1. M 112 Short Course in Calculus Chapter 2 – Rate of Change: The Derivative Sections 2.1 – Instantaneous Rate of Change V. J. Motto

2. Instantaneous Rate of Change 8/13/20152

3. 3 So we have the function; s(t) = -16t 2 + 100 t + 6

4. Figure 2.1: Average velocities over intervals on either side of t = 1 showing successively smaller intervals Let’s look over smaller and smaller intervals in the neighborhood of t = 1 8/13/20154

5. Some basic definitions 8/13/20155

6. Example 1 (page 89) The quantity (in mg) of a drug in the blood at time t (in minutes) is given by Q = 25(0.8) t. Estimate the rate of change of the quantity at t = 3 and interpret your answer. What kind of function is this? What is the domain and range? 8/13/20156

7. Solution We estimate the rate of change at t = 3 by computing the average rate of change over intervals near t = 3. We can make our estimate as accurate as we like by choosing our intervals small enough. Let’s look at the average rate of change over the interval 3 ≤ t ≤ 3.01: 8/13/20157

8. Solution (continued) A reasonable estimate for the rate of change of the quantity at t = 3 is −2.85 mg/minute. 8/13/20158

9. Another Basic definition 8/13/20159

10. Example 2 (page 90) Estimate f′ (2) if f(x) = x 3. What does the graph of f look like? What is the domain and range of f? 8/13/201510

11. Solution (continued) Since f′(2) is the derivative, or rate of change, of f(x) = x 3 at 2, we look at the average rate of change over intervals near 2. Using the interval 2 ≤ x ≤ 2.001, we see that 8/13/201511

12. Figure 2.2: Visualizing the average rate of change of f between a and b Figure 2.3: Visualizing the instantaneous rate of change of f at a Visualizing the Derivative of a function 8/13/201512

13. 8/13/201513

14. Example 3 (page 90) Use a graph of f(x) = x 2 to determine whether each of the following quantities is positive, negative, or zero: (a)f′(1) (b)f′(−1) (c)f′(2) (d)f′(0) 8/13/201514

15. Solution What is the domain? What is the range? 8/13/201515

16. Solution (continued) Figure 2.5 shows tangent line segments to the graph of f(x) = x 2 at the points x = 1, x = −1, x = 2, and x = 0. Since the derivative is the slope of the tangent line at the point, we have: (a) f′(1) is positive. (b) f′(−1) is negative. (c) f′(2) is positive (and larger than f′(1)). (d) f′(0) = 0 since the graph has a horizontal tangent at x = 0. 8/13/201516

17. Example 2 (page 92) The graph of a function y = f(x) is shown in Figure 2.7. Indicate whether each of the following quantities is positive or negative, and illustrate your answers graphically. 8/13/201517

18. Solution Since there are no y-values, I can’t find a model for this function. I must work with the graph. (a) Since f ′(1) is the slope of the graph at x = 1, we see in Figure 2.8 that f′(1) is positive. 8/13/201518

19. Solution (continued) (b) The difference quotient is the slope of the secant line between x = 1 and x = 3. We see from Figure 2.9 that this slope is positive. 8/13/201519

20. Solution (continued) (c) Since f(4) is the value of the function at x = 4 and f(2) is the value of the function at x = 2, the expression f(4) − f(2) is the change in the function between x = 2 and x = 4. Since f(4) lies below f(2), this change is negative. See Figure 2.10. 8/13/201520

21. Example 4 (Page 92) The total acreage of farms in the US1 has decreased since 1980. See Table 2.2. a)What was the average rate of change in farm land between 1980 and 2000? b)Estimate f′(1995) and interpret your answer in terms of farm land. 8/13/201521

22. Solution (a)Between 1980 and 2000, million acres per year. Between 1980 and 2000, the amount of farm land was decreasing at an average rate of 4.7 million per year. 8/13/201522

23. Solution (continued) (b) We use the interval from 1995 to 2000 to estimate the instantaneous rate of change at 1995: million acres per year. In 1995, the amount of farm land was decreasing at a rate of approximately 3.6 million acres per year. 8/13/201523

24. An Alternative Solution Find a model for this data: 8/13/201524

25. Alternative Solution (continued) So we have the function f(x) = 0.05x 2 – 5.83x + 1039.31 Average rate of change is still the slope of the secant! But the can actually find the instantaneous rate of change --- the first derivative by using the function f and our calculator: That is f′(15) = -4.197. 8/13/201525

26. First Derivative Function TI 83/84 1.Math Button 2.Option 8:nDeriv( 3.When you press ENTER the function appears on the HOME Screen. We need to add the parameters. 4.nDeriv( y1, x, 15). The first parameter is found using the VARS button. Find the derivative with respect to the x variable, and the let x = 15. Don’t forget to close the parenthesis. 5.Press the Enter key to calculate 8/13/201526

27. First Derivative Function TI-89 1.From the HOME Screen press F3 to get to the calculus functions 2.Choose option 1 d( differentiate 3.Add the parameters y1(x) by typing all the characters. 4.Then add a comma followed by the variable x, and the close the parenthesis. 5.Add the characters |x=15 6.When you press ENTER to calculate 8/13/201527