## Related questions with answers

In the problem below, the function f is differentiable and $f_x(2,1)=-3, f_y(2,1)=4$, and $f(2,1)=7$.

Find $f_r(2,1)$ and $f_\theta(2,1)$, where $r$ and $\theta$ are polar coordinates, $x=r \cos \theta$ and $y=r \sin \theta$. If $\vec{u}$ is the unit vector in the direction $2 \vec{i}+\vec{j}$, show that $f_{\vec{u}}(2,1)=f_r(2,1)$ and explain why this should be the case.

Solution

VerifiedWe first note that for a function $f(x,y)$ where $x$ and $y$ are functions of $r$ and $\theta$, its partial with respect to $r$ and $\theta$ is given by:

$f_r(x,y)=\dfrac{\partial f}{\partial r}=\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial x}{\partial r}+\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial y}{\partial r }=f_x\cdot\dfrac{\partial x}{\partial r}+f_y\dfrac{\partial x}{\partial r}$

$f_{\theta}(x,y)=\dfrac{\partial f}{\partial \theta}=\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial x}{\partial \theta}+\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial y}{\partial \theta}=f_x\cdot\dfrac{\partial x}{\partial \theta}+f_y\dfrac{\partial x}{\partial \theta}$

Therefore, with $f_x(2,1)=-3, f_y(2,1)=4, x=r\cos\theta$ and $y=r\sin\theta$, we have

$\begin{align*} f_r(2,1)&=-3\cdot\dfrac{\partial }{\partial r}(r\cos \theta)+4\cdot \dfrac{\partial }{\partial r}(r\sin\theta)\\ &=-3\cos\theta+4\sin\theta\\ f_{\theta}(2,1)&=-3\cdot\dfrac{\partial}{\partial \theta}(r\cos\theta)+4\cdot\dfrac{\partial}{\partial \theta}(r\sin\theta)\\ &=3r\sin\theta+4r\cos\theta \end{align*}$

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