5.1 Calculating the Slope of a Curve and Instantaneous Rate of Change 181 Therefore the slope of the tangent line is 10/27; the equation of the tangent line is of the form y − y 1 = 10 27 (x − x 1 ). The tangent line goes through the point (−3, f(−3)) = (−3, 5/9), so when x =−3, y must be 5/9. y − 5 9 = 10 27 (x + 3) or y = 10 27 (x + 3) + 5 9 . Alternatively, the equation of the tangent line can be written y = 10 27 x + 5 3 . ◆ In the next section we look at f , the derivative function. PROBLEMS FOR SECTION 5.1 The limiting process enables us to get a handle on the slope of a curve and the instantaneous rate of change. Problems 1 through 4 ask for approximations before arriving at an exact answer; this is to remind you of the process. 1. Suppose a ball is thrown straight up from a height of 48 feet and given an initial upward velocity of 8 ft/sec. Then its height at time t, t in seconds, is given by h(t) =−16t 2 + 8t + 48, for t ∈ [0, 2]. In this problem we will look at the ball’s velocity at t = 1. (a) At t = 1, is the ball heading up, or down? Explain your reasoning. (b) By calculating the average rate of change of height with respect to time, h t ,on the intervals [0.9, 1] and [1, 1.1], give bounds for the ball’s velocity at t = 1. (c) Improve your bounds by using the intervals [0.99, 1] and [1, 1.01]. (d) Use the limit definition of f (1) to find the ball’s instantaneous velocity at t = 1. 2. Let f(x)= 1 x . In this problem we will look at the slope of the tangent line to f(x) at point P = ( 1 2 ,2). (a) Is the slope of the tangent line to f at P positive, or negative? (b) By calculating the slope of the secant line through P and a nearby point on the graph of f , approximate f ( 1 2 ). First choose the point with an x-coordinate of 0.49. Next choose the point with an x-coordinate of 0.501. Now produce an approximation that is better than either of the previous two. (c) By calculating the limit of the difference quotient, find f ( 1 2 ). (d) Find the equation of the tangent line to f at P . 3. Let f(x)=x 3 and P be the point (1, 1) on the graph of f . (a) Approximate the slope of the line tangent to f at P by looking at the slope of the secant line through P and Q, where Q = (1 + h, f(1+h)). Calculate the difference quotient for various values of h, both positive and negative. See if your calculator or computer will produce a table of values. (b) Calculate f (1) by computing the limit of the difference quotient. 182 CHAPTER 5 The Derivative Function (c) For what values of h is the difference quotient greater than f (1)? For what values of h is the difference quotient less than f (1)? Make sense out of this by looking at the graph of x 3 . 4. Let f(x)=x 3 and P be the point (0, 0) on the graph of f . (a) Approximate f (0) by looking at the slope of the secant line through P and a nearby point Q on the graph of f . Use a calculator or computer to get a sequence of successive approximations corresponding to allowing the point Q = (h, f (h)) to slide along the graph of f toward P . Choose both positive and negative values of h. (b) Use the results of part (a) to guess the slope of the line tangent to x 3 at x = 0. (c) Calculate f (0) by computing the limit of the difference quotient f(0+h)−f(0) h . (d) Challenge question: In the previous three problems, by varying the sign of h in the difference quotient f(c+h)−f(c) h we obtain upper and lower bounds for f (c). In this problem, the difference quotient computed in part (a) is always larger than f (0). Explain what is going on by looking at the graphs of the various functions. In Problems 5 through 8, estimate f (c) by calculating the difference quotient f(c+h)−f(c) h for successively smaller values of |h|. Use both positive and negative values of h. 5. f(x)= √ x. Approximate f (9). 6. f(x)= 1 √ x . Approximate f (4). 7. f(x)=x 4 .Approximate f (1). 8. f(x)= 3 √ x. Approximate f (8). 9. Let f(x)= 6 x+1 . Let P and Q be points on the graph of f with x-coordinates 3 and 3 + h, respectively. (a) Sketch the graph of f and the secant lines through P and Q for h =1 and h =0.1. (b) Find the slope of the secant line through P and Q for h =1, h =0.1, and h =0.01. (c) Find the slope of the tangent line to f at point P by calculating the appropriate limit. (d) Find the equation of the line tangent to f at point P . 