Contents 10. Graphs of Trigonometric Functions

Transcription

1 Contents 10. Graphs of Trigonometric Functions Sine and Cosine Curves: Horizontal and Vertical Displacement Example Composite Sine and Cosine Curves Calculator Graphics Example Example Example Graphs of the Other Trigonometric Functions Example Example Parametric Equations Using a Calculator to Graph Parametric Equations Example Example Example Example Polar Coordinates Using a Calculator to Graph Polar Equations Example Example

2 Peterson, Technical Mathematics, 3rd edition Sine and Cosine Curves: Horizontal and Vertical Displacement Example The data in the Table 10.1 gives the number of hours between sunrise and sunset in Seattle, Washington on the 1st and 16th of each month for Table 10.1: Length of Day in Seattle, WA, 2002 Month January February March April Date Hours Month May June July August Date Hours Month September October November December Date Hours Determine a sinusoidal regression for this data. Solution Before we enter the data in a calculator we need to decide on a numbering system for the x-axis. One system would be to let January 1 = 1, January 16 = 16, February 1 = 32, February 28 = 59,..., December 31 = 365. Using this procedure we get the data in Table Table 10.2: Length of Day in Seattle, WA, 2002 Month January February March April Date Day Hours Month May June July August Date Day Hours Month September October November December Date Day Hours The data is entered into a calculator using the same procedures we have used before. A graph of the data is in Figure 10.17a. To determine the sinusoidal regression curve we press APPS 6 2 and open the variable where the above 1 Source:

3 Peterson, Technical Mathematics, 3rd edition 3 data was entered. The press F5 [CALC] and scroll down until B SinReg is highlighted and press ENTER. Tell the calculator where to get the x-and yvalues and where you want the regression equation saved. Press ENTER twice and a few seconds you should get the regression values for y1(x) = a sin(bx + c) + d. After you have written the regression values press ENTER F3 [GRAPH] to get the result in Figure 10.17b. FIGURE 10.17a FIGURE 10.17b As we see, the regression equation for this data is hours of daylight on day t of y sin(0.0167t )

4 Peterson, Technical Mathematics, 3rd edition Composite Sine and Cosine Curves Example Calculator Graphics Graphing these curves by hand takes a great deal of time. Using a calculator or a computer to draw the graphs of trigonometric curves would save time. We will briefly describe how to use a graphing calculator to graph trigonometric functions. All of the directions and illustrations are specific to a Texas Instruments TI-89 graphics calculator. If you have another brand or model then you may have to vary the directions slightly. While you can graph trigonometric functions with your calculator in degree mode, it is easier if you first set it in radian mode. Use a graphing calculator to graph y = 1 2 cos(3x + π). Solution Make sure that your calculator is in radian mode. Now, press F1 [Y=] to enter the function you want to graph. Press the following sequence of keys: 0.5 2nd COS 3 X + 2nd π ) ENTER The calculator screen should look like the one shown in Figure 10.25a. (Note that we could have typed ( 1 2 ) instead of 0.5.) Press F2 [ZOOM] 6 [6:ZoomStd] for the standard viewing window and you should obtain the graph shown in Figure 10.25b. Compare this graph to the one in Figure FIGURE 10.25a FIGURE 10.25b This graph is very difficult to see. We need to change the window settings. We know that the amplitude of this function is 1 2 = 0.5 and that it has a period of 2π Press F2 [WINDOW] and set the scale as follows: xmin = 2π 3, xmax = π, xscl = π 4, ymin = 1, ymax = 1, yscl = 0.5. Why did we select these values for xscl and xscl? The x-values were selected to allow us to view two complete cycles of the graph. We selected yscl = 0.5 to help us see where the highest and lowest values of the graph are. Now press F3 [GRAPH]. The result is shown in Figure 10.25c. FIGURE 10.25c

