Question

If the instantaneous rate of change of a population is `50t^2 - 100t^(3/2)` (measured in individuals per year) and the initial population is 25000 then what is the population after t years?

Answer 1

`P(t)= 50/3t^3-40t^(5/2)+25000`

To answer this question, you need to find a particular antiderivative.

Let `P(t)` be the population `t` years after the initial population was counted. (So `P(0)=25000`)

We are told that the population is changing at a rate -- stop! that means we're going to be told something about the derivative (the rate of change).

The population is changing at a rate of `50t^2-100t^(3/2)`.

This tells us that
`P'(t)=50t^2-100t^(3/2)`.

Our task is to find `P(t)`

If `f'(x)=x^n`, the `f(x)=x^(n+1)/(n+1)`

So `P(t)=50t^3/3-100*2/5 t^(5/2)+C` . . (See below if needed)

`P(t)=50/3t^3-40 t^(5/2)+C`

We also know that `P(0)=25000`, so we can substitute to find `C`

`25000=50/3(0)^3-40 (0)^(5/2)+C`, so `C=25000`
`P(t)= 50/3t^3-40t^(5/2)+25000`

I think writing all of the following clutters up the work:
To find the antiderivative of `t^(3/2)`:

`t^(3/2+1)/(3/2+1)=t^(5/2)/(5/2)=2/5 t^(5/2)`

(dividing is multiplying by the reciprocal.)