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calculus

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find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)

  • answer only - can not post derivation - ,

    Once again I can not post my answer but in the end I get
    3 y = -2 x + 5

  • screaming frustration - ,

    Can other teachers post all right ?

  • Trying again - ,

    no luck - hope someone can help you.
    the idea is to do f(d+dx) - f(x)
    where f(x+dx) = f(x) + (dy/dx) dx

  • calculus - ,

    I've sent your posts to Bob and Leo. I'll send this one too.

  • Thanks - ,

    Thanks - I can not really answer this because it will not let me write out derivative method.

  • calculus - ,

    Bob knows. He has cleaned out the banned list, and said he'd check more specifically later.

    I've texted Leo ... I'm surprised there's been no answer.

    =(

  • calculus - ,

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

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

    so y-1 = (-3/2)(x-1)
    2y - 2 = -3x + 3
    2y = -3x + 5
    y = (-3/2)x + 5/2

    Damon, I had the same problem, now it seems to work

  • calculus - ,

    tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)
    -----------------------------------
    (1/3) / x^(2/3) + (1/2) dy/dx /y^(1/2)=0
    so
    (1/2)dy/dx = -(1/3) y^(1/2)/x^(2/3)
    at (1.1)
    dy/dx = -(2/3)
    then
    y = -2x/3 + b
    1 = -2/3 + b
    b = 5/3
    so
    y = -2x/3 + 5/3
    3 y = -2 x + 5

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