Answers and Solutions to Section 2.1 Homework Problems S. F. Ellermeyer

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1 Answers and Solutions to Section 2.1 Homework Problems S. F. Ellermeyer 1. A tank holds 00 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume, V, of water remaining in the tank (in gallons) after t minutes. t (min) V (gal) a. If P is the point 15,250 on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with t 5,, 20, 25, and 30. i. The slope of the secant line passing through the points P 15,250 and Q 5,694 is ii. iii. The slope of the secant line passing through the points P 15,250 and Q,444 is The slope of the secant line passing through the points 1

2 P 15,250 and Q 20,111 is iv. The slope of the secant line passing through the points P 15,250 and Q 25,28 is v. The slope of the secant line passing through the points P 15,250 and Q 30,0 is b. Estimate the slope of the tangent line at P by averaging the slopes of two secant lines. The slope of the secant line passing through the points P 15,250 and Q,444 is 38.8 and the slope of the secant line passing through the points P 15,250 and Q 20,111 is The average of these slopes is We can thus estimate that the slope of the tangent line at the point P 15,250 is However, we certainly don t know that this is the exact value of the slope of the tangent line and, furthermore, there is no way to find the exact value because we are only given a discrete set of data points. In order to find the exact value of the slope of the tangent line at P 15,250, we would need to be given the values of V at all values of t in some (at least small) time interval containing t 15. c. Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank at the instant t 15 min. In other words, it is the instantaneous flow rate at t 15 min.) Once again (just as in part b), our estimate is bound to be rather crude because it relies on guessing from incomplete information. Using the hand drawn guess at a tangent line shown in the picture below, we arrive at m This is remarkably close to the answer that we got in part b! 2

3 3. The point P 1, 1 lies on the curve y x. 2 1 x a. Use your calculator to find the slope of the secant line passing x through P and the point Q x, where x is each of the following 1 x values given in parts i through viii below. i. We will do this one by hand without using a calculator: If x 1/2, then y so the point Q is Q 1, 1. The slope of the secant line 2 3 passing through P and Q (pictured below) is

4 ii. iii. iv. If x 0.9, then 9 y so the point Q is Q 9, 9. The slope of the secant line 19 passing through P and Q is If x 0.99, then 99 y so the point Q is Q 99, 99. The slope of the secant line passing through P and Q is If x 0.999, then 999 y so the point Q is Q , The slope of the secant line passing through P and Q is 4

5 v. If x 1.5, then vi y so the point Q is Q 15, 3. The slope of the secant line 5 passing through P and Q is If x 1.1, then 11 y so the point Q is Q 11, 11. The slope of the secant line 21 passing through P and Q is vii. If x 1.01, then 1 y so the point Q is Q 1, 1. The slope of the secant line passing through P and Q is viii. If x 1.001, then 01 y so the point Q is Q 01, 01. The slope of the secant line passing through P and Q is b. Using the results in part (a), guess the slope of the tangent line to the curve y x at the point P 1, 1. 1 x 2 Based on these calculations, I guess that the slope of the tangent line at P is 0.25 (which is the same as 1/4) c. The point in question is P 1, 1 and I am guessing that the slope of 2 the tangent line is 1/4. Therefore (assuming that my guess for the slope is correct), an equation of the tangent line is 5

6 y 1 x 1 4 which can also be written (in slope intercept form) as y 1 4 x 1 4. When I graph this line along with the curve below, it looks to be reasonable! 5. If a ball is thrown into the air with a velocity of 40 ft/sec, its height (in feet) after t seconds is given by y 40t 16t 2. a. Find the average velocity for the time period beginning when t 2 and lasting i. 0.5 sec When t 2, we have y This means that 2 seconds after the ball is thrown, it is 16 feet above the ground. When t 2.5, we have y This means that 2.5 seconds after the ball is thrown, it is on the ground. The average velocity of the ball during the time period 2 t 2.5 is 0ft 16ft 2.5s 2s 16ft 32 ft/s. 0.5s 6

