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Thread: Calc- Word Problem

  1. #1
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    Calc- Word Problem

    The function $\displaystyle C(t) = e^{-2t} - e^{-5t}, t \geq 0$ can be used to model the concentration at the time $\displaystyle t$ (where $\displaystyle t$ 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?

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

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

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

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


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

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

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

    If I took the derivative correctly, I don't know how to simplify it further
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Macleef View Post
    The function $\displaystyle C(t) = e^{-2t} - e^{-5t}, t \geq 0$ can be used to model the concentration at the time $\displaystyle t$ (where $\displaystyle t$ 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?

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

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

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

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


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

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

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

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

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

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

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

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

    we see that

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