10. A pool is being drained. The amount of water in the pool at time t, t in hours, is given by w(t) =−16t 2 + 256 gallons. (a) Find the average rate of change of water in the pool over the time interval [2, 2.5]. Include units in your answer. Why is the value w t negative? (b) Use the dropped-rock problem results from this section to determine the rate of change of water in the pool at time t = 2. Include units in your answer. In Problems 11 and 12, estimate the slope of the tangent line at point P on the graph of f(x)by choosing a nearby point Q on the graph of f , Q = (a, f(a))and finding 5.1 Calculating the Slope of a Curve and Instantaneous Rate of Change 183 the slope of the secant line through P and Q. By choosing Q successively closer to P , guess the slope of the tangent line at P . 11. f(x)= √ x. P = (4, 2). Fill in the table. x-coordinate of Q Slope of PQ 3 3.8 3.95 3.998 4.001 4.02 4.3 5 12. f(x)=x 5 .P =(2, 32). Use your calculator to construct a table. x-coordinate of Q Slope of PQ 13. The point P = (3, f(3)) lies on the graph of f(x).Suppose Q = (w, f(w))is another point on the graph of f(x). (a) Write an expression for the slope of the secant line through P and Q. (b) Which one of the following corresponds to the slope of the tangent line to f at point P ? i. f(3)−f(w) 3−w ii. lim w→0 f(3)−f(w) 3−w iii. lim w→3 f(3)−f(w) 3−w (c) Explain, using pictures, why the answer you chose in part (b) corresponds to the process of taking point Q closer and closer to P and computing the slope of PQ. 184 CHAPTER 5 The Derivative Function 14. If we have a formula for f we can get quite good numerical estimates of the slope of the tangent line to f at a particular point. In this exercise we will do that. Below is the graph of y = f(x)=x 2 −4. –3 –2 –1 1 2 3 – 4 –3 –2 –1 1 x f(x) (a) Goal: We want to estimate the value of f (2), i.e., to approximate the slope of the tangent line to the graph at the point (2, 0). To this end, do the following: i. Sketch the tangent line to the graph at (2, 0). Finding the slope of this line is problematic, since we need two points on a line to find its slope. The only point we know for sure is on the tangent line is (2, 0). ii. Draw a line through the points (2, 0) and (2.1, f(2.1)). The slope of this line is close to the slope of the tangent line. Is it bigger than the slope of the tangent line, or less than the slope of the tangent line? Compute the slope of this line through points (2, 0) and (2.1, f(2.1)). iii. Draw a line through the points (2, 0) and (1.9, f(1.9)). The slope of this line is close to the slope of the tangent line. Is it bigger than the slope of the tangent line, or less than the slope of the tangent line? Compute the slope of this line through points (2, 0) and (1.9, f(1.9)). We are now in pretty good shape: We have an upper and lower bound for the slope of the tangent line. If you have done the problem correctly, you should now have the slope of the tangent line to within 0.2. Suppose we want to get an even better estimate. Do the following: iv. Find the slope of the line through (2, 0) and (2.01, f(2.01)). Is this slope bigger than the slope of the tangent line at (2, 0), or smaller? v. Find the slope of the line through (2, 0) and (1.99, f(1.99)). Is this slope bigger than the slope of the tangent line at (2, 0), or smaller? vi. We will ask you a few more questions of this sort. You can save your- self time and energy by computing the slope of the line through (2, 0) and (2 + h, f(2+h)). Then you will be able to simply evaluate your an- swer for different values of h. Make your life easier by computing this slope. vii. Find the slope of the line through (2, 0) and (2.002, f(2.002)). viii. Find the slope of the line through (2, 0) and (2.0001, f(2.0001)). 5.1 Calculating the Slope of a Curve and Instantaneous Rate of Change 185 ix. Find the slope of the line through (2, 0) and (1.998, f(1.998)). x. Find the slope of the line through (2, 0) and (1.9999, f(1.9999)). xi. At this point, you should have a fairly accurate idea of the slope of the graph of f at (2, 0). Give the best bounds you can for f (2) given the work you’ve done in parts vii through x. If you simplified your answer to part vi, then you can see what happens as h gets arbitrarily small. What do you think is the value of f (2)? We now turn our attention to another point. (b) Goal: We want to estimate the value of f (1); that is, we want to approximate the slope of the tangent line to the graph at the point (1, −3). i. Sketch the tangent line to the graph at (1, −3). ii. Find the slope of the line through (1, −3) and (1 + h, f(1+h)). Use this to answer parts iii through vi. (The more you simplify your answer the less time this problem will take.) iii. Find the