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Math Help - Differentiation/rate of chaange

  1. #1
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    Differentiation/rate of chaange

    A certain industrial machine depreciates so that its value t years after purchase is given by
    Q(t) = 75,000 e -0.05t

    i) What was its value at the time of purchase?

    ii) What will its value be 10 years after purchase?
    iii) At what rate is the value of the machine depreciating after t years? Hint: differentiate Q(t).

    iv) Does this rate depend on t or is it a constant?

    v) At what rate is the value of the machine depreciating after 10 years

    help needed please.
    Last edited by shizelle; March 24th 2010 at 01:36 PM. Reason: forgot to put in something
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  2. #2
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    Quote Originally Posted by shizelle View Post

    i) What was its value at the time of purchase?
    make  t = 0

    Quote Originally Posted by shizelle View Post

    ii) What will its value be 10 years after purchase?
    help needed please.
    make  t = 10
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  3. #3
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    answer to question part 1

    Q(0)=75000 e^-0.05*0
    =75000
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  4. #4
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    Quote Originally Posted by shizelle View Post
    Q(0)=75000 e^-0.05*0
    =75000



    Now do the second one

    75000e^{-0.05\times 10}= \dots
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  5. #5
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    part 2 solution

    Q(10)=75000 e^-0.5
    =45489.8
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  6. #6
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    Differentiation & integration

    A manufacturer finds that the total cost of producing a product is given by the cost function c(q) = 0.05q2 + 5q + 500. At what level of output will average cost per unit be a minimum?


    Evaluate the integral below between the limits of 1 and 10

    ( vx + 3 x -1 2ex) dx
    HELP NEEDED PLEASE.
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  7. #7
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    Quote Originally Posted by shizelle View Post


    Evaluate the integral below between the limits of 1 and 10

    ( vx + 3 x -1 2ex) dx
    HELP NEEDED PLEASE.
    Is this the integral?

    \int_0^{10} vx+3x-1-2e^x~dx

    You should start new questions in a separate thread.
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  8. #8
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    I am not sure, that's how they gave us the question but I believe it is suppposed to be that way. I cant get through to posting a new thread.
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  9. #9
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    I'm gonna guess that -.05t is the exponent. Based on that assumption, the value at the time of purchase, in other words--at t=0, is 75000e^-.05x0, which is 75000, because .05 times 0 = 0, and anything to the xero power is at t=10, you just plug 7500o(e^-.05*10) into your calculator.

    the derivative of any function gives the function's instantaneous rate of change at a point. SO to find the rate of depreciation at time t, meaning to find an equation that will give instantaneous rate of change at any value of t, find d/dt(f(x)), or the first derivative of the equation. That is -3750e^(-.05t).

    You can logically determine that the rate of change depends on t, because t is in the exponent. If it were constant the equation would be constant , like y=2, or linear, like y=mx+b, right?

    Okie dokie. Plug 10 in for t in the first derivative equation, and you'll get you're rate of change at t=10.

    Happy deriving! Don't forget your units in your solutions!
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  10. #10
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    Quote Originally Posted by pickslides View Post
    Is this the integral?

    \int_0^{10} vx+3x-1-2e^x~dx

    You should start new questions in a separate thread.
    Yes that is how it is supposed to be, i guess.
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