5 Peterson, Technical Mathematics, 3rd edition 5 ( ) sin x Example Use a graphing calculator to graph y = (sin 2x). x Solution This is the same function we graphed in Example We will work this so we can separately graph y 1 = sin 2x and y 2 = sin x and then graph their product x y = y 1 y 2. Based on our work in Example 10.18, we choose window settings of xmin = 0, xmax = 10, xscl = π 4, ymin = 1.1, ymax = 1.1, and yscl = 0.5. Now press F1 [Y=] and make sure the cursor is on the line for y1 and then press CLEAR 2nd SIN 2 X ) ENTER. This should appear on line Y1. To enter y 2 = sin x, make sure the cursor is on line y2 and press x 2nd SIN X ) X ENTER We will put y = y 1 y 2 on line y3. Rather than entering both y 1 and y 2 again, we will let the calculator do this for us. The cursor should be on line y3. If it is not move it there and press the following key sequence: Y 1 ( X ) * Y 2 ( X ) ENTER FIGURE 10.26a Line Y3 now reads y3= y1(x)* y2(x) and the calculator screen should look like that in Figure 10.26a. Press F3 [GRAPH]. You should see each of the functions y 1, y 2, and y graphed in sequence, with the final result like that shown in Figure 10.26b. This graph is too cluttered. Let s change it so we see only the graph of y = y 1 y 2. Press the to move the cursor to the line for y2 and press F4. This will remove the check mark at the left of this line. Removing the check mark has the effect of deselecting the function so it will not be graphed. Press the again to move the cursor to the line for y1 and press F4 to deselect line y1. Press F3 [GRAPH]. The result, shown in Figure 10.26c, is the graph of y = y 1 y 2 = ( ) sin x (sin 2x). x FIGURE 10.26b FIGURE 10.26c

6 Peterson, Technical Mathematics, 3rd edition 6 Example Use a graphing calculator to graph two cycles of y = 3 cos 2x sin 4x + 5. Solution We need to analyze this function to help determine the period and the range. To help our analysis we let y = y 1 + y 2 with y 1 = 3 cos 2x and y 2 = sin 4x + 5. The function y 1 = 3 cos 2x has amplitude 3, period π, and vertical displacement 0. The function y 2 = sin 4x + 5 has amplitude 1, period π 2, and vertical displacement 5. Because the least common multiple of π and π 2 is π, the period for y = y 1 + y 2 is π. The amplitudes are 3 and 1, so the largest possible amplitude is 4. With a vertical displacement of 5, the range of this function is in the interval [5 4, 5 + 4] = [1, 9]. Note that this is probably not the actual range of the function. This range is just an estimate to help us determine what values we will need on the y-axis. This estimate will be very important when we do calculator graphing. (The actual range is about (1.515, 8.485).) Since the period is π, an interval from x = 0 to x = 2π will include two cycles. We will set the viewing window to [ 0, 2π, π 8 ] [0, 9, 1]. Even though the range is no more than [1, 9], we set the y-values in the viewing window to [0, 9] so the x-axis would be included in the graph. The graphs of y 1, y 2, and y 3 = y 1 + y 2 are in Figure 10.27a. Notice that the graph of y 3 is thicker than the other two graphs. The graph of just y = 3 cos 2x sin 4x + 5, using the viewing window we selected, is in Figure 10.27b. h 0, 2π, π i [ 3, 9, 1] 8 h 0, 2π, π i [0, 9, 1] 8 FIGURE 10.27a FIGURE 10.27b

7 Peterson, Technical Mathematics, 3rd edition Graphs of the Other Trigonometric Functions Calculator Graphics Example Use a graphing calculator to sketch the graph of y = 1 3 tan ( 2x π 4 ). Solution Make sure your calculator is in radian mode. From Table 10.5, we see that the period is π We would like to see at least two periods of this graph, so we set xmin = π 4, xmax = 9π 8, xscl = π 8, ymin = 4, ymax = 4, and yscl = 1. Now press Y= and clear any existing functions from the screen. Press the following sequence of keys: ( ) ( 1 3 ) 2nd TAN 2 X + ( 2nd π 4 ) ) ENTER The calculator screen should now look like the one shown in Figure 10.34a. Press F3 [GRAPH] and you should obtain the graph in Figure 10.34b. FIGURE 10.34a FIGURE 10.34b FIGURE Notice that the calculator seems to have drawn in the vertical asymptotes. Look closely. These are not vertical lines. The portion above the x-axis is not directly above the portion below the x-axis. These apparent vertical asymptotes are put in by some graphing calculators and computer graphing programs. Example Use a graphing calculator to sketch y = 1.5 sec (x + 2). Solution We set xmin = 1, xmax = 7, xscl = 1, ymin = 5, ymax = 5, and yscl = 1. Press Y= and clear any functions that are there. A calculator does not have a secant key, so before you graph this function you will need to use the reciprocal identity sec x = 1 cos x to rewrite this function as y = 1.5 cos (x + 2). Now, press the following sequence of keys: 1.5 ( 2nd COS X + 2 ) ) ENTER Press F3 [GRAPH] and you should obtain the graph in Figure Once again, you can see that the calculator seems to have drawn in the vertical asymptotes.

