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Math Help - rate of change

  1. #1
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    rate of change

    Hey, I'm doing some homework and I'm a little confused with this question.

    Find the rate of change of the volume V of a cube with respect to
    (a) the length w of a diagonal on one of the faces.
    (b) the length z of the one of the diagonals of the cube.

    So i'm trying to find  \frac{dv}{dw}
    I know that the Volume of a cube is  v=x^3

    What i need to do is find a way of expressing V in terms of w.

    I know that  x^2 + x^2 = w^2 and solving for x I get  x= \frac{w}{2} . Unless my math is incorrect (which is very possible).

    The next step is the plug that into  v=x^3 so i get  v=(\frac{w}{2})^3

    Do i just find the derivative of that to get answer? I know that's wrong somewhere since the answer is  (\frac{3 (2^(\frac{1}{2})}{4})w^2

    Can anyone tell me what i'm doing wrong? thank you very much
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  2. #2
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    Quote Originally Posted by discobob View Post
    Hey, I'm doing some homework and I'm a little confused with this question.

    Find the rate of change of the volume V of a cube with respect to
    (a) the length w of a diagonal on one of the faces.
    (b) the length z of the one of the diagonals of the cube.
    V = x^3

    w^2 = x^2 + x^2

    w^2 = 2x^2

    w = \sqrt{2} \cdot x

    x = \frac{w}{\sqrt{2}}

    V = \frac{w^3}{2\sqrt{2}}

    find \frac{dV}{dw}



    z^2 = w^2 + x^2

    z^2 = w^2 + \frac{w^2}{2}

    z^2 = \frac{3w^2}{2}

    w = \sqrt{\frac{2}{3}} \cdot z

    use the chain rule ...

    \frac{dV}{dz} = \frac{dV}{dw} \cdot \frac{dw}{dz}
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