1(y) = 3+√17+8y − 17 8 }. 4 on the domain {y|y ≥ Answer 9.4 Only (B) is one-to-one on the entire domain. Answer 9.5 Answer 9.6 (a) h = f(x) = −2x2 + 124x. 31 − 1 2 √3844 − 2h. (c) x = g(h) = -3 -2 -1 1 2 x 3 Answer 9.7 (a) (1) -0.5 -1 Answer 8.2 (a) y = f(g(x)), if f(x) = x5, g(x) = x − 11. (b) y = f(g(x)), if f(x) = 3√x, g(x) = 1 + x2. (c) y = f(g(x)), if f(x) = 2x5 − 5x2 + (1/2)x + 11, g(x) = x − 3. (d) y = f(g(x)), if f(x) = (1/x), g(x) = x2 + 3. (e) y = f(g(x)), if f(x) = √x, if f(x) = 2 − √5 − x2, g(x) = √x + 1. g(x) = 3x − 1. (f) y = f(g(x)), y x Answer 8.4 (c) x = −1 and x = 2 (2) x = f−1(y) = y+2 bers. (3) 3 has domain and range all real num- 6 6 296 APPENDIX B. ANSWERS x Answer 11.4 (c) The 500cosh( x 500 feet. = ) − 440.536 and the minimum height is 59.46 is modeled by y curve y Answer 12.1 (a) 0.6826; 2.3979; 3.3030; 3.3219;0.3010. (c) x = log10 y; x = log10(3y); (b) 3.555; 19.8; 0.0729. x = (1/3) log10(y). Answer 12.2 (a) I(d) = I ◦ (0.94727)d. (b) 85 meters. Answer 12.3 (a) y = 13e1.09861t. (b) y = 2e−2.0794t. Answer 12.4 (a) 9.9 years. (b) 34.66%. ) = 3( y+2 (4) f(f−1(y)) = f( y+2 3 3 f−1(f(x)) = f−1(3x − 2) = 3x−2+; = x. 3 t 30 hours; Range: 0 Answer 9.8 (a) w = f(t) = 2 q w 0 (c) t = f−1(w) = 30 − 6 )2. Domain: 0 Range: 0 ≤ 25 − ( w 2 q 30 hours. 25 − ( t 6 ≤ ≤ ≤ t − 5)2. Domain: 10 ft. (b) t = 6 hours. ≤ ≤ w ≤ ≤ 10 ft.; Answer 12.5 (a) 8.8 cm. Maximum possible length of the halibut. (b) 135 cm. (c) 4.84 yrs. (d) Answer 9.9 (a) 2 nanoseconds. (f) domain φ={0 range φ={0 6}. y ≤ ≤ t ≤ ≤ 10}; Answer 12.6 (a) C(t) = 1.03526t. Answer 10.1 (a) 31.5443; (b) 355.1134; (c) 36.4622; (d) 0.0616; (e) 51,168; (f) 0.009794; Answer 10.2 (a) y = 3( 1 2 )x (b) y = ( 1 2 )x. (d) y = 1 27 ( 1 √3 )x. Answer 12.7 (c) v(x) = 55000(1.0315)x. $200,000 during 1993. Valued at Answer 12.8 (a) x = 2.7606. (b) 1.392. x = e8. (e) x = 11.513. x = −0.7658 + 2kπ or x = 3.9074 + 2kπ. (f)-0.61. (d) (c) 0.3552. (g) sin(x) = −0.6931; Answer 10.3 (a) 1.64 formulas are identical. × 106 cells. (b) True. (c) The two Answer 12.9 (d) 8.617 weeks. Answer 10.4 (a) 261.31 Hz. (b) 440 Hz. (c) 27.5 Hz. (d) 16.35 Hz. Answer 12.10 31.699250014 days Answer 10.5 (a) 29 = 512. (b) 2n−1. (c) 263 = 9.2 9.2 1015 meters. × Answer 10.6 (a) M(p) is the higher curve. (d) 1018. (d) × Answer 12.11 (a) 2015 (b) 2048. Answer 12.12 (a) 46.701735 years (b) 137.3113631016 years after 1980 fraction Answer 13.1 The graph is given below: 0.8 0.6 0.4 0.2 20 40 60 80 p 100 Answer 11.1 (a) w(t) = $1.113(1.046408)t . (b) $1.11. (c) It is below; should be $5.70 by the model. Answer 11.2 (a) p(x) = 860(1.070674)x, l(x) = 70x + 860. (b) p(10) = 1702, l(10) = 1560. Answer 11.3 If we use 1989 and 2000 in a(t), we get two data points and a corresponding