slope of the line through (1, −3) and (1.1, f(1.1)). iv. Find the slope of the line through (1, −3) and (1.001, f(1.001)). v. Find the slope of the line through (1, −3) and (0.9, f(0.9)). vi. Find the slope of the line through (1, −3) and (0.99, f(0.99)). vii. What happens to the slope of the secant line through (1, −3) and (1 + h, f(1+h)) as h gets arbitrarily small? What do you think is the value of f (1)? (c) Goal: We want to estimate the value of f (0); i.e., we want to approximate the slope of the tangent line to the graph at the point (0, −4). i. Sketch the tangent line to the graph at (0, −4). ii. Find the slope of the line through (0, −4) and (0 + h, f(0+h)). iii. Use your answer to part ii to estimate the slope of the tangent line. (d) Goal: To estimate the value of f (c) for c a constant. i. Find the slope of the line through (c, f(c)) and (c + h, f(c+h)). ii. Use your answer to the previous question to find the slope of the tangent line to f at x = c. iii. Sketch the graph of f (x). 15. Let f(x)=2x 2 −5x−1. Let P and Q be points on the graph of f with x-coordinates 3 and 3 + h, respectively. (a) Sketch the graph of f and the secant lines through P and Q for h = 1 and h = 0.1. (b) Find the slope of the secant line through P and Q for h = 1, h = 0.1, and h = 0.01. (c) Find the slope of the tangent line to f at point P by calculating the appropriate limit. (d) Find the equation of the line tangent to f at point P . 16. Find the equation of the line tangent to y = 2 x+1 at the point (1, 1). 17. Let g be a function of t. Assume g is locally linear. (a) Label the coordinates of each of the six points shown on the following page. 186 CHAPTER 5 The Derivative Function w 4– 4 w 4+ s s+p g t A B C D E F r (b) Match the following expressions on the left with one of those on the right. The expressions on the right can be used more than once. (We use m sec as shorthand for “the slope of the secant line” and m tan as shorthand for “the slope of the tangent line.”) i) g(4+w)−g(4) w A. m sec through points D and F ii) g(4)−g(4−w) w B. m sec through points B and C iii) lim t→0 g(4+t)−g(4) t C. m sec through points A and B iv) g(r)−g(s) r−s D. m sec through points D and E v) lim t→s g(t)−g(s) t−s E. g (4) = m tan at point B vi) lim t→r g(r)−g(t) r−t F. g (r) = m tan at point F vii) g(s+p)−g(s) p G. g (s) = m tan at point D viii) lim h→0 g(s+h)−g(s) h 18. Let g be a function that is locally linear. We know that g (a) is the slope of the tangent line to the graph of g at point A = (a, g(a)). Let Q = (t, g(t)) be an arbitrary point on the graph of g, Q distinct from A. g A Q (a,g(a)) (t, g(t)) t (a) Write a difference quotient (i.e., an expression of the form ?−? ?−? , the quotient of two differences) that gives the slope of the secant line through points A and Q. (b) Take the appropriate limit of the difference quotient in part (a) to arrive at an expression for g (a). 19. Let f(x)= 3 x−5 . (a) Using the limit definition of derivative, find f (2). 5.2 The Derivative Function 187 (b) Find two ways of checking whether or not your answer is reasonable. These methods should not involve simply checking your algebra. They can be numerical or graphical—use your ingenuity. 20. Let f be a function that is locally linear. Let A = (5, f(5)) and B = (5 + h, f(5+h)). A is a fixed point on the graph of f while B is mobile; as h varies, B moves along the graph of f . A B (5, f(5)) (5+h, f(5+h)) x f The following statements are equivalent. Replace each question mark with the appropriate expression. (We use m sec as shorthand for “the slope of the secant line” and m tan as shorthand for “the slope of the tangent line.”) As B → A, (m sec through points A and B) → (m tan at point A). (a) As h → ? , ? ? → f ? (b) lim h→ ? ? ? = f ? For Problems 21 through 24 use the limit definition of f (a) to find the derivative of f at the point indicated. 21. f(x)= 3 2−x at x = 1 22. f(x)=x(x + 3) at x = 2 23. f(x)= x 2 + 2 x at x = 1 24. f(x)=(x − 3) 2 at x = 3 5.2 THE DERIVATIVE FUNCTION Definition The derivative function f (x) is given by f (x) = lim h→0 f(x +h) − f(x) h . 