# Multivariable Calc Chain Rule proof problem

• May 27th 2009, 06:55 PM
Susaluda
Multivariable Calc Chain Rule proof problem
If u=f(x,y), where x = e^s cos(t) and y = e^s sin(t), show that:

(d^2 u)/(dx^2) + (d^2 u)/(dy^2) = e^(-2s) ((d^2 u)/(ds^2)+(d^2 u)/(dt^2))

Sorry that's really hard to read...don't know how to use the math tags yet. Those are second partial derivatives. How do I do this problem? It has something to do with the chain rule since it's in that section of the book.
• May 27th 2009, 07:02 PM
Sampras
Quote:

Originally Posted by Susaluda
If u=f(x,y), where x = e^s cos(t) and y = e^s sin(t), show that:

(d^2 u)/(dx^2) + (d^2 u)/(dy^2) = e^(-2s) ((d^2 u)/(ds^2)+(d^2 u)/(dt^2))

Sorry that's really hard to read...don't know how to use the math tags yet. Those are second partial derivatives. How do I do this problem? It has something to do with the chain rule since it's in that section of the book.

So $\displaystyle u = f(x,y)$ where $\displaystyle x = e^{s} \cos t$ and $\displaystyle y = e^{s} \sin t$ and you want to show that $\displaystyle \frac{\partial^{2}u}{\partial x^{2}} + \frac{\partial^{2}u}{\partial y^2} = e^{-2s} \left(\frac{\partial^{2}u}{\partial s^2} + \frac{\partial^{2}u}{\partial t^2} \right)$? Compute the LHS, and then the RHS to show that they are equal.

So $\displaystyle \frac{\partial u}{\partial s} = e^{s} \cos t \frac{\partial u}{\partial x}+ e^{s} \sin t \frac{\partial u}{\partial y}$ and $\displaystyle \frac{\partial u}{\partial t} = e^{s} \sin t \frac{\partial u}{\partial x}+ e^{s} \cos t \frac{\partial u}{\partial y}$.
• May 27th 2009, 07:11 PM
Susaluda
Yes, that's right.
• May 27th 2009, 07:38 PM
Susaluda
Ok so then do I take the derivative of those again? After that what do I do?

Sorry, what did you get when you take the derivative again? For some reason it's not working right...
• May 27th 2009, 07:45 PM
Sampras
Quote:

Originally Posted by Susaluda
Ok so then do I take the derivative of those again? After that what do I do?

Then you add them together and multiply by $\displaystyle e^{-2s}$. You should get $\displaystyle (\cos^{2}t + \sin^{2}t) \left( \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2} \right) = \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2}$.
• May 27th 2009, 07:59 PM
Sampras
Quote:

Originally Posted by Susaluda
Ok so then do I take the derivative of those again? After that what do I do?

Sorry, what did you get when you take the derivative again? For some reason it's not working right...

So $\displaystyle \frac{\partial^{2}u}{\partial s^2} = e^{s} \cos t \frac{\partial u}{\partial x} +e^{s} \cos t \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \right)+e^{s} \sin t \frac{\partial u}{\partial y} + e^{s} \sin t \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \right)$. We know that $\displaystyle \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \right) = \frac{\partial^{2}u}{\partial x^{2}} \frac{\partial x}{\partial s} + \frac{\partial^{2} u}{\partial y \partial x} \frac{\partial y}{\partial s} = e^{s} \cos t \frac{\partial^{2}u}{\partial x^{2}}+ e^{s} \sin t \frac{\partial^{2}u}{\partial y \partial x}$. Also $\displaystyle \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \right) = \frac{\partial^{2}u}{\partial y^{2}} \frac{\partial y}{\partial s} + \frac{\partial^{2} u}{\partial x \partial y} \frac{\partial x}{\partial s} = e^{s} \sin t \frac{\partial^{2}u}{\partial y^{2}}+ e^{s} \cos t \frac{\partial^{2}u}{\partial x \partial y}$. By continuity of partials, $\displaystyle \frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial^{2} u}{\partial y \partial x}$.

So the same thing for $\displaystyle \frac{\partial^{2} u}{\partial t^2}$.
• May 27th 2009, 08:10 PM
Susaluda
Ok so then I just add them together after that?
• May 27th 2009, 08:13 PM
Sampras
yes and multiply the quantity by $\displaystyle e^{-2s}$.