Multivariable Calc Chain Rule proof problem

**Susaluda** · May 27th 2009, 06:55 PM

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.

**Sampras** · May 27th 2009, 07:02 PM

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

So $\frac{\partial u}{\partial s} = e^{s} \cos t \frac{\partial u}{\partial x}+ e^{s} \sin t \frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial t} = e^{s} \sin t \frac{\partial u}{\partial x}+ e^{s} \cos t \frac{\partial u}{\partial y}$ .

**Susaluda** · May 27th 2009, 07:11 PM

Yes, that's right.

**Susaluda** · May 27th 2009, 07:38 PM

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...

**Sampras** · May 27th 2009, 07:45 PM

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 $e^{-2s}$ . You should get $(\cos^{2}t + \sin^{2}t) \left( \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2} \right) = \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2}$ .

**Sampras** · May 27th 2009, 07:59 PM

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

So the same thing for $\frac{\partial^{2} u}{\partial t^2}$ .

**Susaluda** · May 27th 2009, 08:10 PM

Ok so then I just add them together after that?

**Sampras** · May 27th 2009, 08:13 PM

yes and multiply the quantity by $e^{-2s}$ .

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