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Math Help - Simple chain rule problem

  1. #1
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    Simple chain rule problem

    The question:

    Let z = \sqrt{x^2 + y^2}, x = t^2, y = t^3. Use a chain rule to find dz/dt.

    My attempt:
    \frac{dz}{dt} = \frac{\partial{z}}{\partial{x}}\frac{dx}{dt} + \frac{\partial{z}}{\partial{y}}\frac{dy}{dt}

    \frac{\partial{z}}{\partial{x}} = \frac{x}{\sqrt{x^2 + y^2}}
    \frac{\partial{z}}{\partial{y}} = \frac{y}{\sqrt{x^2 + y^2}}
    \frac{dx}{dt} = 2t
    \frac{dy}{dt} = 3t^2

    So we get:
    \frac{t^2}{\sqrt{t^4 + t^6}}(2t) + \frac{t^3}{\sqrt{t^4 + t^6}}(3t^2)
    = \frac{t^2(2t + 3t^3)}{t^2\sqrt{1 + t^2}}
    = \frac{2t + 3t^3}{\sqrt{1 + t^2}}

    However, the answer is apparently:
    \frac{2t^2 + 3t^4}{\sqrt{1 + t^2}}

    What have I done incorrectly? Thanks.
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  2. #2
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    You're overcomplicating this...

    \displaystyle z = \sqrt{t^4 + t^6}.


    Let \displaystyle u = t^4 + t^6 so that \displaystyle z = u^{\frac{1}{2}}.

    Go from here.
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  3. #3
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    I'd say you are right. An easy way to check is to substitute x(t) and y(t) into the original equation at the start and find dz/dt directly.
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  4. #4
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    Ahh yes, that is a lot easier. I guess I got caught up with all this partial derivative stuff at uni.
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