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exponential model: E(t) = 15.918(1.243301)( t − 1980). The exponential model grows faster than the cubic model and eventually exceeds a(t). y 2 1.5 1 0.5 -2 -1.5 -1 -0.5 0.5 1 1.5 x 2 -0.5 -1 -1.5 -2 Answer 13.2 (a) 6 4 2 0 -3 -2 -1 0 1 2 -2 -4 -6 1.5 1 0.5 297 y 30 25 20 15 10 5 -2 -1 1 2 3 x 4 Answer 13.5 (a) y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 -1 -0.5 0 0 0.5 1 1.5 2 (b) No. (c) -0.5 -1 -1.5 10 5 0 -12 -10 -8 -6 -4 -2 0 2 4 -5 -10 0 if x ≤ if −2 if 0 if x ≤ ≥ −2x + 2 0 y = f(−x) = y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 y = −f(x) -1 0 −2x − 2 x − 2 0 y 1 if x ≤ if −1 if 0 if x ≤ ≥ −b) (c) Answer 13.3 − 1 3 ≤ 0. x ≤ Answer 13.4 (a5) (1) horizontally dilate (compress) by a factor of 2; (2) horizontal shift right by 1 2 ; (3) vertical dilate (expand) by a factor of 3; vertical shift up by 5. -1 -2 -3 -4 -4 -3 -2 -1 1 2 3 x 4 f(x) = −6x + 8 6x + 2 if x < 1 2 1 if x 2 ≥ (d) Here are the graphs of y = 2f(x) and y = 1 tively: 2 f(x), respec- 298 APPENDIX B. ANSWERS y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 -1 y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 -1 (e) Here are the graphs of y = f(2x) and y = f( 1 tively: 2 x), respec- y 4 3 2 1 Answer 13.10 y = 1 (2x) is obtained by vertically dilating 3 y = 2x; it is vertically compressed. y = 2x/3 is a horizontal dilation of y = 2x; it is horizontally stretched. Answer 14.1 (a) domain={x|x = 2}; zero at x = 0; horizontal asymptote y = 2; vertical asymptote x = 1; graph below: = 1}; range={y|y -4 -2 y 15 10 5 -5 -10 -15 2 4 x Answer 14.2 (a) y = 0.2x−10.4 ft. (c) 10 ft. x−10 . (b) x = 9.143 ft.; x = 9.916 Answer 14.3 (a) m(t) = 35t + 200. 1987. (d) r(t) = 35t+200 30t−50 . (e) 7 6 . (b) k(t) = 30t − 50. (c) Answer 14.4 (a) y = 0.03x2 −1.1x+14 (b) w = 0.03x−1.1+ 14 x (c) 70 or 6 2 3 Answer 14.5 f(x) = asymptote is y = 41 11 . 41 11 x + 35 x + 65 11 11 = 41x + 35 11x + 65 . The horizontal -4 -3 -2 -1 1 2 3 x 4 Answer 14.6 f(x) = 6x + 10 x + 1 -1 y 4 3 2 1 -4 -2 2 4 x -1 (f) c = 11 2 . (g) c = 5 2 . (h) c = d = 1 3 . Answer 13.6 (a) a(x) = 2(6 − x)√3x − 9 with domain 3 x 8√3 + 1 (when x = 1 0 ≤ 6. (b) Max of a(x) is 4√3 and max of 2a(3x + 3) + 1 is (c) 1 + 36(1 − x)√x on the domain 1. The range is all y values between 1 and 8√3 + 1. 3 ). ≤ x ≤ ≤ Answer 14.7 You should study for 11.25 hours. Answer 14.8 (a) k = 400 (b) I(t) = t = 1.5 (d) t = 1.05 and 1.95. 400 484t2 −1452t+1189 (c) Answer 14.9 Answer 14.10 (a)y = f(x) = 400x+20000 (d)x = f−1(y) = 200y−20000 ≤ 388.46}, Range f−1 = {0 5000}. The inverse function x ≤ takes the number of customers per day as an input value and gives the amount the shop spent on advertising as an output value. . Domain f−1 = {100 (b) x = $400. 