Written Homework 7 Solutions

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1 Written Homework 7 Solutions MATH 0 - CSM Assignment: pp 5-56 problem 6, 8, 0,,, 5, 7, 8, Find formulas for the derivatives of the following functions; that is, differentiate them. Solution: (a) f (x) = 2x 6.2x + πx 2 (b) g (x) = 2 x 5 x 6 (c) h (w) = 6w 7 cos(w) 2 w (d) R (u) = 4 sin(u) sec 2 (u) + u 2/ (e) V (s) = 4s /4 (f) F (z) = ln(2) 7 2 z (g) P (t) = a t + v 0 8. In each case below, find a function f(x) whose derivative f (x) is: (a) f (x) = 2x ; Solution: f(x) = x 2 (b) f (x) = 5x 7 ; Solution: f(x) = 5x8 8 (c) f (x) = cos(x) + sin(x) ; Solution: f(x) = sin(x) cos(x) (d) f (x) = ax 2 + bx + c ; Solution: f(x) = ax + bx2 2 + cx (e) f (x) = 0 ; Solution: f(x) = (f) f (x) = 5 ; Solution: f(x) = 0 x x 0. (a) For which values of x is the function x x increasing? Solution: If we want to know when x x is increasing, we should find its derivative, set it equal to zero to find the critical points, and then test the intervals created by the critical points to see when the derivative is positive. Remember that if the derivative is positive, we know that the function is increasing. f(x) = x x, so f (x) = x 2

2 Now set f (x) = 0 x 2 = 0 = x 2 = x2, hence x = ± So we need to test the following intervals: (, ), (, ) and (, ). By plugging in points from these intervals into f (x), we can figure out the behavior of f(x). These calculations have been omitted, but f (x) > 0 for x (, ), thus f(x) is increasing on this interval. (b) Where is the graph of y = x x rising most steeply? Solution: We want toknow when y = x x is rising most steeply. We know it has to be in the interval (, ) since this is the only time when f is increasing. One way to figure out when f is rising most steeply is to look at the second derivative. If f (x) = x 2, then f = 6x. Now we set f (x) = 0 to find its critical points and we see that f (x) = 0 when x = 0. Notice f (0) = > 0 so x = 0 is a maximum for f (x). So to answer our original question, f is rising most steeply when x = 0. (c) At what points is the graph of y = x x horizontal? Solution: We have already seen that the critical points of f are at x = ±. Recall that f obtains its critical points when f (x) = 0. So f(x) is horizontal, or of zero slope, or f (x) = 0, when x = ± (d) Make a sketch of the graph of y = x x that reflects all these results. Solution:

3 . (a) Sketch the graph of the function y = 2x + 5 on the interval 0.2 x 4. x Solution: (b) Where is the lowest point on that graph? Give the value of the x-coordinate exactly. Solution: The lowest point on this graph occurs when 0.2 x 4 and f (x) = 0. Since f(x) = 2x + 5 x, f (x) = 2 5, and after some simple algebra, we see that x2 f 5 (x) = 0 when x = 2. Thus the lowest point on this graph is ( 5 2, 6.246).. This problem was not graded and is very similar to (a) Write the microscope equation for y = x at x = 600. Solution: The microscope equation is y y = x. (b) Use the microscope equation to estimate 628 and 592. How far are these estimates from the values given by a calculator? Solution: Using the above microscope equation, we will first estimate 628. To figure out our estimate for y, we first need to find x: x = = 28. Now we can plug into y 60 x. So y 60 ( 28) If we plug 628 into a calculator we get , so there is very little discrepancy. Now we can estimate 592. We first need x: x = = 8. Thus y 60 x. So y (8) If we plug 628 into a 20 calculator we get , so again there is very little discrepancy. 7. A ball is held motionless and then dropped from the top of a 200 foot tall building. After t seconds have passed, the distance from the ground to the ball is d = f(t) = 6t feet.

4 (a) Find a formula for the velocity v = f (t) of the ball after t seconds. Check that your formula agrees with the given information that the initial velocity of the ball is 0 feet/second. Solution: Since we re given that v = f (t) and f(t) = 6t , it s easy to see that v(t) = 2t. Now we can check that v 0 = 0: v(0) = 2 0 = 0. (b) Draw graphs of both the velocity and the distance as functions of time. What time interval makes physical sense in this situation? (For example, does t < 0 make sense? Does the distance formula make sense after the ball hits the ground?) Solution: The time interval that makes the most sense is 0 t 4 since we think of time starting at zero and the ball hits the ground shortly before t = y = -6x² y = -2x -60 (c) At what time does the ball hit the ground? What is its velocity then? Solution: The ball hits the ground when d = f(t) = 6t = 0 which happens when t.54. When t =.54, the ball s velocity is v(.54) = 2.54 = A second ball is tossed straight up from the top of the same building with a velocity of 0 feet per second. After t seconds have passed, the distance from the ground to the ball is d = f(t) = 6t 2 + 0t feet. (a) Find a formula for the velocity of the second ball. Does the formula agree with given information that the initial velocity is +0 feet per second? Compare the velocity formulas for the two balls; how are they similar, and how are they different? Solution: Again, we know f(t) and that v = f (t), so we get that v = 2t + 0. (b) Draw graphs of both the velocity and the distance as functions of time. What time interval makes physical sense in this situation? Solution: Again, the time interval that makes the most sense is 0 t 4. 4

5 60 80 y = -6x² + 0x y = -2x (c) Use your graph to answer the following questions. During what period of time is the ball rising? During what period of time is it falling? When does it reach the highest point of its flight? Solution: The ball is roughly rising for 0 < t < 0. and falling for 0. < t <.85. The highest point is reached when v = 2t + 0 = 0 which occurs when t 0.. (d) How high does the ball rise? Solution: By plugging t = 0. into our distance equation, we see that the ball reaches f(0.) = 6(0.) 2 + 0(0.) = feet, or.56 feet above the top of the building. 20. A steel ball is rolling along a 20-inch long straight track so that its distance from the midpoint of the track (which is 0 inches from either end) is d = sin(t) inches after t seconds have passed. (Think of the track as aligned from left to right. Positive distances mean the ball is to the right of the center; negative distances mean it is to the left.) (a) Find a formula for the velocity of the ball after t seconds. What is happening when the velocity is positive; when it is negative; when it equals zero? Write a sentence or two describing the motion of the ball. Solution: v(t) = d (t) = cos(t). When the velocity goes from positive to negative and visa versa it is simply switching directions on the track from right to left and left to right respectively. When the velocity equals zero, the ball is at its maximum distance from the center. The ball is moving back and forth along the track reaching its higher speeds towards the center of the track and pausing briefly when its distance from the center is maximized. (b) How far from the midpoint of the track does the ball get? How can you tell? Solution: The maximum distance occurs when d = sin(t) is maximized which 5

6 occurs when v(t) = cos(t) = 0 or when t = π 2. sin( π 2 ) = = inches. Thus the maximum distance is (c) How fast is the ball going when it is at the midpoint of the track? Does it ever go faster than this? How can you tell? Solution: The first time the ball is in the center of the track after it has started to move is when d = sin(t) = 0 which occurs when t = π. The speed at t = π is v(π) = cos(π) = ( ) = =, so the speed of the ball at the midpoint of the track is inches per second which is the maximum speed. 6