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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Find an equation of the normal line to the curve y = √x  that is parallel to the line 6x + y = 1.?

3 Answers

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  • alex
    Lv 7
    2 months ago

    hint:

    gradient of the normal to the curve = -6

    --->gradient of the tangent = 1/6

  • ?
    Lv 6
    2 months ago

    y = sqrt(x)

    y' = -0.5/sqrt(x)

    6x + y = 1

    Gradient -6

    Gradient of normal 1/6

    1/6 = -0.5/sqrt(x)

    x = (6*-0.5)^2

     = 9

  • ?
    Lv 7
    2 months ago

    The slope of the given line is -6, so you need a point on y = sqrt(x) where the derivative is +1/6.

    The derivative is dy/dx = 1/[2*sqrt(x)], and if you set this equal to 1/6, the value of x must be 9.

    So the line you're looking for goes through the point (9,3), and its equation is 6x + y = 57.

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