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

Question

17 days ago

15

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.

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-03-31T00:16:20+03:00

Setting both partial derivatives to zero results in a single critical point at $(x,y)=\left(\dfrac12,0\right)$ , located within the unit disk.

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

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

and the second-order partial derivative with respect to $x$ yields

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

This suggests that the critical point represents a local minimum, marking it as the coldest area on the plate with a temperature of $T\left(\dfrac12,0\right)=-\dfrac14$ .

To find the hottest area on the plate, it must be located along the boundary. Let $x=\cos\theta$ and $y=\sin\theta$ , so that

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

Thus, the plate's boundary (the circle $x^2+y^2=1$ ) is treated as a single variable function $\theta$ examined over $\theta\in[0,2\pi)$ . A single differentiation gives

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

You will discover that $T(\theta)$ achieves three extrema on the interval $(0,2\pi)$ , with relative maxima occurring at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , and a relative minimum at $\theta=\pi$ (and $\theta=0$ , if you wish to include that).

Our minimum has already been identified inside the plate - which you can check to have a lower temperature than at the points noted by $T(\theta)$ - and we identify two maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , both showing a maximum temperature of $T=\dfrac94$ .

Reverting to Cartesian coordinates, these points match up with $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$ .