lucasjoystick
contestada

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.

Respuesta :

LammettHash
[tex]T(x,y)=x^2+2y^2-x[/tex]
[tex]\implies\nabla T(x,y)=(T_x,T_y)=(2x-1,4y)[/tex]

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

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

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

while the value of the second-order partial derivative with respect to [tex]x[/tex] is

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

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 [tex]T\left(\dfrac12,0\right)=-\dfrac14[/tex].

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

[tex]T(x,y)=T(\theta)=\cos^2\theta+2\sin^2\theta-\cos\theta[/tex]
[tex]T(\theta)=\dfrac32-\cos\theta-\dfrac12\cos2\theta[/tex]

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

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

You'll find that [tex]T(\theta)[/tex] attains three extrema on the interval [tex](0,2\pi)[/tex], with relative maxima at [tex]\theta=\dfrac{2\pi}3[/tex] and [tex]\theta=\dfrac{4\pi}3[/tex] and a relative minimum at [tex]\theta=\pi[/tex] (and [tex]\theta=0[/tex], 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 [tex]T(\theta)[/tex] - and we find two maxima at [tex]\theta=\dfrac{2\pi}3[/tex] and [tex]\theta=\dfrac{4\pi}3[/tex], each giving a maximum temperature of [tex]T=\dfrac94[/tex].

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

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

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

  • [tex]\mathbf{T_{x,y}=x^2+2y^2-x}[/tex]  

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

  • [tex]\mathbf{T_{x} = 2x -1 \ \\ ; \ \ T_y = 4y}[/tex]

By equating, [tex]\mathbf{T_x = 0}[/tex], we have:

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

By equating, [tex]\mathbf{T_y= 0}[/tex], we have

  • 4y = 0
  • y = 0/4
  • y = 0

∴

The stationary points are [tex]\mathbf{\Big(\dfrac{1}{2}, 0\Big)}[/tex]

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

[tex]\implies \mathbf{x^2+2y^2 -x}[/tex]

[tex]\implies \mathbf{(\dfrac{1}{2})^2+2(0)^2 -\dfrac{1}{2}}[/tex]

[tex]\mathbf{\implies \dfrac{1}{2}}[/tex]

Therefore, the temperature at the coldest point is [tex]\mathbf{\dfrac{1}{2}}[/tex]

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);

  • [tex]\mathbf{T_{((x,y)} = x^2 +2y^2 -x }[/tex]
  • [tex]\mathbf{T_{((x)} = x^2 +2(1-x)^2 -x }[/tex]
  • [tex]\mathbf{T_{(x)} = x^2 +2-2x^2 -x }[/tex]
  • [tex]\mathbf{T_{(x)} = -x^2 -x+2 \ --- (3) }[/tex]

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 = 1
  • x = -1/2

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

  • y² = 1 - x²
  • [tex]\mathbf{y^2 = 1 - (-\dfrac{1}{2})^2}[/tex]
  • [tex]\mathbf{y^2 = (1 -\dfrac{1}{4})}[/tex]
  • [tex]\mathbf{y^2 = ( -\dfrac{3}{4})}[/tex]
  • [tex]\mathbf{y = \dfrac{\pm \sqrt{3}}{2}}[/tex]

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

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

∴

  • [tex]\mathbf{x^2 +2y^2 -x }[/tex]

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

  • [tex]\mathbf{x^2 +2y^2 -x }[/tex]

  • [tex]\mathbf{(-\dfrac{1}{2} )^2 +2(\dfrac{\sqrt{3}}{2})^2 -(-\dfrac{1}{2}) }[/tex]

  • [tex]\mathbf{(\dfrac{1}{4} ) +2(\dfrac{3}{4}) +(\dfrac{1}{2}) }[/tex]

  • [tex]\mathbf{(\dfrac{1}{4} ) +(\dfrac{6}{4}) +(\dfrac{1}{2}) }[/tex]

  • [tex]\mathbf{=\dfrac{1+6+2}{4} }[/tex]

  • [tex]\mathbf{=\dfrac{9}{4} }[/tex]  

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

Learn more about differentiation here:

https://brainly.com/question/14620493?referrer=searchResults

Otras preguntas