1. ## integral calculas???

r(t)=dr/dt=0.05e^.02t (200-t)

r(0)=1000

using integral calculas i need to show that r(t)=2.5e^.02t (250-t)+375

How is this found???

But I could see that you need to solve a differential equation with initial condition.

3. ## is this a little easier to understand???

If the rate of growth of the number of wasps at time t days is given by -

r(t)=dr/dt=0.05e^.02t (200-t), r(0)=1000

show that the number of reproductive wasp r(t) at time t equals -

2.5e^.02t (250-t)+375

4. Originally Posted by softdrink_jr
If the rate of growth of the number of wasps at time t days is given by -

r(t)=dr/dt=0.05e^.02t (200-t), r(0)=1000

show that the number of reproductive wasp r(t) at time t equals -

2.5e^.02t (250-t)+375
is this what you want?

$r(t) = \frac{dr}{dt} = 0.05e^{0.02t}(200 - t)$
where $\, r(0)=1000$

and show that
$r(t) = 2.5e^{.02t} (250-t)+375$

but why is that $\, r(t) = \frac{dr}{dt}$?