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Chapter 17: Problem 78

A square plate \(R=\\{(x, y):\) \(0 \leq x \leq 1,0 \leq y \leq 1\\}\) has a temperature distribution \(T(x, y)=100-50 x-25 y\) a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature \(\nabla T(x, y)\) c. Assume the flow of heat is given by the vector field \(\mathbf{F}=-\nabla T(x, y) .\) Compute \(\mathbf{F}\) d. Find the outward heat flux across the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\) e. Find the outward heat flux across the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\)

Short Answer

Expert verified
Question: Sketch two level curves of the temperature in the plate and find the outward heat flux across the boundaries x=1 and y=1. Solution: Step 1: For the given temperature function \(T(x,y)=100-50x-25y\), we find two level curves corresponding to the temperature values \(T_1=75\) and \(T_2=50\). Solve for x and y: For \(T_1=75\): \(100-50x-25y=75\) For \(T_2=50\): \(100-50x-25y=50\) Step 2: Compute the gradient of the temperature function: \(\nabla T(x,y)=\begin{bmatrix} -50 \\ -25 \end{bmatrix}\) Step 3: Calculate the heat flow vector field: \(\mathbf{F}= \begin{bmatrix} 50 \\ 25 \end{bmatrix}\) Step 4: For the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\), the outward heat flux is 50. Step 5: For the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\), the outward heat flux is 25.

Step by step solution

01

Find Two Level Curves of the Temperature

Let's find two possible values for the temperature, say \(T_1=75\) and \(T_2=50\). Now, we need to solve the equation for x and y coordinates for these two temperature values. For \(T_1=75\), we have: \(T(x,y)=100-50x-25y=75\) For \(T_2=50\), we have: \(T(x,y)=100-50x-25y=50\)
02

Compute Gradient of Temperature

The gradient of a function is defined as the vector of its partial derivatives with respect to the variables. Calculate the partial derivatives of the temperature function \(T(x,y)\) with respect to x and y: \(\frac{\partial T}{\partial x}= -50\) \(\frac{\partial T}{\partial y}= -25\) So the gradient is: \(\nabla T(x,y)=\begin{bmatrix} -50 \\ -25 \end{bmatrix}\)
03

Compute Heat Flow Vector Field

Now, we can compute the heat flow vector field using \(\mathbf{F}=-\nabla T(x,y)\). Therefore, \(\mathbf{F}=- \begin{bmatrix} -50 \\ -25 \end{bmatrix}= \begin{bmatrix} 50 \\ 25 \end{bmatrix}\)
04

Find Outward Heat Flux Across the Boundary x=1

The outward normal vector for the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\) is \(\mathbf{n}=\begin{bmatrix} 1\\ 0 \end{bmatrix}\). Compute the dot product of \(\mathbf{F}\) and \(\mathbf{n}\): \(\mathbf{F} \cdot \mathbf{n}=\begin{bmatrix} 50 \\ 25 \end{bmatrix}\cdot \begin{bmatrix} 1\\ 0 \end{bmatrix} = 50\) Now, integrate the dot product over the given boundary: \(\int_{0}^{1} 50 dy=\left[50y\right]_0^1=50\) Therefore, the outward heat flux across the boundary x=1 is 50.
05

Find Outward Heat Flux Across the Boundary y=1

The outward normal vector for the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\) is \(\mathbf{n}=\begin{bmatrix} 0\\ 1 \end{bmatrix}\). Compute the dot product of \(\mathbf{F}\) and \(\mathbf{n}\): \(\mathbf{F} \cdot \mathbf{n}= \begin{bmatrix} 50 \\ 25 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 1 \end{bmatrix} = 25\) Now, integrate the dot product over the given boundary: \(\int_{0}^{1} 25 dx=\left[25x\right]_0^1=25\) Therefore, the outward heat flux across the boundary y=1 is 25.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Level Curves in Calculus
In calculus, level curves are an essential concept to visualize the behavior of a function of two variables on a plane. They represent a set of points where the function has the same value. Picture contour lines on a geographic map that indicate regions of equal altitude and you've grasped the idea behind level curves. When applied to the temperature distribution of a surface, such as a square plate in our exercise, these curves can show areas of equal temperature.

For instance, if you're given the temperature distribution function like T(x, y) = 100 - 50x - 25y, you can sketch level curves for specific temperature values. For temperatures T1 = 75 and T2 = 50, you get two different equations representing curves on the plate where the temperature is constant at 75 and 50 degrees, respectively. These help us visualize the temperature gradient in a tangible way.
Vector Field in Calculus
A vector field in calculus is a tool that assigns a vector to every point in space. Think of it as a map that tells you how the wind blows at different spots in the atmosphere. In the context of temperature and heat flow, which is our topic of interest, it's used to describe the direction and magnitude of heat transfer at different points on a surface.

