# Math Help - gradients/directional derivatives

This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.

2. Originally Posted by asw-88
This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.
One approach of many:

Polar coordinates: $x = r \cos \theta$ and $y = r \sin \theta$.

Therefore:

$r^2 = x^2 + y^2$ .... (1)

$\tan \theta = \frac{y}{x}$ .... (2)

Results that will be needed:

From (1):

$2 r \frac{\partial r}{\partial x} = 2x \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \theta$ .... (A)

$2 r \frac{\partial r}{\partial y} = 2y \Rightarrow \frac{\partial r}{\partial y} = \frac{y}{r} = \sin \theta$ .... (B)

From (2):

$sec^2 \theta \, \frac{\partial \theta}{\partial x} = -\frac{y}{x^2} = -\frac{r \sin \theta}{r^2 \cos^2 \theta} \Rightarrow \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r}$ .... (C)

$sec^2 \theta \, \frac{\partial \theta}{\partial y} = \frac{1}{x} = \frac{1}{r \cos \theta} \Rightarrow \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}$ .... (D)

Then:

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial x} = .....$ .... (E)

$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial y} = .....$ .... (F)

where the results (A) - (D) are to be substituted into (E) and (F).

Then substitute (E) - (F) into $\nabla f$, do a little bit of re-arranging and then the job's done.