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Math Help - equation of the normal line at a given point

  1. #1
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    equation of the normal line at a given point

    I am not getting the same answer as the book for this question. it is to find the equation of the normal line at the given point

    y = 2x^(3) + x^(-3) where x = -1

    I solved for y getting 2 giving the point (-1,2)

    taking the derivative I got y' = 6x^2 -3x^-4 and for f'(-1) I got 3

    giving me the equation

    y-2=3(x+1)

    so for the slope of the normal line I take the negative reciprocal being -1/3

    so I have

    y-2 = [-1/3][x+1] I multiply everything by 3 to get [3y-6] = -1[x+1]

    which gives me

    x + 3y + 5 = 0

    but the book has the answer down as being x+3y + 10 = 0

    thank you!
    Last edited by Jonroberts74; September 28th 2013 at 04:13 PM.
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  2. #2
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    Re: equation of the normal line at a given point

    Looks like y(-1)=-3
    Thanks from Jonroberts74 and topsquark
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  3. #3
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    Re: equation of the normal line at a given point

    2x^3 + x^-3 with -1 for x

    2(-1)^3 = -2 and (-1)^-3 = 1/[-1]

    -2/[1/-1] is 2

    or do you mean y'?

    I still get 3 for y'
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  4. #4
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    Re: equation of the normal line at a given point

    Quote Originally Posted by Jonroberts74 View Post
    I am not getting the same answer as the book for this question. it is to find the equation of the normal line at the given point
    y = 2x^(3) + x^(-3) where x = -1
    I solved for y getting 2 giving the point (-1,2)
    taking the derivative I got y' = 6x^2 -3x^-4 and for f'(-1) I got 3
    so for the slope of the normal line I take the negative reciprocal being -1/3
    which gives me x + 3y + 5 = 0

    but the book has the answer down as being x+3y + 10 = 0
    Your point is incorrect. It should be (-1,-3).
    Thanks from Jonroberts74 and topsquark
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  5. #5
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    Re: equation of the normal line at a given point

    Oh, duh haha. Silly. Thank you
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