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Given the curve x^2-xy+y^2=9

A) write a general expression for the slope of the curve.
B) find the coordinates of the points on the curve where the tangents are vertical
C) at the point (0,3) find the rate of change in the slope of the curve with respect to x

I don't even know where to start!! Please help!

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5 answers

  1. A and B are right. C is the second derivative

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  2. Does that mean for 7 whole years, people got #3 incorrect...?

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  3. A)
    2x - x dy/dx - y + 2y dy/dx = 0
    dy/dx(2y - x) = y - 2x
    dy/dx = (y-2x)/(2y-x)

    B) for a vertical tangent dy/dx is undefined
    For that to happen , the denominator has to be zero
    2y - x = 0
    or x = 2y
    sub back into original:
    (2y)^2 - (2y)(y) + y^2 = 9
    4y^2 - 2y^2 + y^2 = 9
    3y^2 = 9
    y^2 = 3
    y = ± √3
    if y =+√3 , then x = 2√3 ---> point (2√3, √3)
    if y = -√3 , then x = -2√3 ---> point (-2√3, -√3)

    C) at (0,3)
    dy/dx = (3 - 0)/(6-0) = 1/2

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  4. woah

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  5. correct way to do part c

    c) To find the rate of change of the slope, you take the second derivative, so take the derivative of (y-2x) / (2y-x) with respect to x:

    [(2y-x)(dy/dx - 2)-(y-2x)(2dy/dx - 1)] / (2y-x)^2

    To find dy/dx, simply plug in (0,3) into (y-2x) / (2y-x) to get 1/2, so the answer is:

    [(6)(1/2- 2)] / (6)^2 = -1/4.

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