1. Chapter 17 Applications of differential calculus A Kinematics
B Rates of change
C Optimization

3. Displacement
Displacement - the moving of something from its place or position. s(t) is the displacement function which represents the separation between the origin and point P with respect to time. On the horizontal axis through O: if s(t) > 0, P is located to the right of O if s(t) = 0, P is located at O if S(t) < 0, P is located to the left of O.

4. Motion Graph
A motion graph is a horizontal graph showing the location of the object in motion with respect to time.
s(t) = t2 + 2t – 3
t = 0
s(0) = -3 cm
t = 1
s(1) = 0 cm
t = 2
s(2) = 5 cm
t = 3
s(3) = 12 cm
t = 4
s(4) = 21 cm

5. Velocity
The average velocity of an object moving in a straight line in the time interval from t = t1 to t = t2 is the ratio of the change in displacement to the time taken. If s(t) is the displacement function then average velocity is If s(t) is the displacement function of an object moving in a straight line, then v(t) = s’(t) is the instantaneous velocity or velocity function of the object at time t.

6. Acceleration
If an object moves in a straight line with velocity function v(t) then its average acceleration for the time interval from t = t1 to t = t2 is the ratio of the change in velocity to the time taken. If a particle moves in a straight line with velocity function v(t), then the instantaneous acceleration at time t is a(t) = v’(t).

7. Units
Units of velocity is m/s or ms-1 Units of acceleration is m/s2 or ms-2

8. Example 1

12. Exercise 17A.1 #1

15. The instantaneous velocity at t = 1 is 5 m/s

16. The instantaneous velocity of object with respect to time

17. Velocity and Acceleration Function
If a particle P moves in a straight line and its position is given by the displacement function s(t), t > 0, then:
1st derivative
2nd derivative

18. Sign Interpretation
Suppose a particle P moves in a straight line with displacement function s(t) relative to an origin O. Its velocity function is v(t) and its acceleration function is a(t). We can use sign diagrams to interpret: a. where the particle is located relative to O b. the direction of motion and where a change of direction occurs c. when the particle’s velocity is increasing or decreasing.

19. Sign of s(t)
Location of the object with respect to the original location.

20. Sign of v(t)
The direction of the velocity of the object.

21. Sign of a(t)
Increasing or decreasing velocity. (getting faster or slowing down)

22. Frame of reference of the “0”

23. Speed
Velocity of a vector quantity which means it has both magnitude and direction. Velocity may have positive or negative values depending on its direction. Speed is the magnitude of the velocity which means it has no direction and can never be negative.

24. Speed
To determine when the speed S(t) of an object P with displacement s(t) is increasing or decreasing, we need to employ a sign test.

25. Example 2

29. B. Rate of Change
Examples of other real world rates of change temperature varies continuously the height of a tree varies as it grows the prices of stocks and shares vary with each day’s trading.

30. Example 3

33. EXERCISE 17B #1

34. Example 4

36. Exercise 17B. #5

37. Example 5

38. Area = ½ bc sin A
(chapter 9 part A)

39. C. Optimization
There are many problems for which we need to find the maximum or minimum value of a function. The solution is often referred to as the optimum solution and the process is called optimization. We can find optimum solutions in several ways: using technology to graph the function and search for the maximum or minimum value using analytical methods such as the formula x =-b/2a for the vertex of a parabola using differential calculus to locate the turning points of a function. These last two methods are useful especially when exact solutions are required.

40. Warning!!
The maximum or minimum value does not always occur when the first derivative is zero. It is essential to also examine the values of the function at the endpoint(s) of the interval under consideration for global maxima and minima.

41. OPTIMIZATION PROBLEM SOLVING METHOD

42. Example 6
A rectangular cake dish is made by cutting out squares from the corners of a 25 cm by 40 cm rectangle of tin-plate, and then folding the metal to form the container. What size squares must be cut out to produce the cake dish of maximum volume?

43. Step 1

44. Step 2
Find the volume of the dish.

45. Step 3
Determine the x-value of the stationary points.

46. Step 4
Draw a sign diagram to determine the maximum.

47. Step 5
Summarize the answer in a sentence. The maximum volume is obtained when x = 5, which is when 5 cm squares are cut from the corners. Note: If they ask for the maximum volume, then simply substitute 5 in for the x-value and solve for the volume.

48. Example 7
A 4 liter container must have a square base, vertical sides, and an open top. Find the most economical shape which minimizes the surface area of material needed.

49. Step 1

50. Step 2

51. Step 3

52. Step 4

53. Step 5
The most economical shape has a square base 20 cm x 20 cm, and height 10 cm.

54. Example 8

57. Exercise 17C #2