400−y x+200 ≤ ≤ y . Answer 15.1 (a) 13o24 ′ or 0.233874 rads. (b) 1.0788 degs or .01882 rads. (c) 5.7296 degs or 5o43 ′46.5". Answer 15.2 (a) 6080 ft. (b) 29.95 mph. (c) 15.63 knots. Answer 15.3 (a) 430 sq. in. (b) 2.56◦. (c) 84.47 in. (d) 23.04 in. (e) 7.29 in. Answer 13.7 (c) Horizontally shift the graph of y = x2 to the right 4 units. Answer 15.4 (a) 1413.7 sq. ft. (c) 4.244 sec. Answer 13.8 (a) y = 7|2x + 4| + 2. Answer 15.5 2164.208272472 miles. Answer 13.9 (a1) f(2x) = 4x. f(2x − 1) = 1 2 4x. Answer 15.6 (a) 2.147 hrs. 6236 miles. (d) 13760 miles. (b) 1103 mph. (c) 12.47 hrs., 6 6 Answer 15.7 (b) 84.47 sq. in. (c) 181.5 sq. in. Answer 15.8 0.685078 miles. Answer 15.9 Middle picture: shaded area= 12.537 sq. in. Answer 16.1 (a) 10π/3 rad = 10.47 radians. (b) 4/2π rev * 1 hour/rev = .64 hours = 38.2 minutes. (c) Using (2.2.2), (21)(10π/3) = 219.91 meters. Answer 16.2 194 RPM. 299 Answer 17.12 Top right scenario: (b)θ(t) = 1.2+ 4πt (d) b(1) = (−1.710,1.037). (1.252, − 1.560). b(22) = (1.753,0.962). 9 (c) b(t) = (2 cos(1.2+ 4πt 9 b(0) = (0.725,1.864). ◦ = 1.2 rad. ),2 sin(1.2+ 4πt )). 9 b(3) = (a) θ Answer 18.1 (b) sin2(x) = 1 2 (1 − cos(2x)). 2 1.5 1 0.5 Answer 16.3 (a) ω = 11π v = 2.094 in/sec, ω = 2.5 RPM. 30 rad/sec, v = 121π 15 ft/sec. (b) -6 -4 -2 2 4 6 Answer 16.4 (b) 700 ft. (c) 70 sec. (d) 15000 sq. ft. Answer 16.5 (a) r = 233.427 ft. (b) θ = 0.06283 rad. (c) 2π/5 rad counterclockwise from P. (c) Answer 16.6 (a) 40π ft/sec; 85.68 mph. (b) 400 RPM. (c) 60 RPM; 32π ft/sec. (e) 0.7 sec; 1.4π rad. (d) 1.445 rad = 82.8◦; 28.9 ft. Answer 16.7 4.4 inches. Answer 16.8 r = 2.45 inches. -0.5 -1 -1.5 -2 2 1.5 1 0.5 -6 -4 -2 2 4 6 -0.5 -1 -1.5 -2 2 1.5 1 0.5 Answer 17.1 (a) If you impose coordinates with the center of the wheel at (0,237.427), then ground level coin(a) T (t) = (x(t),y(t)), where x(t) = cides with the x-axis. 233.427 cos( 2π 5 t − 0.06283) + 237.427. (c) First find the slope of a radial line from the wheel center out to Tiff’s launch point. 5 t − 0.06283) and y(t) = 233.427 sin( 2π (b) T (6) = (85.93,454.45). Answer 18.2 (a) Answer 17.2 (a) y = 0.2309(x + 1) + 2 ± Answer 17.3 290 ft. Answer 17.4 (a) 21.14892) (c) (19.07064, -10.89497) (-21.91218, -1.498834) (b) (-5.92564, Answer 17.5 (a) (−1.674,23.942). (d) (23.882,2.375). (19.9,13.42). (b) (22.55,8.21). (c) Answer 17.6 101.496936 feet above the ground. (b) Answer 17.7 The dam is 383 feet high. Answer 17.8 108 ft. Answer 17.9 With the center of the track at the origin, and the northernmost point on the positive yaxis, Charlie’s location after one minue of running is (59.84016,4.37666). Answer 17.10 105.2718216 feet Answer 17.11 (a) 204.74 ft. (b) no. (c) -6 -4 -2 2 4 6 -0.5 -1 -1.5 -2 2 1.5 1 0.5 -6 -4 -2 2 4 6 -0.5 -1 -1.5 -2 300 x 15 10 5 0 0 -6 -4 -2 2 4 6 Answer 18.3 (a) 7/25 or -7/25. (b) -0.6. (c) √40 7 . ± Answer 18.4 In the first case, one period Answer 18.5 π 4 + 2kπ and 5π 4 + 2kπ, k = 0, 1, ± ± 2, ± 3, . . . . Answer 19.1 (a) 1, π, π 2 , 1. (b) 6, 2, 0, -1. Answer 19.2 (d) h(t) = 8 sin( 2π 1.2 (t − 0.3)) + 18. Answer 19.3 (a) b(t) = 0.6 sin( 2π 100 (t − 30)) + 1.2. APPENDIX B. ANSWERS 2 4 6 8 t y 2 1 -1 -2 Answer 19.7 (a)2 volts. (b) Never zero since 2x is positive for all x. (c)p(t) = 3 sin( 2π (t − 3 ) + 1, so A = 3, D = 1, B = 5 2/5 16. (e)t = 0.65635 + k(0.4) and V(t) 2/5, C = 3/5. (d) 0.25 t = 0.74365 + k(0.4), k = 0, 2, . . . are ALL solutions. 