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Math Help - implicit differentiation help

  1. #1
    Junior Member TheMathTham's Avatar
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    implicit differentiation help

    The exact wording of the problem is: "Determine the slope of the normal line to the curve x^{3}+xy^{2} = 10y at the point (2,1)"

    Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

    I got: 3x^{2}+y^{2}y{}'+x2yy{}' = 10y{}'
    Which would mean that \frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}

    However, when I plugged in x and y, I got \frac{12}{5}, which was not among the answer choices of 0, 2, -\frac{7}{3}, -\frac{6}{13}, or \frac{1}{2}.
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    Quote Originally Posted by TheMathTham View Post
    The exact wording of the problem is: "Determine the slope of the normal line to the curve x^{3}+xy^{2} = 10y at the point (2,1)"

    Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

    I got: 3x^{2}+ {\color{red}y^{2}y{}'} +x2yy{}' = 10y{}' Mr F says: The red term is wrong.

    Which would mean that \frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}

    However, when I plugged in x and y, I got \frac{12}{5}, which was not among the answer choices of 0, 2, -\frac{7}{3}, -\frac{6}{13}, or \frac{1}{2}.
    ..
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    Quote Originally Posted by TheMathTham View Post
    The exact wording of the problem is: "Determine the slope of the normal line to the curve x^{3}+xy^{2} = 10y at the point (2,1)"

    Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

    I got: 3x^{2}+y^{2}y{}'+x2yy{}' = 10y{}'
    Which would mean that \frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}

    However, when I plugged in x and y, I got \frac{12}{5}, which was not among the answer choices of 0, 2, -\frac{7}{3}, -\frac{6}{13}, or \frac{1}{2}.
    The slope of the normal to the tangent is given by the equation km = -1.

    So if the slope of the tangent is k= 13/6 then the slope of the tangent is m = -6/13.

    Check your implicit differantiation.

    \frac{d(xy^2)}{dx}= y^2 + 2xy\frac{dy}{dx}
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    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by xalk View Post
    Check your implicit differantiation.

    \frac{d(xy^2)}{dx}= y^2 + 2xy\frac{dy}{dx}
    Can you walk me through the steps on how you got that? I'm confused
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    Quote Originally Posted by TheMathTham View Post
    Can you walk me through the steps on how you got that? I'm confused
    \frac{d(xy^2)}{dx}=\frac{d}{dx}\left(xy^2\right).
    Apply the product rule.
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    Quote Originally Posted by TheMathTham View Post
    Can you walk me through the steps on how you got that? I'm confused
    \frac{d(xy^2)}{dx}=\frac{dx}{dx}y^2 + x\frac{d(y^2)}{dx} = 1y^2 +2xy\frac{dy}{dx}
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  7. #7
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by xalk View Post
    \frac{d(xy^2)}{dx}=\frac{dx}{dx}y^2 + x\frac{d(y^2)}{dx} = 1y^2 +2xy\frac{dy}{dx}
    I think I must have learned implicit differentiation incorrectly, because I'm still lost. Why was it left as \frac{d(xy^2)}{dx}=1y^2 +2xy\frac{dy}{dx} ? I don't recall ever learning how to use \frac{d(xy^2)}{dx} in implicit differentiation.

    The derivative of x^{3} is 3x^{2}.
    The derivative of xy^{2} is (y^{2} + 2xy)\frac{\mathrm{d} y}{\mathrm{d} x}
    The derivative of 10y is 10\frac{\mathrm{d} y}{\mathrm{d} x}

    I'm assuming that the one highlighted in blue is where I'm messing up. u=x, u{}'=1, v=y^{2}, v{}'=2y
    vu{}'+uv{}'=y^{2}+2xy

    EDIT:
    Never mind, I realized that dy/dx isn't included in the y^2 portion of the product rule
    Last edited by TheMathTham; February 11th 2010 at 08:20 AM. Reason: understand it now
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    Quote Originally Posted by TheMathTham View Post
    I think I must have learned implicit differentiation incorrectly, because I'm still lost. Why was it left as \frac{d(xy^2)}{dx}=1y^2 +2xy\frac{dy}{dx} ? I don't recall ever learning how to use \frac{d(xy^2)}{dx} in implicit differentiation.

    The derivative of x^{3} is 3x^{2}.
    The derivative of xy^{2} is (y^{2} + 2xy)\frac{\mathrm{d} y}{\mathrm{d} x}
    The derivative of 10y is 10\frac{\mathrm{d} y}{\mathrm{d} x}

    I'm assuming that the one highlighted in blue is where I'm messing up. u=x, u{}'=1, v=y^{2}, v{}'=2y
    vu{}'+uv{}'=y^{2}+2xy

    EDIT:
    Never mind, I realized that dy/dx isn't included in the y^2 portion of the product rule
    v{}' = 2yy{}' and not  v{}' =2y

    because  v{}' = \frac{dv}{dx} and according to chain rule

    \frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx}
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  9. #9
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by xalk View Post
    v{}' = 2yy{}' and not  v{}' =2y

    because  v{}' = \frac{dv}{dx} and according to chain rule

    \frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx}
    I was referring to {\color{red} y^{2}} + 2xy\frac{\mathrm{d} y}{\mathrm{d} x} actually, but thanks anyways
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