8 Peterson, Technical Mathematics, 3rd edition Parametric Equations Using a Calculator to Graph Parametric Equations Before you can use a graphing calculator to draw the graphs of parametric equations, you need to put the calculator in parametric mode. On a TI-89 this is done by pressing MODE. The screen on a TI-89 should look like the one shown in Figure 10.49a. To put the calculator mode you need to press 2 ENTER. ENTER Example Use a graphing calculator to sketch the curve represented by the parametric equations x = 2t and y = t 2 4. Solution These are the same parametric equations we graphed in Example We will use the following table from that example to help graph these equations. t x y FIGURE To enter the window settings, first press F2 [WINDOW]. On a TI-89, you are first asked for tmin, tmax, and tstep. Based on the table, we will let tmin = 4 and tmax = 4. The value of tstep determines how much the value of t should be increased before calculating new values for x and y. We will pick tstep = 0.5. You may want to try different values. Based on this table, let xmin = 8, xmax = 8, xscl = 1, ymin = 4, ymax = 12, and yscl = 1. Now press F! [Y=]. On the first line, enter the right-hand side of the parametric equation for x. Press 2 T. Next, enter the parametric equation for y. Press ENTER to move the cursor to the line labeled y1t and press the key sequence T 2 4 ENTER. The result in displayed in Figure 10.50a. Now press F3 [GRAPH] and you should see the result in Figure 10.50b. FIGURE 10.50a FIGURE 10.50b Example Use a graphing calculator to sketch the curve represented by the parametric equations x = 2 cos t and y = sin t.

9 Peterson, Technical Mathematics, 3rd edition 9 Solution Because the periods of sin and cos are 2π, we will let tmin = 0, tmax = 6.3, and tstep = 0.1. Since the range of x = 2 cos t is [ 2, 2], we choose xmin = 2, xmax = 2, and xscl = 1. The range of y = sin t is [ 1, 1]. We let ymin = 1.5, ymax = 1.5, and yscl = 1. Now press F1 Y= and enter the parametric equations by pressing 2 2nd COS T ) ENTER 2nd SIN T ) ENTER F3 [GRAPH] The result is displayed in Figure FIGURE Gravity-influenced Trajectories When a communications satellite or the Space Shuttle is launched, it is given enough velocity to enable it to go into Earth s orbit instead of falling back to Earth. How much velocity is enough? The answer depends on the weight of the rocket and where you want the orbit to be. The Space Shuttle, for example, must be sent off at 17,500 mph in order to reach its orbit. If a projectile moves with an initial velocity that is below its escape velocity, it returns to Earth without going into orbit. We ll now return to motion which does not reach orbit, and is therefore called sub-orbital motion. Summary of Sub-Orbital Projectile Motion The parametric equations that describe the position of an object thrown into the air with velocity less than the Earth s escape velocity are { x(t) = vx0 t + x 0 y(t) = 1 2 gt2 + v y0 t + y 0 where t is the time after the toss; x(t) the horizontal distance traveled; y(t) the vertical distance traveled; v x0 the initial horizontal component of velocity (v x0 is constant during flight); x 0 the horizontal distance from zero point at release; g = 32 if the units are feet and seconds and g = 9.81 if the units are meters and seconds; v y0 is the initial vertical component of velocity; and y 0 is the height above ground at release. Example A ball is thrown upward at an angle of 50 from a height of 30 ft with a velocity of 92 ft/sec. (a) Write the parametric equations of the position of the ball at any time t. (b) Plot the trajectory on your calculator. (c) Estimate to two decimal places the time when the ball reaches the ground. (d) Estimate to the nearest foot the maximum height the ball reaches. (e) Estimate to the nearest foot the horizontal distance the ball traveled. Solutions: We are given that y 0 = 30, and we can set x 0 = 0. Since v 0 = 92, we have v x0 = 92 cos ft/sec, and v y0 = 92 sin ft/sec. Since y 0 and v 0 are both in terms of feet, we use g = 32.

