Homework Help: Calculus Verification

1. Jan 10, 2005

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. Jan 10, 2005

learningphysics

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

Yes, once you have the formula just substitute.

3. Jan 10, 2005

dextercioby

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.

4. Jan 10, 2005

thanks a lot guys

5. Jan 10, 2005

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)

6. Jan 10, 2005

learningphysics

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?

7. Jan 10, 2005

a and b are constants. So i get the second line.

8. Jan 10, 2005

learningphysics

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