1, ± Four of these lie in the domain 0 1. (f) Maxima have coordinates (0.3 + k(0.4),16) and minima have coordinates (0.1 + k(0.4),0.25), where k = 0, 2, . . . . (g) If we restrict V(t) to 0.5 1, 0.7, the inverse function has rule = arcsin( ln(y)−ln(2) 3 ln(2) 5π ) + 3π . If we restrict V(t) to 0.5+k(0.4) function has rule: t ≤ ≤ 0.7+k(0.4), the inverse t = k(0.4) + arcsin( ln(y)−ln(2) 3 ln(2) 5π ) + 3π . In particular, if we restrict V(t) to 0.1 function has rule: t ≤ ≤ 0.3, the inverse t = −0.4 + arcsin( ln(y)−ln(2) 3 ln(2) 5π ) + 3π . If we restrict V(t) to 0.7 rule: t 0.9, the inverse function has ≤ ≤ − arcsin( ln(y)−ln(2) ) + 4π . 3 ln(2) 5π t = Answer 19.4 (a) h(t) = 5 sin( π 6 hours after midnight. (b) 7.5 ft. above low tide. (t−10)+5, where t indicates If we restrict V(t) to 0.7+k(0.4) function has rule: t ≤ ≤ 0.9+k(0.4), the inverse Answer 19.5 (a) A = 25, B = 5 seconds, C = 1.75, D = 28. (b) t=1.75 and 4.25 seconds t = k(0.4) + − arcsin( ln(y)−ln(2) 3 ln(2) 5π ) + 4π . Answer 19.6 First scenario: x(t) = 2 sin( 2π 4.5 where Cx = −(1.2 + π )( 9 4π 2 Cy = −(1.2)( 9 ). Plots are below: 4π ); y(t) = 2 sin( 2π 4.5 (t − Cx)), (t − Cy)), where x 2 1 -1 -2 2 4 6 8 t In particular, if we restrict V(t) to 0.3 function has rule: t ≤ ≤ 0.5, the inverse t = −0.4 + − arcsin( ln(y)−ln(2) 3 ln(2) 5π ) + 4π . Answer 19.8 (c) On domain t 0, ≥ B = B(t) = 2 cos(6πt) + 36 − 4(sin(6πt))2. q (e) 0.1023, 0.2310. Answer 20.1 (a1) 0, 1.5708, −1.5708, 1.0472, 0.7168, −0.2762, not defined. Answer 20.2 (a) Principal solution: x = 1.9106, symmetry solution: x = 4.3726; graph below with these two solutions graphically indicated: y 2 1.5 1 0.5 -0.5 -1 -1.5 -2 -10 -5 5 10 x Answer 20.3 (a) 9.39, 13.63, 11.09. (b) Feb. 3, Nov. 4. Answer 20.4 (a) A = 15, D = 415, B = 10, C = 15/2. Note that C = 10k + 15/2, k = 0, 2, . . . are also all valid ± (b) maximum temperature= choices for the phase shift. 430oF. (c) minimum temperature= 400oF. (d) 12.1635 minutes. (e) 14.6456 minutes. (f) 6.80907 minutes. ± 1, Answer 20.5 The cake should be in the oven for 67.572141357 minutes. Answer 20.6 6.42529 hours Answer 20.7 8.9286 hours of dry time each day. this: to use over and over Answer 20.8 The key fact is = (100 cos(0.025t),100 sin(0.025t)); T (t)= T’s location after t seconds = (100 cos(0.03t + π),100 sin(0.03t + π)). location after M(t)=M’s seconds t 301 (b) α = Answer 20.9 (a)α = arcsin[3960/(3960 + t)]. 6.696degs. The interior angle is 166.608 degs and so one satellite covers 46% of the circumference. Thus you need 3 (not two-point-something) satellites to cover the earth’s (c)α = 52.976deg. The interior angle is circumference. 