If $ \nabla f (1,-1,\sqrt{2})=\langle 1,2,-2 \rangle$ find $\frac{\partial f}{\partial \theta}$ at this point.

Question

If $ \nabla f (1,-1,\sqrt{2})=\langle 1,2,-2 \rangle$ find $\frac{\partial f}{\partial \theta}$ at this point.

Here's what I have in mind,

$$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \theta}$$

And I know $\langle f_x,f_y,f_z \rangle=\langle 1,2,-2 \rangle$ at my point so now I just need to find $x_{\theta}, y_{\theta}z_{\theta}$ and not sure how.

Answer 1

Cylindrical coordinates:

$x = r \cos \theta$.

$y = r \sin \theta$.

$z = z$.

So $z_{\theta}=0$. Then $f_{\theta}=x_{\theta}+2y_{\theta}=-r\sin \theta+2r\cos \theta $. If $x=1$ and $y=-1$, then $r=\frac{1}{\cos \theta}=-\frac{1}{\sin \theta}$. Thus $f_{\theta}=-r\sin \theta+2r\cos \theta =-(-\frac{1}{\sin \theta})\sin \theta+2(\frac{1}{\cos \theta})\cos \theta=3$.