To find the vector field representing the flow of heat, which is commonly denoted as \(\textbf{F}\), you'll often take the negative gradient of the temperature distribution. The gradient, which is the vector of partial derivatives with respect to the variables, \(abla T(x, y)\), points in the direction of the greatest rate of increase of the function. Therefore, the negative gradient, \(-abla T(x, y)\), reveals the direction in which heat is flowing (towards lower temperatures) across the surface of the plate, and its magnitude shows the rate of heat transfer.
Heat Flux in Calculus
Heat flux is a term that quantifies the rate at which heat energy passes through a surface. In calculus, it's calculated by taking the dot product of the heat flow vector field with an outward normal vector of a designated boundary on the surface. Essentially, it tells us how much heat is moving out of the surface through a certain boundary.

In the exercise, we calculate the outward heat flux across two boundaries of a square plate: one where x = 1 and the other where y = 1. By taking the dot products \(\textbf{F} \text{銉粆 \textbf{n}\), where \(\textbf{n}\) is the outward normal vector for each boundary, and integrating along the boundary, the heat flux can be quantified. For example, across the boundary x = 1, where \(\textbf{n} = [1, 0]\), the heat flux is found to be 50, suggesting that 50 units of heat energy are passing through that side of the plate per unit time.

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Most popular questions from this chapter

The rotation of a threedimensional velocity field \(\mathbf{V}=\langle u, v, w\rangle\) is measured by the vorticity \(\omega=\nabla \times \mathbf{V} .\) If \(\omega=\mathbf{0}\) at all points in the domain, the flow is irrotational. a. Which of the following velocity fields is irrotational: \(\mathbf{V}=\langle 2,-3 y, 5 z\rangle\) or \(\mathbf{V}=\langle y, x-z,-y\rangle ?\) b. Recall that for a two-dimensional source-free flow \(\mathbf{V}=\langle u, v, 0\rangle,\) a stream function \(\psi(x, y)\) may be defined such that \(u=\psi_{y}\) and \(v=-\psi_{x} .\) For such a two-dimensional flow, let \(\zeta=\mathbf{k} \cdot \nabla \times \mathbf{V}\) be the \(\mathbf{k}\) -component of the vorticity. Show that \(\nabla^{2} \psi=\nabla \cdot \nabla \psi=-\zeta\) c. Consider the stream function \(\psi(x, y)=\sin x \sin y\) on the square region \(R=\\{(x, y): 0 \leq x \leq \pi, 0 \leq y \leq \pi\\} .\) Find the velocity components \(u\) and \(v\); then sketch the velocity field. d. For the stream function in part (c), find the vorticity function \(\zeta\) as defined in part (b). Plot several level curves of the vorticity function. Where on \(R\) is it a maximum? A minimum?

Find the upward flux of the field \(\mathbf{F}=\langle x, y, z\rangle\) across the plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) in the first octant where \(a, b,\) and \(c\) are positive real numbers. Show that the flux equals \(c\) times the area of the base of the region. Interpret the result physically.

Conservation of energy Suppose an object with mass \(m\) moves in a region \(R\) in a conservative force field given by \(\mathbf{F}=-\nabla \varphi,\) where \(\varphi\) is a potential function in a region \(R .\) The motion of the object is governed by Newton's Second Law of Motion, \(\mathbf{F}=m \mathbf{a},\) where a is the acceleration. Suppose the object moves from point \(A\) to point \(B\) in \(R\) a. Show that the equation of motion is \(m \frac{d \mathbf{v}}{d t}=-\nabla \varphi\) b. Show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\) c. Take the dot product of both sides of the equation in part (a) with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\) and integrate along a curve between \(A\) and B. Use part (b) and the fact that \(\mathbf{F}\) is conservative to show that the total energy (kinetic plus potential) \(\frac{1}{2} m|\mathbf{v}|^{2}+\varphi\) is the same at \(A\) and \(B\). Conclude that because \(A\) and \(B\) are arbitrary, energy is conserved in \(R\).

Find the exact points on the circle \(x^{2}+y^{2}=2\) at which the field \(\mathbf{F}=\langle f, g\rangle=\left\langle x^{2}, y\right\rangle\) switches from pointing inward to pointing outward on the circle, or vice versa.

Green's Second Identity Prove Green's Second Identity for scalar-valued functions \(u\) and \(v\) defined on a region \(D:\) $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$ (Hint: Reverse the roles of \(u\) and \(v\) in Green's First Identity.)
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