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Math Help - Equ. of normal

  1. #1
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    Post Equ. of normal

    Could you check this question to see if I have done it right:

    Question: Find the equation of the normal to the curve y=x+ 4/x at the point p where x=4. If the normal meets the curve agian at N, find the co-ordinates of N.

    This is what I have done:
    y= x+4x^-1
    dy/dx= 1 - 4x^-2
    x=4 y=(4)*4(4)^-1 = 5
    dy/dx= 1 - 4(4)^-2 = 0.75 or 3/4
    Normal gradient = -4/3
    (y - 5) = -4/3(x - 4)
    y - 5 = -4/3x + 16/3
    y = -4/3x + 31/3 or 3y = - 4x + 31

    This is where I got up to, i'm not to sure what you have to do when it says 'If the normal meets the curve agian at N, find the co-ordinates of N'

    If you could please explain and check if I have done the first bit right.

    Thank you
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  2. #2
    TD!
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    Looks good so far, at first sight.

    To find all intersection points, solve the equation you get by expressing that the function values have to be equal, i.e. solve:

    <br />
x + \frac{4}<br />
{x} =  - \frac{4}<br />
{3}x + \frac{{31}}<br />
{3}<br />

    Of course, one of the solutions is x = 4, but you are asked to find possible others (there is one more).
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  3. #3
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    Quote Originally Posted by dadon
    Could you check this question to see if I have done it right:

    Question: Find the equation of the normal to the curve y=x+ 4/x at the point p where x=4. If the normal meets the curve agian at N, find the co-ordinates of N.

    This is what I have done:
    y= x+4x^-1
    dy/dx= 1 - 4x^-2
    x=4 y=(4)*4(4)^-1 = 5
    dy/dx= 1 - 4(4)^-2 = 0.75 or 3/4
    Normal gradient = -4/3
    (y - 5) = -4/3(x - 4)
    y - 5 = -4/3x + 16/3
    y = -4/3x + 31/3 or 3y = - 4x + 31

    This is where I got up to, i'm not to sure what you have to do when it says 'If the normal meets the curve agian at N, find the co-ordinates of N'

    If you could please explain and check if I have done the first bit right.

    Thank you
    If y=x+\frac{4}{x}
    Then, y'=1-4x^{-2}
    At x=4 you have,
    y'=3/4
    But you need normal which is perpendicular to tangent thus its slope is,
    m=-4/3 and passes through point (4,5) (because at x=4 we have y=5).
    Thus, using slope-intercept formula you have,
    y-5=-\frac{4}{3}(x-4)
    Thus,
    y=-\frac{4}{3}x+\frac{31}{3}
    Look like you are right.
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    Post

    To find all intersection points, solve the equation you get by expressing that the function values have to be equal, i.e. solve:

    <br />
x + \frac{4}{x} = - \frac{4}{3}x + \frac{{31}}{3}<br />

    Of course, one of the solutions is x = 4, but you are asked to find possible others (there is one more).
    Hey! thanks for replying and checking my answer

    so to solve the equ.

    3x + 12/x =-4x + 31

    3x + 12 = x(-4x + 31)

    -4x^2 -28x - 12 = 0

    4x^2 + 28x + 12 = 0

    x^2 + 7x + 3 = 0

    x = -0.45 x = -6.54

    Is this right? as you said that one is meant to be x = 4
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  5. #5
    TD!
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    Quote Originally Posted by dadon
    3x + 12/x =-4x + 31

    3x + 12 = x(-4x + 31)
    You miss a square in the first term, it should say 3x. Further one you also made a sign mistake.
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  6. #6
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    Post Re:

    You miss a square in the first term, it should say 3x. Further one you also made a sign mistake.


    Hey! So the answer should be:

    <br />
3x + 12/x = -4x + 31<br /> <br />
3x^2 + 12 = x(-4x + 31)<br /> <br />
3x^2 + 4x^2 - 31x + 12 = 0<br /> <br />
7x^2 - 31x + 12 = 0<br /> <br />
x = 4, x = 3/7<br />
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  7. #7
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    Quote Originally Posted by dadon

    Hey! So the answer should be:

    <br />
3x + 12/x = -4x + 31<br /> <br />
3x^2 + 12 = x(-4x + 31)<br /> <br />
3x^2 + 4x^2 - 31x + 12 = 0<br /> <br />
7x^2 - 31x + 12 = 0<br /> <br />
x = 4, x = 3/7<br />
    Yes
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  8. #8
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    Post Re:

    If the normal meets the curve agian at N, find the co-ordinates of N.

    So when x = 3/7

    y = (3/7) + 4(3/7)^-1

    y = 205/21

    (3/7, 205/21)

    Thanks guys!
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  9. #9
    TD!
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    You're welcome
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