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Math Help - Finding an equation of a curve?

  1. #1
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    Finding an equation of a curve?

    I've never come across a question like this before, perhaps someone will be able to point me in the right direction.
    Find the equation of a curve with x-intercept equal to 2 and whose tangent line at any point (x,y) has slope  xe^{-y}.

    Watch this be something really easy I skipped over in my text book, anyways thanks in advance for the help.
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  2. #2
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    Hello,

    Quote Originally Posted by doughnuts View Post
    I've never come across a question like this before, perhaps someone will be able to point me in the right direction.
    Find the equation of a curve with x-intercept equal to 2 and whose tangent line at any point (x,y) has slope  xe^{-y}.

    Watch this be something really easy I skipped over in my text book, anyways thanks in advance for the help.
    If the tangent line at any point has slope xe^{-y}, this means that y'=xe^{-y}, because y' represents the slope of the tanget.

    So you have to solve for y in y'=xe^{-y} and with y(2)=0, because x-intercept is 2.

    y'=xe^{-y}=\frac{x}{e^y} \implies y'e^y=x.

    Integrating both parts, we have :
    e^y=\frac{x^2}{2}+C.

    y=\ln \left(\frac{x^2}{2}+C\right)

    Determine C thanks to the information y(2)=0
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  3. #3
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    The slope of the tangent line of the curve f(x) at the point (x,y) is \frac{dy}{dx}.

    So this question basically says \frac{dy}{dx} = xe^{-y}. Solve it and use the x-intercept value to find the integration constant.
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  4. #4
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    Wow that was way simpler then I thought it would be. The question was grouped in with a bunch of parametric equation questions so I thought it would be something along those lines. Thanks for the help.
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