Find the equation of a line that is perpendicular to the tangent line of x^2 + (y-x)^3 = 9.?

I know i need to do the derivative of that equation but i cant really come up with it since there is a y in there. Can someone do this for me please?

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  • Anonymous
    1 decade ago
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    Take derivative:

    2x + (3(y - x)^2)(dy/dx - 1) = 0

    (3(y - x)^2)dy/dx = 3(y - x)^2 - 2x

    dy/dx = [3(y - x)^2 - 2x]/[3(y - x)^2] = slope of tangent at any point

    Take the arbitrary point (1,3) which lies on the curve. The slope of the tangent line at this point is

    dy/dx = [3(3 - 1)^2 - 2*1]/[3(3 - 1)^2] = 5/6

    slope of perpendicular = negative reciprocal = -6/5

    Equation of perpendicular at (1,3):

    (y - 3)/(x - 1) = -6/5

    y = -6x/5 + 21/5

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