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Math Help - Another Geographic Derivative

  1. #1
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    Another Geographic Derivative

    I'm stuck on this question:

    In the curve to which the equation is y^2+x^2=4, find the value of x at those points where the slope = 1.
    I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for y, and then take the derivative?

    y^2+x@=4

    y^2=4-x^2

    y=\pm\sqrt{4-x^2}

    is this right so far? Thanks.
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  2. #2
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    Quote Originally Posted by dbakeg00 View Post
    I'm stuck on this question:



    I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for y, and then take the derivative?

    y^2+x@=4

    y^2=4-x^2

    y=\pm\sqrt{4-x^2}

    is this right so far? Thanks.
    You can use implicit differentiation to find \frac{dy}{dx}

    2y\frac{dy}{dx}+2x=0

    \frac{dy}{dy}=-\frac{x}{y}

    Set the derivative equal to 1

    -\frac{x}{y}=1

    y=-x

    Substitute back into the original equation and solve for x
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  3. #3
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    Thank you...guess I haven't learned howto do that yet.
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  4. #4
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    Quote Originally Posted by dbakeg00 View Post
    I'm stuck on this question:



    I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for y, and then take the derivative?

    y^2+x@=4

    y^2=4-x^2

    y=\pm\sqrt{4-x^2}

    is this right so far? Thanks.
    Yes. Find the derivative, set it equal to 1 and solve for x
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  5. #5
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    Quote Originally Posted by dbakeg00 View Post
    I'm stuck on this question:



    I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for y, and then take the derivative?

    y^2+x@=4

    y^2=4-x^2

    y=\pm\sqrt{4-x^2}

    is this right so far? Thanks.
    Yes. Just find the derivative, set it equal to 1 and solve for x
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  6. #6
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    I dont even know how to go about setting that equal to 1 and solving...any tips on how to start?
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  7. #7
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    Quote Originally Posted by dbakeg00 View Post
    I dont even know how to go about setting that equal to 1 and solving...any tips on how to start?
    Break y up into 2 functions

    y=\sqrt{4-x^2} and y=-\sqrt{4-x^2}

    For y=\sqrt{4-x^2}=(4-x^2)^{\frac{1}{2}}

    y'=\frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)

    Set it equal to 1 and solve for x

    \frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)=1

    \frac{-x}{\sqrt{4-x^2}}=1

    [tex]-x=\sqrt{4-x^2}[/math

    x^2=4-x^2

    Finish it up
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  8. #8
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    got it solved. thanks for your help!
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