7 ii. iii. This means that, on average, the ball was travelling downward at a speed of 32 ft/s during the final 1/2 second of its flight. Note that the velocity of the ball is always changing. The velocity that we have computed is an average velocity over a particular time interval. 0.1 sec When t 2.1, we have y This means that 2.1 seconds after the ball is thrown, it is ft above the ground. The average velocity of the ball during the time period 2 t 2.1 is 13.44ft 16ft 2.1s 2s ft s sec When t 2.05, we have y This means that 2.05 seconds after the ball is thrown, it is ft above the ground. The average velocity of the ball during the time period 2 t 2.05 is 14.76ft 16ft 2.05s 2s ft s. iv sec When t 2.01, we have y This means that 2.05 seconds after the ball is thrown, it is ft above the ground. The average velocity of the ball during the time period 2 t 2.01 is ft 16ft s 2s ft s. b. Find the instantaneous velocity when t 2. Based on the calculations done above, I guess that the instantaneous velocity of the ball at the instant t 2 is 24 ft s. This means that exactly 2 seconds after the ball is thrown, it is travelling downward at a speed of 24 ft s. Below is shown a graph of the function y 40t 16t 2 along with the line y t 2, which is the tangent line at the point P 2,16. 7

8 7. The displacement (in feet) of a certain particle moving in a straight line is given by s t 3 /6, where t is measured in seconds. a. Find the average velocity over the following time periods: i. 1,3 The average velocity of the particle over the time period from t 1 to t 3 is ii. iii. iv ft/sec ,2 The average velocity of the particle over the time period from t 1 to t 2 is ft/sec ,1.5 The average velocity of the particle over the time period from t 1 to t 1.5 is ft/sec ,1.1 The average velocity of the particle over the time period from t 1 to t 1.1 is 8

9 ft/sec b. Find the instantaneous velocity when t 1. It is hard to guess the instantaneous velocity when t 1 based on the few calculations done above. I guess that it is 0.5 ft/sec., but a few more numerical calculations as done above would needed to convince me that this is correct (and then I still might not be convinced). c-d. Draw the graph of s as a function of t and then draw the secant lines whose slopes are the average velocities found in part (a). Also, draw the tangent line whose slope was guessed in part (b). 9. The point P 1,0 lies on the curve y sin /x. a. If Q is the point x,sin /x, find the slope of the secant line passing through P and Q for x 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. For x 2, we have y sin /2 sin 5 0 and the slope of the secant line PQ is For x 1.5, we have y sin /1.5 sin 20 /3 3 /2 and the slope of the secant line PQ is 9

10 For x 1.4, the slope of the secant line PQ is sin / For x 1.3, the slope of the secant line PQ is sin / For x 1.2, the slope of the secant line PQ is sin / For x 1.1, the slope of the secant line PQ is sin / For x 0.5, the slope of the secant line PQ is sin / For x 0.6, the slope of the secant line PQ is sin / For x 0.7, the slope of the secant line PQ is sin / For x 0.8, the slope of the secant line PQ is sin / For x 0.9, the slope of the secant line PQ is sin / Note that these computations seem to be of no help in trying to guess the slope of the tangent line to the curve y sin /x at the point P 1,0 because these slopes seem to be jumping around erratically. b. Use a graph of the curve y sin /x to explain why the slopes of the secant lines computed in part (a) are not close to the slope of the tangent line to this curve at the point P 1,0. Here is the graph of y sin /x drawn on the interval 0.5 x 2.

11 x Can you see why we got such different slopes for the secant lines in part (a)? Try to draw the secant lines whose slopes were computed in part (a) on the picture above. c. By choosing appropriate secant lines, estimate the slope of the tangent line at the point P 1,0. By looking at the graph above, it is obvious that we will need to choose x values very close to 1 in order to be able to do this. For example, x 0.9 and x 1.1 (which were used in part (a)) are not close enough. For x 0.99, the slope of the secant line PQ is sin / For x 0.999, the slope of the secant line PQ is sin / For x 1.01, the slope of the secant line PQ is sin / For x 1.001, the slope of the secant line PQ is sin / Based on these latest calculations (and perhaps a few more that you might do), we guess that the slope of the tangent line to the 11

12 curve y sin /x at the point P 1,0 is 31. This is a good guess, but later in this course we will learn that the exact value of this slope is actually