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Thread: Help with supposedly simple implicit differentiation

  1. June 10th 2009, 11:31 AM #1
    Saibot
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    Help with supposedly simple implicit differentiation

    I have a problem with a supposedly easy question...

    Find the points on the hyperbola \frac {x^2} {25} - \frac {y^2} {9} = 1 where the slope of the tangent is equal to 1.

    i end up with y' = \frac{9x}{25y} where y' becomes 1, and i get another function........

    the listed answer should be ( \frac{-25}{4} , \frac{-9}{4} ) and ( \frac{25}{4} , \frac{9}{4} )
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  2. June 10th 2009, 11:46 AM #2
    Amer
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    Quote Originally Posted by Saibot View Post
    I have a problem with a supposedly easy question...

    Find the points on the hyperbola \frac {x^2} {25} - \frac {y^2} {9} = 1 where the slope of the tangent is equal to 1.

    i end up with y' = \frac{9x}{25y} where y' becomes 1, and i get another function........

    the listed answer should be ( \frac{-25}{4} , \frac{-9}{4} ) and ( \frac{25}{4} , \frac{9}{4} )

    \frac{x^2}{25} - \frac{y^2}{9} =1

    find y with respect to x

    \frac{x^2}{25} -1 =\frac{y^2}{9}

    \sqrt{9\left(\frac{x^2}{25} -1\right)} =y

    y'=\frac{3}{2}\left(\frac{2x}{25}\right)\left(\fra c{1}{(\frac{x^2}{25}-1)^{.5}}\right)

    y'=\frac{3}{2}\left(\frac{2x}{25\left(\frac{x^2}{2 5}-1\right)^{.5}}\right)=1

    \frac{3}{2}\left(\frac{2x}{25\left(\frac{x^2}{25}-1\right)^{.5}}\right)=1

    \frac{3}{2}(2x)=25\sqrt{(\frac{x^2}{25}-1})

    3(x)=25\sqrt{(\frac{x^2}{25}-1})

    9x^2=25^2\left(\frac{x^2}{25}-1\right)

    9x^2=\left(25x^2 - 25^2 \right)

    25x^2-9x^2=25^2

    16x^2 = 25^2 you continue ........

    the rest for you
    Last edited by Amer; June 10th 2009 at 12:25 PM.
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  3. June 10th 2009, 12:19 PM #3
    Saibot
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    yea... i tried that method as well... and i get stuck here:
    y' = \frac{2x}{25\sqrt{9(\frac{x^2}{25} - 1)}}

    If we sub y' = 1, how is it possible to solve/isolate x with algebra O_O

    edit: nvm, your solution was epic.

    nicely done. thanks.
    Last edited by Saibot; June 10th 2009 at 12:29 PM.
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  4. June 10th 2009, 12:26 PM #4
    Amer
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    Quote Originally Posted by Saibot View Post
    yea... i tried that method as well... and i get stuck here:
    y' = \frac{2x}{25\sqrt{9(\frac{x^2}{25} - 1)}}

    If we sub y' = 1, how is it possible to solve/isolate x with algebra O_O
    see my old post I edited it best wishes ....

    Amer
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