Question

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

Find fr(2,1)f_r(2,1) and fθ(2,1)f_\theta(2,1), where rr and θ\theta are polar coordinates, x=rcosθx=r \cos \theta and y=rsinθy=r \sin \theta. If u\vec{u} is the unit vector in the direction 2i+j2 \vec{i}+\vec{j}, show that fu(2,1)=fr(2,1)f_{\vec{u}}(2,1)=f_r(2,1) and explain why this should be the case.

Solution

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We first note that for a function f(x,y)f(x,y) where xx and yy are functions of rr and θ\theta, its partial with respect to rr and θ\theta is given by:

fr(x,y)=fr=fxxr+fyyr=fxxr+fyxrf_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}

fθ(x,y)=fθ=fxxθ+fyyθ=fxxθ+fyxθf_{\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 fx(2,1)=3,fy(2,1)=4,x=rcosθf_x(2,1)=-3, f_y(2,1)=4, x=r\cos\theta and y=rsinθy=r\sin\theta, we have

fr(2,1)=3r(rcosθ)+4r(rsinθ)=3cosθ+4sinθfθ(2,1)=3θ(rcosθ)+4θ(rsinθ)=3rsinθ+4rcosθ\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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