1. DP Studies Y2

2. A. Rates of change B. Instantaneous rates of change C. The derivative function D. Rules of differentiation E. Equations of tangents F. Normals to curves

3. Valentino is riding his motorbike around a racetrack. A computer chip on his bike measures the distance Valentino has travelled as time goes on. This data is used to plot a graph of Valentino’s progress. Things to think about: a. What is meant by a rate? b. What do we call the rate at which Valentino is travelling? c. What is the difference between an instantaneous rate and an average rate? d. How can we read a rate from a graph? e. How can we identify the fastest part of the racetrack?

4. A rate is a comparison between two quantities of different kinds. For examples: 1. the speed at which a car is travelling in kmh -1 or ms -1. 2. the fuel efficiency of a car in kmL -1 or liters per 100 km travelled. 3. the scoring rate of a basketball player in points per game.

5. Example 1: Josef typed 213 words in 3 minutes and made 6 errors, whereas Marie typed 260 words in 4 minutes and made 7 errors. Compare their performance using rates.

6. Solution to example 1:

7. If the graph which compares two quantities is a straight line, there is a constant rate of change in one quantity with respect to the other. This constant rate is the gradient of the straight line. example:

8. If the graph is a curve, we can find the average rate of change between two points by finding the gradient of the chord or line segment between them. The average rate of change will vary depending on which two points are chosen, so it makes sense to talk about the average rate of change over a particular interval.

9. Example 2: The number of mice in a colony was recorded on a weekly basis. a. Estimate the average rate of increase in population for: i. the period from week 3 to week 6 ii. the seven week period. b. What is the overall trend in the population growth over this period?

10. b. The graph is increasing over the period by larger and larger amounts, so the population is increasing at an ever increasing rate.

11. The instantaneous rate of change of a variable at a particular instant is given by the gradient of the tangent to the graph at that point. Example: the graph alongside shows how a cyclist accelerates away from an intersection. The average speed over the first 8 seconds is 100 m/8 sec = 12.5 ms -1. Notice that the cyclist’s early speed is quite small, but it increases as time goes by.

12. Instantaneous rate of change example continue, To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent to the graph at that time and find its gradient.

13. Finding the tangent gradient algebraically. Consider the curve y = x 2 and the tangent at F(1, 1). Let the moving point M have x-coordinate 1 + h, where h ≠ 0. So, M is at (1 + h, (1 + h) 2 ). As point M gets closer and closer to point F the horizontal distance between the two points, h go to zero and the slope of the chord becomes 2.

14. Example 3: Use the algebraic method to find the gradient of the tangent to y = x 2 at the point where x = 2.

15. Solution to example 3:

16. We can hence describe a gradient function which, for any given value of x, gives the gradient of the tangent at that point. We call this gradient function the derived function or derivative function of the curve. If we are given y in terms of x, we represent the derivative function by dy/dx. If we are given the function f(x), we represent the derivative function by f‘(x).

17. The power rule of derivative

18. Example 4:

19. Solutions to example 4:

20. Example 5:

21. Solutions to example 5:

22. Differentiation is the process of finding a derivative or gradient function.

23. Example 6:

24. Solutions to example 6: a.f(x) = 5x 3 + 6x 2 – 3x + 2 therefore, f’(x) = 5(3x 2 ) + 6(2x) – 3(1) = 15x 2 + 12x – 3

25. Example 7:

26. Solution to example 7:

27. Example 8:

28. Using the Ti-84 calculator, we can determine the derivative of a function by using the “CALCULATE” command on the graphing menu. step 1: input the equation into y = step 2: graph step 3: 2 nd TRACE = CALC step 4: choose “6” step 5: input x-value

29. y= x^3 + 3 Graph 2 nd TRACE 6 -2 dy/dx = 12.000001 So the gradient is 12 at x = -2 for y = x 3 + 3

30. Example 9:

31. Solution to example 9:

32. Example 10:

33. Solution to example 10:

34. Example 11: The tangent to f(x) = 2x 2 – ax + b at the point (2, 7) has a gradient of 3. Find a and b.

35. Solution to example 11:

36. The equation of the tangent at point A(a, b) is A(a, b)

37. Example 12: Find the equation of the tangent to f(x) = x 2 + 1 at the point where x = 1.

38. Solution to example 12:

39. Example 13: Use technology to find the equation of the tangent to y = x 3 – 7x + 3 at the point where x = 2.

40. Solution to example 13: Using Ti-84,y= functionx^3 – 7x + 3 graph 2 nd PRGM choice “5”: Tangent( input the x-value2 read y = ax + b5.000001x + -13.000002 So the equation of the tangent for y = x 3 – 7x + 3 at the point x = 2 is y = 5x – 13

41. Example 14: Consider the curve y = x 3 – 4x 2 – 6x + 8. a. Find the equation of the tangent to this curve at the point where x = 0. b. At what point does this tangent meet the curve again?

42. Solution s to example 14

43. In order to solve for part b, we graph both the original function and the tangent equation from part a. y 1 = x 3 – 4x 2 – 6x + 8 y 2 = -6x + 8 Use the 2 nd TRACE command Choice 5 enter x 3 and we should get x = 4, y = -16. So the tangent meets the curve again at (4, -16)

44. A normal to a curve is a line which is perpendicular to the tangent at the point of contact.

45. Example1 5: Find the equation of the normal to f(x) = x 2 – 4x + 3 at the point where x = 4.

46. Solution to example 15:

47. Example 16: Find the coordinates of the point where the normal to y = x 2 – 3 at (1, -2) meets the curve again.

48. Solution to example 16: Using the calculator and the equation above, we can find the answer by graphing the two equations and find the intersect. The intersect occurs at (-1.5, -0.75)