Calculus Vectors

1a. h=-4.9t^2+450

1b. h(t)=-4.9t^2+450
(h(2)-h(0))/(2-0)
((-4.9(〖2)〗^2+450)-(-4.9(0)^2+450))/2
=(430.4-450)/2
=-19.6
∴The average velocity for the first two seconds was 19.6 metres per second.

c. i) i)

= =-24.5
∴ The average velocity from is 24.5 metres/s. ii)

= -14.7 iii)

= -12.25
∴ The average velocity from is 12.25 metres/s.

d) Instantaneous velocity at 1s:

=-9.8
∴ The instantaneous velocity at 1s is 9.8 metres/s.

2a) = = = = =

b)

=
∴ The average rate of change from is -0.4g/s.

C)

∴The instantaneous rate at t = 2 seconds is -1.6g/s

3)
b)

= = =22
∴ from seconds the car moves at an average of 22m/s
c)
t=4 = =16
∴ The instantaneous rate at 4s is 16m/s

4a) In order to determine the instantaneous rate of change of a function using the methods discussed in this lesson, we would use the formula where h will approach 0, and the closer it gets to 0 the more accurate our answer will be.

4b)

∴

=1
Therefore, = 1

5a)

Therefore the instantaneous rate at x=2 is 0.
5b)

Therefore at t=4 the instantaneous rate is 0 and the particle is at rest.
6a)
Rate of change is positive when: Rate of change in negative when: 6b)
Rate of change is 0 when: X=-1, x=1
6c)
Local Maximum: (-1,2)
Local Minimum: (1,-2)

7a. i) Therefore, the limit as h approaches 0 for when x=0 is 4.

7a. ii) Therefore, the limit as h approaches 0 for when x=1 is -4.

7b) The function at x=0 is increasing since 4 is positive. The function at x=1 is decreasing since 4 is negative.
7c) I expect a local maximum because the function is increasing at x=o and decreasing at x=1.

8a) 8b)

8c)

8d) 9) By using the first principle