Examples

Particle moving through temperature gradient

Particle moving through temperature gradient

Problem: A particle traverses a parametric curve in the plane given by $\mathbf{r}(t)=(\cos (t),e^t)$. The ambient temperature is a function of distance from the $y$-axis: $T(x,y)=\frac{1}{x^2+3}$. How fast is the temperature changing at $t=4$?
Solution: We compute $\frac{\partial T}{\partial x}=\frac{-2x}{(x^2+3{)}^2}$ and $\frac{\partial T}{\partial y}=0$, while $x^{\prime }(t)=-\sin (t)$. Then:

$$\frac{df}{dt}=\frac{2\cos (t)\sin (t)}{({\cos }^2(t)+3{)}^2},{\frac{df}{dt}|}_{t=4}=\frac{2\cos (4)\sin (4)}{({\cos }^2(4)+3{)}^2}\approx \mathrm{0.084.}$$

Polar partials and chain rule

Polar partials and chain rule

Problem: Suppose $f(x,y)$ is a real-valued function of points in the plane. Let $r$ and $\theta$ be the standard polar coordinates on the plane.

• (a) Write $\frac{\partial f}{\partial \theta }$ in terms of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ and $x$ and $y$.
• (b) Suppose $f=x^2{y}^3$. Find $f_{\theta }$ at $(x,y)=(1,1)$.

Solution: (a) First compute $\frac{\partial x}{\partial \theta }=-r\sin (\theta )$ and $\frac{\partial y}{\partial \theta }=r\cos (\theta )$. Then:

$$\begin{array}{rl}\frac{\partial f}{\partial \theta }& =\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta }\\ & =-r\sin (\theta )\frac{\partial f}{\partial x}+r\cos (\theta )\frac{\partial f}{\partial y}\\ & =-y\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y}.\end{array}$$

(b) Compute $f_x=2x{y}^3$, $f_y=3x^2{y}^2$. So $f_{\theta }=-2x{y}^4+3x^3{y}^2$ and $f_{\theta }(1,1)=-2+3=1$.