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???
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Your question is a little unreadable. But I could see that you need to solve a differential equation with initial condition.
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
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? $\displaystyle r(t) = \frac{dr}{dt} = 0.05e^{0.02t}(200 - t)$ where $\displaystyle \, r(0)=1000$ and show that $\displaystyle r(t) = 2.5e^{.02t} (250-t)+375$ but why is that $\displaystyle \, r(t) = \frac{dr}{dt}$?
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