74.047 degs and so one satellite covers 20.57% of the circumference. Thus you need 5 satellites to cover the earth’s circumference. (d) Get an equation for the interior angle in terms of t. Solve 2(90 − arcsin[3960/(3960 + t)]) = 20% of 360 degs = 72 degs. You’ll get t = 934.83 miles. Answer 20.10 (a) Domain: 1 3 5. One solution ; Range: 6 ≤ ≤ 18 ; graph is below: 8 6 4 2 -2 -1.5 -1 -0.5 0.5 1 1.5 x 2 -2 Answer 20.11 (a) 3.852624 miles (a) 14.34lb/in2; no explosion (b) 8.32lb/in2; explosion (c) 54.6 degrees Answer 20.12 (a) Many possible answers; for example: −1.9293, −1.3677, 0.8677, 1.4293. 302 APPENDIX B. ANSWERS Appendix C GNU Free Documentation License Version 1.1, March 2000 2000 Free Software Foundation, Inc. Copyright c 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. Preamble The purpose of this License is to make a manual, textbook, or other written document “free” in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modifications m
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Index x-axis, 11 x-axis,positive, 11 xy-coordinate system, 11, 12 y-axis, 11 y-axis,positive, 12 y-intercept, 39 nth root, 135 adjacent, 222 amplitude, 253 analogue LP’s, 213 angle, 192 angle,central, 192 angle,initial side, 192 angle,standard position, 192 angle,terminal side, 192 angle,vertex, 192 angular speed, 207 arc,length, 199 arc,subtended, 192 arccosine function, 272 arcsine function, 272 arctangent function, 272 area,sector, 199 aspect ratio, 13 axis scaling, 13 axis units, 14 axis,horizontal, 11 axis,vertical, 11 belt/wheel problems, 215 CD’s, 215 central angle, 192 chord, 202 circle, 25 circles, 26, 77 circles,circular function, 230 circles,great, 204 circles,point coordinates on, 230 circles,unit, 28 circular function, 229 circular function,triangles, 229 circular function, 191, 221, 226, 227 circular function,circles, 230 circular function,inverse, 267 circular function,special values, 223 circular motion, 209 compound interest, 146, 148 compounding periods, 146 continuous compounding, 151 converting units, 1 coordinates,imposing, 11 cosecant function, 232 cosine function, 222, 248 cotangent function, 232 curves,intersecting, 28 db, 160 decibel, 160 decreasing function, 76 degree, 194 degree method, 193 degree,minute, 194 degree,second, 194 density, 3 dependent variable, 59 difference quotient, 36 digital compact disc, 215 dilation, 170 dilation,horizontal, 173 dilation,vertical, 171 directed distance, 19 distance, directed, 19 distance,between two points, 17, 19 311 312 domain, 58 e, 149, 150 envelope of hearing, 160 equation,quadratic, 45 equatorial plane, 202 even function, 241 exponential decay, 139 exponential function, 149 exponential growth, 139 exponential modeling, 145 exponential type, 139 function, 58 function,circular, 191, 221, 226, 227, 229 function,cosine, 222, 248 function,decreasing, 76 function,even, 241 function,exponential, 149 function,exponential type, 139 function,logarithm base b, 157 function,logarithmic, 153 function,multipart, 79 function,natural logarithm, 154 function,odd, 241 function,periodic, 240 function,picturing, 55, 