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Math Help - the equation of the normal to the curve

  1. #1
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    the equation of the normal to the curve

    I already update the post, every one look it clear....refer to my post on no 10 & 11

    here my problem
    Determine the equation of the normal to the curve y = x^3-6x^2-2x which are perpendicular to the straight line 4-x+2y=0

    just check my calculation..what wrong??

    i already put the actual ans @ pg 2 of 2 at bottom

    -----------------------------------------------
    actual answers:

    y + x + 12 = 0
    or
    27y + 27x + 4 = 0
    Attached Thumbnails Attached Thumbnails the equation of the normal to the curve-normal-pg-1.bmp  
    Last edited by nikk; September 22nd 2008 at 10:11 AM. Reason: insert ans, info
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  2. #2
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    my ans pg 1 of 2

    pg 1 of 2
    Attached Thumbnails Attached Thumbnails the equation of the normal to the curve-normal-pg-1.jpg  
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  3. #3
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    my ans pg 2 of 2

    pg 2 of 2
    Attached Thumbnails Attached Thumbnails the equation of the normal to the curve-normal-equation2.jpg  
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  4. #4
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    sorry because it small in view...how to make it bigger size
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  5. #5
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    just click on it to get the enlarged version
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  6. #6
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    I would really like to help you!
    But I simply cannot read that image!
    Why donít you simply retype the posting to make it readable?
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  7. #7
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    Quote Originally Posted by Plato View Post
    I would really like to help you!
    But I simply cannot read that image!
    Why donít you simply retype the posting to make it readable?
    tq for your response... i will try use image shack...sorry Mr Ploto...

    i not very familiar to use math simbol here

    thank for your help
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  9. #9
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    Last edited by CaptainBlack; September 23rd 2008 at 03:27 AM.
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  10. #10
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    Quote Originally Posted by nikk View Post
    I already update the post, every one look it clear....refer to my post on no 10 & 11

    here my problem
    Determine the equation of the normal to the curve y = x^3-6x^2-2x which are perpendicular to the straight line 4-x+2y=0

    just check my calculation..what wrong??

    i already put the actual ans @ pg 2 of 2 at bottom

    -----------------------------------------------
    actual answers:

    y + x + 12 = 0
    or
    27y + 27x + 4 = 0
    If those are the answers then the question is wrong.

    1. The lines in the answers are NOT perpendicular to the line 4 - x + 2y = 0.

    2. Furthermore, since the gradient of the lines in the answer is m = -1, the tangent to the cubic must have gradient m = 1. Solving 3x^2 - 12x -2 = 1 does not give an integer solution for x and therefore cannot lead to the answers.

    Re-check the question. If the question is as you say, then the question is wrong. There's nothing more can be done.
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    If those are the answers then the question is wrong.

    1. The lines in the answers are NOT perpendicular to the line 4 - x + 2y = 0.

    2. Furthermore, since the gradient of the lines in the answer is m = -1, the tangent to the cubic must have gradient m = 1. Solving 3x^2 - 12x -2 = 1 does not give an integer solution for x and therefore cannot lead to the answers.

    Re-check the question. If the question is as you say, then the question is wrong. There's nothing more can be done.

    tq for your reply sir..do u think my ans is correct....
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  12. #12
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    Quote Originally Posted by nikk View Post
    tq for your reply sir..do u think my ans is correct....
    I can't see any obvious mistakes.
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  13. #13
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    Quote Originally Posted by mr fantastic View Post
    I can't see any obvious mistakes.
    tq for your support
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