- #1

- 217

- 0

$${y = r \sin \theta}$$, $${r^2 = x^2 + y^2}$$ and $${\theta = \tmop{ \arctan} (y / x)}$$ (with some caveats for the last formula).

Suppose $${u = u (x, y)}$$.

*Show that*

$${\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta}$$

$${\frac{\partial u}{\partial \theta} = - \frac{\partial u}{\partial x} r \sin \theta + \frac{\partial u}{\partial y} r \cos \theta}$$

$${\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta}$$

$${\frac{\partial u}{\partial \theta} = - \frac{\partial u}{\partial x} r \sin \theta + \frac{\partial u}{\partial y} r \cos \theta}$$

Now suppose that $${u = u (r, \theta)}$$. Show that

$${\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \

**cos \theta**- \frac{\partial u}{\partial \theta} \

**frac{\sin \theta}{r}}**$$

$${\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \

**sin \theta**+ \frac{\partial u}{\partial \theta} \

**frac{\cos \theta}{r}}**$$

Now I've proved the italic part but am stuck on the bold parts basically how are they calculated. I've tried the same method as used in the italic part ut am getting different answers.

For e.g in the last bold part instead of cos(theta)/r I got (cos(theta))^2.

Help?