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Math Help - Integral Problem

  1. #1
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    Integral Problem

    I seriously have no idea how to approach this, even after staring at it for an hour.

    A high-tech company purchases a new computing system whose intitial value is V. The system will depreciate at the rate f = f(t) where t is the time measured in months. Suppose that

    f(t) = \frac{V}{15} - \frac{V}{450}t if 0<t\leq{30}. If (note: if t>30, f(t) = 0)

    Determine the length of time T for the total depreciation D(t) = \int^t_0 f(s)ds to equal the initial value V.

    Can somebody please help me get this set up? I'm totally frustrated.

    Thanks.
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  2. #2
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    Quote Originally Posted by joatmon View Post
    I seriously have no idea how to approach this, even after staring at it for an hour.

    A high-tech company purchases a new computing system whose intitial value is V. The system will depreciate at the rate f = f(t) where t is the time measured in months. Suppose that

    f(t) = \frac{V}{15} - \frac{V}{450}t if 0<t\leq{30}. If (note: if t>30, f(t) = 0)

    Determine the length of time T for the total depreciation D(t) = \int^t_0 f(s)ds to equal the initial value V.

    Can somebody please help me get this set up? I'm totally frustrated.

    Thanks.
    initial value - total depreciation = 0

    \displaystyle V - \int_0^t \frac{V}{15} - \frac{V}{450} \cdot s \, ds = 0

    solve for t
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  3. #3
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    Thanks, Skeeter. I think that I got this, but I spent so much time spinning my wheels that it would be great if somebody would please look at the math to make sure that I didn't make a mistake.

    V - \int_0^t (\frac{V}{15} - \frac{V}{450} \cdot s) \, ds = 0

    V - \int_0^t \frac{V}{15}ds + \int_0^t \frac{V\cdot s}{450} ds = 0

    V - \frac{V \cdot t}{15} \, + \frac{V}{450} \int_0^t s\, ds = 0

    V - \frac{V \cdot t}{15} \, + \frac{V}{450} \cdot \frac{t^2}{2} = 0

    \frac{900V - 60Vt + Vt^2}{900} = 0

    \frac{V(t^2 - 60t +900)}{900} = 0

    \frac{V(t-30)^2}{900} = 0

    t = 30
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  4. #4
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    looks fine to me
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