Question

A flat circular plate has the shape of the region x2 + y2≤1. The plate, including the boundary where x2 + y2 = 1, is heated such that the temperature at the point (x, y) is T (x, y) = x2 + 2y2-x. Find the temperatures in the hottest and coldest points on the board.

Answer 1

$T(x,y)=x^2+2y^2-x$
$\implies\nabla T(x,y)=(T_x,T_y)=(2x-1,4y)$

Setting both partial derivatives to 0 gives a single critical point at $(x,y)=\left(\dfrac12,0\right)$, which does fall inside the unit disk.

At this point, the value of the derivative of the Hessian matrix is

$|H|=\begin{vmatrix}T_{xx}&T_{xy}\\T_{yx}&T_{yy}\end{vmatrix}=\begin{vmatrix}2&0\\0&4\end{vmatrix}=8>0$

while the value of the second-order partial derivative with respect to $x$ is

$T_{xx}\bigg|_{(x,y)=(1/2,0)}=2>0$

This means the critical point is the site of a local minimum, so this is the coldest point on the plate with a temperature of $T\left(\dfrac12,0\right)=-\dfrac14$.

The hottest point on the plate must then be found on the boundary. Let $x=\cos\theta$ and $y=\sin\theta$, so that

$T(x,y)=T(\theta)=\cos^2\theta+2\sin^2\theta-\cos\theta$
$T(\theta)=\dfrac32-\cos\theta-\dfrac12\cos2\theta$

Then the boundary of the plate (the circle $x^2+y^2=1$) is a function of a single variable $\theta$ considered over $\theta\in[0,2\pi)$. Differentiating once gives

$T'(\theta)=\sin\theta+\sin2\theta=0$
$\implies\theta=0,\theta=\dfrac{2\pi}3,\theta=\pi,\theta=\dfrac{4\pi}3$

You'll find that $T(\theta)$ attains three extrema on the interval $(0,2\pi)$, with relative maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ and a relative minimum at $\theta=\pi$ (and $\theta=0$, if you want to include that).

We already found our minimum on the inside of our plate - which you can verify to have a lower temperature than at the points given by $T(\theta)$ - and we find two maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$, each giving a maximum temperature of $T=\dfrac94$.

Converting back to Cartesian coordinates, these points correspond to the points $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$.