math

find the eqt. of the tangent to the curves at (1,2).
y^3+y^3=9(xy-1)
[ans: y=5x-5]
could some1 show me the calculation work bcus my ans is y=(6x+4)/5.

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asked by lily
  1. I will guess that you have a typo and one of those first terms is x^3

    x^3 + y^3 = 9xy - 9
    3x^2 + 3y^2 dy/dx = 9x dy/dx + 9y
    dy/dx(3y^2 - 9x) = 9y - 3x^2

    at (1,2)
    dy/dx(12-9) = 18 - 3
    dy/dx = 15/3 = 5

    so equation is y = 5x + b
    plug in (1,2)
    2 = 5(1) + b
    b = -3

    tangent equation : y = 5x - 3

    Their answer is wrong, since (1,2) doesn't even satisfy their answer equation.

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    posted by Reiny
  2. the 1st term is y^3

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    posted by lily
  3. Then your equation would simplify to
    2y^3 = 9xy - 9

    6y^2 dy/dx = 9x dy/dx + 9y
    dy/dx(6y^2 - 9x) = 9y
    for the given point (1,2)
    dy/dx(24-9) = 18
    dy/dx = 18/15 = 6/5

    Now the slope of the line will be 6/5 , but in your answer it is 5, as I had before.
    Now it makes even less sense!

    Please check your typing, it is frustrating to keep guessing what the student meant in solving these questions.

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    posted by Reiny
  4. you get the same ans like me so the ans is wrong.
    Is that right?

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    posted by lily
  5. It appears that way, but I still think no textbook would have an equation looking like

    y^3 + y^3 .....

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    posted by Reiny

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