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Thread: Difference Quotient Problem.

  1. #1
    Nov 2009

    Difference Quotient Problem.

    A carpenter is constructing a large cubical storage shed. The volume of the shed is given by V(x) = x^2[(16-x^2)/4x], where x is the side length, in meteres.

    a) Simplify the expression for the volume of the shed.

    I had no clue how to simplify it, but the answer is V(x)=4x-1/4x^3

    b) Determine the average rate of change of the volume of the shed when the side lengths are between 1.5m and 3m.

    Am I supposed to use the difference quotient to find average rate of change? I assumed so, so this is my work which ended up being wrong:


    delta y/delta x= [f(x+h)-f(x)]/h

    =[[4(x+h)-1/4(x+h)^3]-[4(x)-1/4(x)^3]]/h *is that what i was supposed to do..? so skipping a few steps I got:

    =-1/4x^3 - 3/4x^2h - 3/4xh^2 - 1/4h^3 +4x +4h *skipping a few more steps:

    =-3/4x^2 - 3/4xh - 1/4h^2 + 4

    So I know to sub in 1.5 for h, since 1.5m is the interval between the two points (1.5 and 3). But what would I sub in for x? I chose to sub in 1.5 for it also.. can someone explain why if thats right?

    So when I subbed those numbers in I get -7.75.. the answer is 0.0625m^3/m

    c) Determine the instantaneous rate of change of the colume of the shed when the side length is 3m.

    Would I just sub in 3 for x and 0.000001 (a number really close to 0) for h, since the instantaneous interval is 0? Answer is -2.75m^3/m

    I'm having a bit of trouble doing difference quotient stuff =/
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  2. #2
    Super Member
    Jul 2009
    I skipped over the bolded stuff. If you have learned about "deltas", then surely you have learned how to take the derivative of a function? The rate of change of a quantity with respect to another quantity, is simply the derivative. Here you are meant to calculate the rate at which the volume of an object is changing with respect to its dimension; so, using the simplified expression, find the derivative. Once you have the derivative, it is simply a matter of plugging in the values of the sides they give you, so that you may calculate the rate at which the volume is changing at a given side length.

    If you are meant to take the difference quotient of this, the answer I get is $\displaystyle 16-x^{2}$. When taking the difference quotient, and then applying the limit as it approaches zero, your h's will cancel out, leaving you with an equation in x.
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