# Homework Help: Differentiation, change of variable

1. Oct 5, 2006

### thenewbosco

The question:

if $$x=\rho cos \phi$$ and $$y=\rho sin\phi$$

prove that if U is a twice differentiable function of x and y that

$$\frac{d^2U}{dx^2} + \frac{d^2U}{dy^2} = \frac{d^2U}{d\rho^2} + \frac{1}{\rho}\frac{dU}{d\rho} + \frac{1}{\rho^2}\frac{d^2U}{d\phi^2}$$

i am not sure how to approach this since the function is not given. Is there some assumption i am supposed to make about the function? like f(x,y) = xy or something?
any help on the first couple steps would be appreciated

Last edited: Oct 5, 2006
2. Oct 5, 2006

### arildno

With a rather more transparent notation, let u be a function of the polar coordinates, U the function in Cartesian coordinates, related by:
$$u(\rho(x,y),\theta(x,y))=U(x,y)$$

With this notation, you would have lower case u's on your right-hand side of the identity.

You need to compute the partial derivatives of the radial&angular variables with respect to x and y.

3. Oct 5, 2006

### NateTG

Have you tried the chain rule?

4. Oct 5, 2006

### thenewbosco

so $$\rho (x,y)= ((\frac{x}{cos \phi})^2 + (\frac{y}{cos \phi})^2)^\frac{1}{2}$$?

and $$\phi(x,y)=arctan(y/x)$$ is this correct?

5. Oct 5, 2006

### thenewbosco

to what would i apply the chain rule?

6. Oct 5, 2006

### arildno

No, $$\rho(x,y)=\sqrt{x^{2}+y^{2}}$$

That ought to be obvious for anyone claiming to have understood Pythagoras' second most important theorem.

7. Oct 5, 2006

### thenewbosco

so after i compute these derivatives of the radial and angular components with respect to x and y i will have everything and just have to put it all together so it looks like the identity i am trying to show?

8. Oct 5, 2006

### arildno

No, you must also use the chain rule on u.
To give you a start on that:
$$\frac{\partial{U}}{\partial{x}}=\frac{\partial{u}}{\partial{r}}\frac{\partial{r}}{\partial{x}}+\frac{\partial{u}}{\partial{\theta}}\frac{\partial{\theta}}{\partial{x}}$$
and so on..

9. Oct 5, 2006

### thenewbosco

where your r is actually a rho i assume then? if so thanks for the help i should be able to solve this one

10. Oct 5, 2006

### arildno

Yes, r is rho.