Mathematics, 28.09.2019 10:10 vm204618

Aflat 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.

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Answer from: Asrys

The temperature of the hottest point on the board is $\mathbf{\dfrac{9}{4}}$ and the temperature of the coldest points on the board is $\mathbf{\dfrac{1}{2}}$

The temperature at the point (x,y) on a plate is;

$\mathbf{T_{x,y}=x^2+2y^2-x}$

We need to determine the partial derivatives of T(x,y) with respect to x and y and equate them to zero.

$\mathbf{T_{x} = 2x -1 \ \\ ; \ \ T_y = 4y}$

By equating, $\mathbf{T_x = 0}$ , we have:

2x - 1 = 02x = 1x = 1/2

By equating, $\mathbf{T_y= 0}$ , we have

4y = 0y = 0/4y = 0

∴

The stationary points are $\mathbf{\Big(\dfrac{1}{2}, 0\Big)}$

By substituting the stationary points into the equation x² +2y² -x, we can determine the temperature at the coldest points as follows:

$\implies \mathbf{x^2+2y^2 -x}$

$\implies \mathbf{(\dfrac{1}{2})^2+2(0)^2 -\dfrac{1}{2}}$

$\mathbf{\implies \dfrac{1}{2}}$

Therefore, the temperature at the coldest point is $\mathbf{\dfrac{1}{2}}$

Recall that the plate including the region is given by the boundary:

x² + y² = 1 ----- (1)y² = 1 - x² ----- (2)

Let's replace the value of y in equation (2) into (1);

$\mathbf{T_{((x,y)} = x^2 +2y^2 -x }$ $\mathbf{T_{((x)} = x^2 +2(1-x)^2 -x }$ $\mathbf{T_{(x)} = x^2 +2-2x^2 -x }$ $\mathbf{T_{(x)} = -x^2 -x+2 \ --- (3) }$

By differentiating equation (3) with respect to (x), we have:

T'(x) = -2x - 1

To determine the stationary points, let T(x) = 0;

-2x - 1 = 0-2x = 1x = -1/2

Replacing the above value of x into equation (2), we have:

y² = 1 - x² $\mathbf{y^2 = 1 - (-\dfrac{1}{2})^2}$ $\mathbf{y^2 = (1 -\dfrac{1}{4})}$ $\mathbf{y^2 = ( -\dfrac{3}{4})}$ $\mathbf{y = \dfrac{\pm \sqrt{3}}{2}}$

Thus, the stationary points are $\mathbf{\Big( -\dfrac{1}{2}, \dfrac{\pm \sqrt{3}}{2} \Big)}$

Finally, the temperature at the hottest point $\mathbf{\Big( -\dfrac{1}{2}, \dfrac{\pm \sqrt{3}}{2} \Big)}$ is determined by using the values of the stationary point in equation (1);

∴

$\mathbf{x^2 +2y^2 -x }$

Replacing the values of x and y from the stationary point, we have:

$\mathbf{x^2 +2y^2 -x }$ $\mathbf{(-\dfrac{1}{2} )^2 +2(\dfrac{\sqrt{3}}{2})^2 -(-\dfrac{1}{2}) }$ $\mathbf{(\dfrac{1}{4} ) +2(\dfrac{3}{4}) +(\dfrac{1}{2}) }$ $\mathbf{(\dfrac{1}{4} ) +(\dfrac{6}{4}) +(\dfrac{1}{2}) }$ $\mathbf{=\dfrac{1+6+2}{4} }$ $\mathbf{=\dfrac{9}{4} }$

Therefore, we can conclude that the temperature of the hottest point on the board is $\mathbf{\dfrac{9}{4}}$ and the temperature of the coldest points on the board is $\mathbf{\dfrac{1}{2}}$

Learn more about differentiation here:

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Answer from: graymonky12

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

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

$|H|=\begin{vmatrix}T_{xx}&T_{xy}\\T_{yx}&T_{yy}\end{vmatrix}=\begin{vmatrix}2&0\\0&4\end{vmatrix}=80$

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

is

$T_{xx}\bigg|_{(x,y)=(1/2,0)}=20$

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 $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$ .

Answer from: Quest

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Answer from: Quest

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