57 function,rational, 181 function,sine, 222, 248 function,sinusoidal, 191, 247, 251 function,tangent, 222, 248 function,trigonometric, 247 function,cos(θ), 222 function,cos(x), 248 function,cos−1(z), 271 function,sin(θ), 222 function,sin(x), 248 function,sin−1(z), 271 function,tan(theta), 222 function,tan(x), 248 function,tan−1(z), 271 graph, 1, 26, 27, 38 graph,circular function, 242 graph,sin(θ), 244 INDEX graphing, 1 great circle, 202, 204 horizon circle, 282 horizontal line, 26 horizontal axis, 11 identity,composition, 272 identity,even/odd, 241 identity,key, 240 identity,periodicity, 240 imposing coordinates, 11, 15 independent variable, 59 interest, 146 intersecting curves, 28 intervals, 60 inverse circular function, 267, 270, 271 inverse function, 267 knot, 205 latitude, 202 line,horizontal, 26 line,vertical, 26 linear speed, 208 linear functions, 64 linear modeling, 33 lines, 33, 39 lines,horizontal, 25 lines,parallel, 44 lines,perpendicular, 44 lines,point slope formula, 38 lines,slope intercept formula, 39 lines,two point formula, 38 lines,vertical, 25 logarithm conversion formula, 158 logarithm function base b, 157 logarithmic function, 153 longitude, 203 loudness of sound, 159 LP’s, 213 mean, 252 meridian, 203 meridian,Greenwich, 203 modeling, 1 INDEX 313 RPM, 208 rules of exponents, 135 scaling, 13 secant function, 232 sector,area, 199 semicircles, 77 shifting, 168 shifting,principle, 170 sign plot, 75 sine function, 222, 248 sinusoidal function, 247 sinusoidal function, 191, 251 sinusoidal modeling, 251 slope, 36 solve the triangle, 267 sound pressure level, 159 speed,angular, 207 speed,circular, 209 speed,linear, 208 standard position, 192 standard angle, 192 standard fo
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rm, 27 tangent function, 222, 248 triangle,sides, 221 trigonometric function, 247 trigonometric ratios, 223 uniform linear motion, 47 unit circle, 28 units, 1 vertical axis, 11 vertical line test, 63 vertical lines, 26 modeling,exponential, 145 modeling,linear, 33 modeling,sinusoidal, 251 motion,circular, 209 mulitpart function, 80 multipart functions, 79 natural logarithm, 153 natural logarithm function, 154 natural logarithm function, prop- erties, 154 nautical mile, 205 navigation, 202 odd function, 241 origin, 11 parabola,three points determine, 98 parametric equations, 47 period, 253 periodic, 240, 247 periodic rate, 146 phase shift, 252 piano frequency range, 140 picturing a function, 55, 57 positive x-axis, 11 positive y-axis, 12 principal, 146 principal domain, 271 principal domain, cosine, 271 principal domain, sine, 271 principal domain, tangent, 271 principal solution, 270 Pythagorean Theorem, 18 quadrants, 13 quadratic formula, 45 radian, 198 radian method, 196 range, 59 rate, 4, 39 rate of change, 4 rational function, 181 reflection, 166 restricted domain, 59 right triangles, 229
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