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Math Help - Tangent and Normal Lines

  1. #1
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    Exclamation Tangent and Normal Lines

    Question:
    Find the equation of the normal lines to the curve  y=x^4 which is parallel to the line  2x+y=3

    Please answer in steps..
    Thank you
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  2. #2
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    First you need to find the derivative:

    y' = 4x^3

    Then, from the equation of the line you are given, you see that the slopes of the normal line is -2 ; and since you want the slope of the tangets, you get the opposite reciprocal of -2, which is m=1/2.

    Now you have y', just plug it in the derivative function and find your x.
    Then with that, go back to the original equation and find y.
    And you have all the ingredients necessary to set a line equation of the normals.
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  3. #3
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    Quote Originally Posted by Arturo_026 View Post
    First you need to find the derivative:

    y' = 4x^3

    Then, from the equation of the line you are given, you see that the slopes of the normal line is -2 ; and since you want the slope of the tangets, you get the opposite reciprocal of -2, which is m=1/2.

    Now you have y', just plug it in the derivative function and find your x.
    Then with that, go back to the original equation and find y.
    And you have all the ingredients necessary to set a line equation of the normals.

    So From what i understood?

     4x^3=\frac{1}{2}
     x=\frac{1}{2}

     y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}
     x-2y+\frac{3}{8}

    Is my answer correct??

    If not please answer in steps
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    So From what i understood?

     4x^3=\frac{1}{2}
     x=\frac{1}{2}

     y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}
     x-2y+\frac{3}{8}

    Is my answer correct??

    If not please answer in steps
    HI

    your values of x and y are both correct . The gradient of normal is -2 because the normal line is parallel to the line y=-2x+3
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  5. #5
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    Hello mj.alawami
    Quote Originally Posted by mj.alawami View Post
    So From what i understood?

     4x^3=\frac{1}{2}
     x=\frac{1}{2}

     y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2}
     x-2y+\frac{3}{8}

    Is my answer correct??

    If not please answer in steps
    Your answer is partly correct. You have the right values of x and y, but the wrong gradient. The gradient of the normal is -2, so the equation is:
    y-\frac{1}{16}=-2\left(x - \frac12\right)

    i.e. 16y-1=-32x + 16

    i.e. 32x+16y-17 =0
    Grandad
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