1. Velocity, Speed, and Rates of Change
AP Calculus
Mrs. Mongold

2. Definitions
Instantaneous Rate of Change: the instantaneous rate of change of f with respect to x at a is the derivative
Motion along a line
Position Function
Displacement
Average Velocity

3. Instantaneous Velocity- tells how fast an object is moving as well as the direction of motion
Velocity is the derivative of the position function

4. Velocity is the derivative of the position function
Instantaneous Velocity- tells how fast an object is moving as well as the direction of motion
Velocity is the derivative of the position function
When moving forward (s is increasing) and velocity is positive
When moving backward (s is decreasing) and velocity is negative

5. Speed

6. Speed- is the absolute value of velocity and tells how fast regardless of direction

7. Acceleration -

8. Acceleration – the derivative of velocity

9. Consider a graph of displacement (distance traveled) vs. time.
Average velocity can be found by taking:
time (hours)
distance
(miles) B A
The speedometer in your car does not measure average velocity, but instantaneous velocity.
(The velocity at one moment in time.)

10. Velocity is the first derivative of position.

11. Speed is the absolute value of velocity.
Example:
Free Fall Equation
Gravitational
Constants:
Speed is the absolute value of velocity.

12. Acceleration is the derivative of velocity.
example:
If distance is in:
Velocity would be in:
Acceleration would be in:

13. It is important to understand the relationship between a position graph, velocity and acceleration:
time
distance
acc neg
vel pos &
decreasing
acc neg
vel neg &
decreasing
acc zero
vel neg &
constant
acc zero
vel pos &
constant
acc pos
vel neg &
increasing
velocity
zero
acc pos
vel pos &
increasing
acc zero,
velocity zero

14. Rates of Change:
Average rate of change =
Instantaneous rate of change =
These definitions are true for any function.
( x does not have to represent time. )

15. Instantaneous rate of change of the area with respect to the radius.
Example 1:
For a circle:
Instantaneous rate of change of the area with
respect to the radius.
For tree ring growth, if the change in area is constant then dr must get smaller as r gets larger.

16. from Economics:
Marginal cost is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit.

17. Suppose it costs: to produce x stoves.
Example 13:
Suppose it costs:
to produce x stoves.
If you are currently producing 10 stoves, the 11th stove will cost approximately:
Note that this is not a great approximation – Don’t let that bother you.
The actual cost is:
marginal cost
actual cost

18. Note that this is not a great approximation – Don’t let that bother you.
Marginal cost is a linear approximation of a curved function. For large values it gives a good approximation of the cost of producing the next item.

19. Things to Remember
Velocity is the first derivative (how fast an object is moving as well as the direction of motion)
Speed is the absolute value of the first derivative (tells how fast regardless of direction)
Acceleration is the second derivative

20. Examples
Find the rate of change of the area of a circle with respect to its radius
A=πr2
Evaluate the rate of change at r = 5 and r = 10
What units would be appropriate if r is in inches

21. Examples

22. Examples

23. Examples

24. Examples Find the rate of change of volume of a cube with respect to s
V=s3
Evaluate rate of change at s=1 , s=3, and s=5
What would units be if measured in inches

25. Examples

26. Examples

27. Examples

28. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

29. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

30. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

31. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

32. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

33. Example
Suppose a ball is dropped from the upper observation deck, 450 m above ground. What is the velocity of the ball after 5 seconds?

34. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go?

35. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0

36. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0
v = 160 – 32t

37. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0
v = 160 – 32t
0 = 160 – 32t

38. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0
v = 160 – 32t
0 = 160 – 32t
32t=160

39. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0
v = 160 – 32t
0 = 160 – 32t
32t=160
t = 5
Max height at t = 5 so max position 160(5)-16(25)

40. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How high does it go? Highest when velocity=0
v = 160 – 32t
0 = 160 – 32t
32t=160
t = 5
Max height at t = 5 so max position 160(5)-16(25)
The object reaches a height of 400 ft.

41. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256

42. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256
– 16t2+160t = 0

43. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256
– 16t2+160t = 0
16(t2 -10t +16)=0

44. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256
– 16t2+160t = 0
16(t2 -10t +16)=0
(t - 2)(t - 8)=0

45. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256
– 16t2+160t = 0
16(t2 -10t +16)=0
(t - 2)(t - 8)=0
t = 2 and t = 8

46. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
160t – 16t2 = 256
– 16t2+160t = 0
16(t2 -10t +16)=0
(t - 2)(t - 8)=0
t = 2 and t = 8
V(2)= 160 – 32(2) = 96 ft/sec
V(8)= (8) = -96 ft/sec
Speed at both instances are 96

47. Example

48. Example

49. Example Note the acceleration is always downward…when the rock is
Note the acceleration is always downward…when the rock is
Rising, it is slowing down and when it is falling it is speeding up.

50. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?

51. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0

52. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0
160t - 16t2 = 0

53. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0
160t - 16t2 = 0
16t t = 0

54. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0
160t - 16t2 = 0
16t t = 0
16t(t – 10) = 0

55. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0
160t - 16t2 = 0
16t t = 0
16t(t – 10) = 0
t = 0 and 10

56. Example
A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft. after t seconds.
How long does it take for the rock to return to the ground?
Hits the ground when position = 0
160t - 16t2 = 0
16t t = 0
16t(t – 10) = 0
t = 0 and 10
The rock returns to the ground after 10 seconds

57. Homework
worksheet