Question

The distance traveled, D in feet, is a function of time, t, in seconds, with

D=f(t)= t 3 +1
​Find f(10), f'(10), and the relative rate of change f'/f at t=10. Interpret your answers in terms of distance travele

Answer 1

1001 m/s

300
`m/s^2`

0.299.

Explanation:

Given that:

`D=f(t)=t^3+1`

`f(10)` means putting the value of
`t=10` in
`f(t)`.

`f(10) = 10^3+1=1001`
`m/s`

To find
`f'(10)`, we need to find
`f'(t)` first.

Differentiating the equation of Distance
`D` w.r.to
`t`:

`f'(t) = (d)/(dt)(t^3+1)= 3t^(3-1)+1 = 3t^2`

`f'(10) = 3(10)^2=300`
`m/s^2`

Relative rate of change:

`(f')/(f)\ at\ t = 10 \Rightarrow (300)/(1001) \approx 0.299`