10 Peterson, Technical Mathematics, 3rd edition 10 (a) Using the above information, the equations are { x = 59t y = 16t t + 30 FIGURE 10.52a FIGURE 10.52b (b) The trajectory is shown in Figure 10.52a. (c) The ball will hit the ground when y = 0. Using the quadratic formula to solve 16t t + 30 = 0, we see that t 0.39 sec or t 4.77 sec. The first time is unacceptable since it was before the ball was thrown. Thus, it took about 4.77 seconds for the ball to hit the ground. (d) In the last chapter we learned that the vertex of a parabola f(t) = at 2 + bt + c occurs when t = b 2a. Here we have y(t) = 16t2 + 70t + 30, so the vertex is at t = 70 = Using the TRACE feature of the calculator, we see in 2( 16) Figure 10.52b that when t = 2.2, then y = So, the maximum height is about 107 ft. (e) At the time it hits the ground (when t = 4.77), the ball has traveled x(4.8) feet horizontally. Since x(4.77) = , the ball lands feet away from the point where it was thrown. The next example uses the same ideas in a different setting. Example A plane is attempting to drop a package of food near a campsite in the wilderness. The plane is moving horizontally in level flight at 120 mph at a height of 1000 ft. (a) Write the parametric equations that model the position of the plane and the package. (b) Plot the trajectories on your calculator. (c) Estimate to the nearest tenth of a second the time when the food package reaches the ground. (d) Estimate to the nearest foot the horizontal distance from where the food was dropped to where it landed. Solutions: In this example v y0 = 0 because the package is being dropped, not thrown. However v x0 = 120 mph 176 ft/sec. Finally, y 0 = 1000 ft. (a) The equations for the position of the package are { x = 176t y = 16t The equations for the position of the plane are { x = 176t y = 1000

11 Peterson, Technical Mathematics, 3rd edition 11 FIGURE (b) With the settings tmin= 0, tmax= 12, tstep= 0.1, xmin= 0, xmax= 1500, xscl= 100, ymin= 0, ymax= 1200, and yscl= 100, the two trajectories are plotted in Figure (c) Solving y = 16t = 0, we see that the package reaches the ground at about t = 7.9 sec. (d) The horizontal distance the food package traveled is approximately ft before it hit the ground.

12 Peterson, Technical Mathematics, 3rd edition Polar Coordinates Using a Calculator to Graph Polar Equations You can use a graphing calculator or a computer to help you graph many of these curves. We will describe how it is done using a graphics calculator. As before, we will use a TI-89 for these examples. Example Use a TI-89 graphing calculator to graph r = sin 2 3θ. FIGURE 10.62a The θ is above the key. Solution First, put your calculator in polar mode. On a TI-89 this is done by pressing MODE. The screen on a TI-89 should look like the one shown in Figure 10.62a. Here the row 3:POLAR is highlighted. Since this is the mode we want press 3 ENTER. Press F2 [WINDOW] to set the size of the viewing window. The function r = sin 2 3θ has a period of 2π, so it will take no more than from θ = 0 to θ = 2π to completely sketch this graph, so set θmin = 0, θmax = 2π, and θstep = π/24. Since the function has an amplitude of 1, a window of [ 1.5, 1.5] [1, 1] should be right. You are now ready to enter the function into the calculator. Now press F1 Y=. On the first line, enter the right-hand side of the parametric equation for r. Press ( 2nd sin 3 θ ) ) 2 ENTER Remember, that sin 2 θ is entered into a calculator or computer as (sin θ) 2. The result in displayed in Figure 10.62b. Now press F3 [GRAPH] and you should see the result in Figure 10.62c. FIGURE 10.62b FIGURE 10.62c Example Use a TI-89 graphing calculator to graph r = 5 sin 2θ 2 cos 3θ. Solution First, set the size of the viewing window. The period of sin 2θ is π; the period of cos 3θ is 2 3 π. Since the least common multiple of π and 2 3π is 2π, the period of r is a multiple of 2π. We will set θmin = 0, θmax = 2π, and θstep = Since the ranges of sine and cosine are both [ 1, 1], we know that 7 r 7. Because the screen is longer horizontally than it is vertically, we set xmin 10, xmax = 10, xscl = 1, ymin = 6, ymax = 6, and yscl = 1. Press Y= and erase and existing functions. Then press the following:

13 Peterson, Technical Mathematics, 3rd edition nd SIN 2 θ ) 2 2nd COS 3 F3 [GRAPH] θ ) ENTER The result is a butterfly-shaped curve as shown in Figure FIGURE 10.63