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Calculus Verification

  1. Jan 10, 2005 #1
    Hello all

    If x^3 + y^3 - 3xy = 0, dx/dt = -1 , x = 3/2. y = 3/2 what is dy/dr?

    So 3x^2 (dx/dt) + 3y^2 (dy/dt) - 3[x(dx/dt) + y(dy/dt)] = 0. So I just substitute values. Is this correct?

    Thanks
     
  2. jcsd
  3. Jan 10, 2005 #2

    learningphysics

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    You made a mistake calculating out derivative of 3xy...

    Yes, once you have the formula just substitute.
     
  4. Jan 10, 2005 #3

    dextercioby

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    No,the bracket is wrong...

    [tex] 3x^{2}\frac{dx}{dt}+3y^{2}\frac{dy}{dt}-3[x\frac{dy}{dt}+\frac{dx}{dt}y]=0 [/tex]

    Daniel.
     
  5. Jan 10, 2005 #4
    thanks a lot guys
     
  6. Jan 10, 2005 #5
    Also if we have

    b^2 * x^2 + x^2*y^2 = a^2 * b^2, is:

    dy/dt = 2b^2 *x (dx/dt) + 2a^2*y (dy/dt) = 0

    Thanks (a and b are constants)
     
  7. Jan 10, 2005 #6

    learningphysics

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    That's not right... You are given:

    b^2 * x^2 + x^2*y^2 = a^2 * b^2

    How did you get to your next line?
     
  8. Jan 10, 2005 #7
    a and b are constants. So i get the second line.
     
  9. Jan 10, 2005 #8

    learningphysics

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    b^2 * x^2 + x^2*y^2 = a^2 * b^2

    taking the derivative of both sides, I get

    (2b^2)x(dx/dt) + (2x)(dx/dt)(y^2) + (x^2)(2y)(dy/dt)=0
     
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