calculus

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Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=1 where A and B are constants to be determined. Given that this curve passes through the point (3,2) and that its tangent at this point has slope -1, find A and B.

  • calculus -

    take the derivative:

    2x + (Ax)(2y)dy/dx + Ay^2 + 3By^2 dy/dx = 0
    .....
    ...
    dy/dx = (-2x - Ay^2)/(2Axy + 3By^2)
    at (3,2) dy/dx = -1
    -1 = (-6 - 4A)/(12A + 12B)
    ..
    4A + 6B = 3

    also in original,
    9 + 12A + 8B = 1
    3A + 2B = -2

    triple the last one, and subtract ....
    5A=-9
    A=-9/5

    sub back in to get B = 17/10

    (Was expecting "nicer" numbers. I might have made an arithmetic error, check my arithmetic)

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