What is the derivative of (b^2*x^2 + x^2*y^2 = a^2*b^2) with given values?

  • Thread starter Thread starter courtrigrad
  • Start date Start date
  • Tags Tags
    Calculus
AI Thread Summary
The discussion revolves around finding the derivative of the equation b^2*x^2 + x^2*y^2 = a^2*b^2, given specific values for x, y, and dx/dt. Participants highlight the importance of correctly applying the product rule in differentiation and point out mistakes in earlier calculations. The correct approach involves differentiating both sides of the equation and substituting the known values to find dy/dt. Clarifications are made regarding the constants a and b, emphasizing their role in the differentiation process. Accurate application of calculus principles is essential for solving the problem correctly.
courtrigrad
Messages
1,236
Reaction score
2
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
 
Physics news on Phys.org
courtrigrad said:
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

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

Yes, once you have the formula just substitute.
 
courtrigrad said:
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
No,the bracket is wrong...

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

Daniel.
 
thanks a lot guys
 
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)
 
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?
 
a and b are constants. So i get the second line.
 
courtrigrad said:
a and b are constants. So i get the second line.

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
 
Back
Top