188 CHAPTER 5 The Derivative Function The function f (x) can be thought of as the slope function of f(x). An Alice-in- Wonderland exercise is useful. Imagine the smooth curve given by y = f(x) enlarged enormously. Shrink yourself to the size of a tiny bug and crawl along the curve from left to right. At each point you should feel as if you are crawling along a line. 7 As you crawl along the curve the slope changes, but from your bug’s perspective, it looks like a line with a given slope at each point. The slope is a function of x. The derivative gives a bug’s-eye view of the world. Let’s start by considering two simple cases. If f is a linear function, f(x)=mx + b, then f (x) = m; the slope function is the slope. If f is a constant function, f(x)=k, then f (x) = 0; the slope of a horizontal line is zero. EXERCISE 5.1 (a) Use the limit definition of f (x) to show that if f(x)= mx + b, then f (x) = m.In other words, verify that the derivative of a linear function gives the slope of the line. (b) Conclude that if f is the constant function, f(x)=k,then f (x) = 0. Answers are provided at the end of the section. In the example of the falling rock in Section 5.1, the rock’s velocity is a function of time. This is equivalent to saying that the slope of the position versus time graph is a function of t. From this graphical perspective you might observe that the slope is always negative; the graph is decreasing. As t increases, the slope (corresponding to the instantaneous velocity) gets more and more negative. The derivative function, f (t), which is both the velocity function and the slope function, is negative and decreasing. 1 2 3 4 t s the slope is more negative at t=3 than at t=2. (3, s(3)) (2, s(2)) Figure 5.11 7 As an analogy for those uncomfortable with the metamorphosis into a bug, if you were standing in the middle of a huge cornfield in a Midwestern plain, you would feel as if you were standing on a plane—completely oblivious to the roundness of the earth. 5.2 The Derivative Function 189 Given a function f(x),wecan interpret f in two ways: as the slope function that associates to each value of x the slope of the graph of f at (x, f(x)),orasthe function whose output is the instantaneous rate of change of f with respect to x. If y = f(x) we can think of f (x) both as lim h→0 f(x+h)−f(x) h and as lim x→0 y x where x = (x + h) − x = h and y is defined to be f(x +h) − f(x). x f ∆ y=f(x+h) – f(x) ∆ x y=f(x) (x, f(x)) (x+h, f(x+h)) Figure 5.12 ◆ EXAMPLE 5.3 Consider the graph of the function h = f(V)drawn below. This is a height versus volume graph for the bucket in the first set of exploratory problems. bucket height height versus volume for a bucket volume Figure 5.13 Characterize the function f (V ) and explain the implications of the answers in plain English. i. Is f positive, zero, or negative? ii. Is f increasing, or decreasing? iii. Graph f . SOLUTION i. f (V ) is positive. Graphical Reasoning: f (V ) is the slope of the graph and the graph always has positive slope. Implication: f (V ) = lim V →0 h V = lim V →0 height volume . It is the instantaneous rate of change of height with respect to volume. As the volume increases, the height increases. 190 CHAPTER 5 The Derivative Function ii. f (V ) is decreasing. Graphical Reasoning: The slope of the graph of f decreases as V increases. Implication: Because the bucket is widest at the top, as the volume increases the rate of change of the height with respect to volume decreases. iii. f ' volume Figure 5.14 ◆ Calculating the Derivative Function The limit definition of the derivative function is f (x) = lim h→0 f(x +h) − f(x) h . In the examples below we use this definition to calculate derivative functions. ◆ EXAMPLE 5.4 Suppose f(x)=x 2 +3x. (a) Find f (x). (b) Where does the graph of f have a horizontal tangent line? SOLUTION f (x) = lim h→0 f(x +h) − f(x) h = lim h→0 (x + h) 2 + 3(x + h) − (x 2 + 3x) h We know f(x)=x 2 +3x. = lim h→0 x 2 + 2hx + h 2 + 3x + 3h − x 2 − 3x h Multiply out in order to simplify. = lim h→0 2hx + 3h + h 2 h Simplify. = lim h→0 h(2x + h + 3) h Again, since h = 0, h h = 1. = lim h→0 2x + 3 + h = 2x + 3 Notice that knowing f (x) = 2x + 3 we now know that f (1) = 5, f (2) = 7 (as shown in Example 5.1), f (7) = 17, etc.; by finding f (x) we have a formula for determining the slope of the graph of f at any x-value. . Let P and Q be points on the graph of f with x-coordinates 3 and 3 + h, respectively. (a) Sketch the graph of f and the secant lines through P and Q for h =1 and h =0.1. (b) Find the slope of the. Let P and Q be points on the graph of f with x-coordinates 3 and 3 + h, respectively. (a) Sketch the graph of f and the secant lines through P and Q for h = 1 and h = 0.1. (b) Find the slope of. one of those on the right. The expressions on the right can be used more than once. (We use m sec as shorthand for “the slope of the secant line” and m tan as shorthand for “the slope of the tangent line.”) i) g(4+w)−g(4) w A.