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Math Help - Depreciation/Rate of Change (Calculus I)

  1. #1
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    Angry Depreciation/Rate of Change (Calculus I)

    Problem: The value of a piece of machinery depreciates exponentially. When purchased new, it was worth $70,000. After 4 years, it was worth $55,200. Find the rate of change of the value of the equipment after 4 years.

    I started with:

    4r = ln (55,200/70,000)
    so..
    4r = -0.2375322...
    divide by 4 on both sides..
    r = -.05938...
    multiply by initial value (70000)
    I got:
    $4,156.60

    which was not an answer choice, lol

    I'm obviously setting this up wrong, any help is appreciated
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  2. #2
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    Well if it is depreciation in business it is the same every year so

    70,000-55200= 14800

    so divide by 4

    14800/4= $3700
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  3. #3
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    Lexington, MA (USA)
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    Hello, DirtMcGirt!

    The value of a piece of machinery depreciates exponentially.
    When purchased new, it was worth $70,000. After 4 years, it was worth $55,200.
    Find the rate of change of the value of the equipment after 4 years.

    We have: . V \;=\;70,\!000\;\!e^{-kt}


    When t = 4,\;V = 55,\!200

    Then: . 55,\!200 \:=\:70,\!000\;\!e^{-4k} \quad\Rightarrow\quad e^{-4k} \:=\:\tfrac{55,\!200}{70,\!000} \:=\:\tfrac{138}{175}

    Hence: . -4k \:=\:\ln\left(\tfrac{138}{175}\right) \quad\Rightarrow\quad t \:=\:-\tfrac{1}{4}\ln\left(\tfrac{138}{175}\right) \;\approx\; 0.059

    . . The function is: . V \;=\;70,\!000\;\!e^{-0.059t}


    The rate of change is: . \frac{dV}{dt} \;=\;70,\!000(-0.059)\;\!e^{-0.059t} \;=\;-4130\;\!e^{-0.059t}

    When t = 4\!:\;\;\frac{dV}{dt} \;=\;-4130\;\!e^{-0.059(4)} \;\approx\;\boxed{-3261.79}


    At the end of 4 years, it is depreciating at the rate of $3261.79 per year.

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