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calculus

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the curve with equation 2(x^2+y^2)^2=9(x^2-y^2) is called a lemniscate. find the points on the lemniscate where the tangent is horizontal.

  • calculus -

    just plug and chug:

    4(x^2 + y^2)(2x + 2yy') = 9(2x - 2yy')

    Now, the tangent is horizontal when y' = 0
    4(x^2 + y^2)(2x) = 9(2x)
    8x^3 + 8xy^2 = 18x
    2x(4x^2 + 4y^2 - 9) = 0

    So, either x=0
    substitute back into original equation:
    2y^4 = -9y^2
    y=0

    or,

    4x^2 + 4y^2 = 9
    x^2 = (9 - 4y^2)/4
    Substitute that back into the original equation
    2((9 - 4y^2)/4 + y^2)^2 = 9((9 - 4y^2)/4 - y^2)
    Expand the binomial and solve for y^2

  • calculus -

    Tan^3(xy^2+y)=x
    Use derivative

  • calculus -

    Find .... Tan^3(xy^2+y)=x
    Use derivative

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