# 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?

Mar 26, 2015

$P \left(t\right) = \frac{50}{3} {t}^{3} - 40 {t}^{\frac{5}{2}} + 25000$

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

Let $P \left(t\right)$ be the population $t$ years after the initial population was counted. (So $P \left(0\right) = 25000$)

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 $50 {t}^{2} - 100 {t}^{\frac{3}{2}}$.

This tells us that
$P ' \left(t\right) = 50 {t}^{2} - 100 {t}^{\frac{3}{2}}$.

Our task is to find $P \left(t\right)$

If $f ' \left(x\right) = {x}^{n}$, the $f \left(x\right) = {x}^{n + 1} / \left(n + 1\right)$

So $P \left(t\right) = 50 {t}^{3} / 3 - 100 \cdot \frac{2}{5} {t}^{\frac{5}{2}} + C$ . . (See below if needed)

$P \left(t\right) = \frac{50}{3} {t}^{3} - 40 {t}^{\frac{5}{2}} + C$

We also know that $P \left(0\right) = 25000$, so we can substitute to find $C$

$25000 = \frac{50}{3} {\left(0\right)}^{3} - 40 {\left(0\right)}^{\frac{5}{2}} + C$, so $C = 25000$
$P \left(t\right) = \frac{50}{3} {t}^{3} - 40 {t}^{\frac{5}{2}} + 25000$

I think writing all of the following clutters up the work:
To find the antiderivative of ${t}^{\frac{3}{2}}$:

${t}^{\frac{3}{2} + 1} / \left(\frac{3}{2} + 1\right) = {t}^{\frac{5}{2}} / \left(\frac{5}{2}\right) = \frac{2}{5} {t}^{\frac{5}{2}}$

(dividing is multiplying by the reciprocal.)