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?

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Answer 1 · 2015-03-26T23:56:12

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

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.)

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