# Calc- Word Problem

• Jun 4th 2008, 05:04 PM
Macleef
Calc- Word Problem
The function $C(t) = e^{-2t} - e^{-5t}, t \geq 0$ can be used to model the concentration at the time $t$ (where $t$ is measured in hours), of a drug injected into the bloodstream. Determine when the concentration level in the body is a max.

I'm not sure if I took the 1st derivative of the polynomial correctly?

$C'(t) = e^{-2t}(-2) - e^{-5t}(-5)$

$C'(t) = -2e^{-2t} + 5e^{-5t}$

$0 = -2e^{-2t} + 5e^{-5t}$

$0 = \frac {-2}{e^{2t}} + \frac {5}{e^{5t}}$

$0 = \frac {-2e^{5t} + 5e^{2t}}{e^{10t}}$

$0 = -2e^{5t} + 5e^{2t}$

$0 = -2e^{5t} + 5e^{2t}$

If I took the derivative correctly, I don't know how to simplify it further
• Jun 4th 2008, 05:15 PM
Mathstud28
Quote:

Originally Posted by Macleef
The function $C(t) = e^{-2t} - e^{-5t}, t \geq 0$ can be used to model the concentration at the time $t$ (where $t$ is measured in hours), of a drug injected into the bloodstream. Determine when the concentration level in the body is a max.

I'm not sure if I took the 1st derivative of the polynomial correctly?

$C'(t) = e^{-2t}(-2) - e^{-5t}(-5)$

$C'(t) = -2e^{-2t} + 5e^{-5t}$

$0 = -2e^{-2t} + 5e^{-5t}$

$0 = \frac {-2}{e^{2t}} + \frac {5}{e^{5t}}$

$0 = \frac {-2e^{5t} + 5e^{2t}}{e^{10t}}$

$0 = -2e^{5t} + 5e^{2t}$

$0 = -2e^{5t} + 5e^{2t}$

If I took the derivative correctly, I don't know how to simplify it further

you need to solve that

$0=5e^{2t}-2e^{5t}$

Factoring out a $e^{2t}$ we get

$0=e^{2t}(5-2e^{3t})$

Now applying the zero product property and the fact that $e^{u(x)}\ne{0}$

we see that

$5-2e^{3t}=0\Rightarrow{e^{3t}=\frac{5}{2}}\Rightarro w{t=\frac{\ln(\frac{5}{2})}{3}}$