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Math Help - [SOLVED] Normal line to a curve

  1. #1
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    [SOLVED] Normal line to a curve

    The original equation is y=sqrtx/(x+4)
    and the question asks to find the equation of tangent and normal line at point (9,3/13)

    so I derived and found the tangent line to be y=-5/1020x + 243/884 and that is the correct answer, but I keep getting the normal line wrong

    So for the normal line's slope I went -1/m and I got so the slope 1020/5 which is 204. and then I plugged it in to y-3/13=204(x-9)

    and that gives y=204x-23865/13

    I've redone it over and over again, but it still says the answer is wrong.

    I'm just wondering where I went wrong.
    Thank you for helping me!!!
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  2. #2
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    Quote Originally Posted by electricsparks View Post
    The original equation is y=sqrtx/(x+4)
    and the question asks to find the equation of tangent and normal line at point (9,3/13)

    so I derived and found the tangent line to be y=-5/1020x + 243/884 and that is the correct answer, but I keep getting the normal line wrong

    So for the normal line's slope I went -1/m and I got so the slope 1020/5 which is 204. and then I plugged it in to y-3/13=204(x-9)

    and that gives y=204x-23865/13


    \color{red}\mbox{I get here in the fraction }\,\frac{23868}{13}=1836

    Tonio

    I've redone it over and over again, but it still says the answer is wrong.

    I'm just wondering where I went wrong.
    Thank you for helping me!!!
    .
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  3. #3
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    Quote Originally Posted by electricsparks View Post
    The original equation is y=sqrtx/(x+4)
    and the question asks to find the equation of tangent and normal line at point (9,3/13)

    so I derived and found the tangent line to be y=-5/1020x + 243/884 and that is the correct answer, but I keep getting the normal line wrong

    That isn't the correct equation for the tangent. Differentiating and plugging in x=9 gives the tangent slope as \frac{-5}{1014} which gives the tangent as: y = \frac{-5x}{1014} + \frac {93}{338}

    So for the normal line's slope I went -1/m and I got so the slope 1020/5 which is 204. and then I plugged it in to y-3/13=204(x-9)

    So the slope of the normal is now \frac{1014}{5}. I think you can work it out from there.

    and that gives y=204x-23865/13

    I've redone it over and over again, but it still says the answer is wrong.

    I'm just wondering where I went wrong.
    Thank you for helping me!!!
    ..
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  4. #4
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    Oh! ok, I see where I went wrong. That was a dumb mistake. Thanks!
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