/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Suppose a solid object in \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T(\text {the Laplacian of } T) .\) Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-x^{2}+y^{2}+z^{2}}$$

Short Answer

Expert verified
The heat flow vector field is given by: $$\mathbf{F} = \begin{bmatrix} 200 kx e^{-x^2 + y^2 + z^2} \\ -200 k y e^{-x^2 + y^2 + z^2} \\ 200 kz e^{-x^2 + y^2 + z^2} \end{bmatrix}$$ And the divergence of the heat flow vector field is: $$\nabla \cdot \mathbf{F} = -k((-400 + 800x^2)e^{-x^2 + y^2 + z^2} + (400 - 800y^2)e^{-x^2 + y^2 + z^2} + (-400 + 800z^2)e^{-x^2 + y^2 + z^2})$$

Step by step solution

01

Find the gradient of the temperature distribution function

First, we need to find \(\nabla T\). So, we will find the partial derivatives of \(T(x, y, z)\) with respect to \(x, y, z\) respectively, i.e., \(\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\) and then form the gradient vector. Given \(T(x, y, z) = 100 e^{-x^2 + y^2 + z^2}\), we can compute the partial derivatives: - \(\frac{\partial T}{\partial x} = -200 x e^{-x^2 + y^2 + z^2}\) - \(\frac{\partial T}{\partial y} = 200 y e^{-x^2 + y^2 + z^2}\) - \(\frac{\partial T}{\partial z} = -200 z e^{-x^2 + y^2 + z^2}\) Now we can form the gradient vector \(\nabla T\): $$\nabla T = \begin{bmatrix} -200 x e^{-x^2 + y^2 + z^2} \\ 200 y e^{-x^2 + y^2 + z^2} \\ -200 z e^{-x^2 + y^2 + z^2} \end{bmatrix}$$
02

Compute the heat flow vector field \(\mathbf{F}\)

Now that we have calculated the gradient, we will compute the heat flow vector \(\mathbf{F} = -k\nabla T\). We are given: $$\mathbf{F} = -k\nabla T = -k\begin{bmatrix} -200 x e^{-x^2 + y^2 + z^2} \\ 200 y e^{-x^2 + y^2 + z^2} \\ -200 z e^{-x^2 + y^2 + z^2} \end{bmatrix}= \begin{bmatrix} 200 kx e^{-x^2 + y^2 + z^2} \\ -200 k y e^{-x^2 + y^2 + z^2} \\ 200 kz e^{-x^2 + y^2 + z^2} \end{bmatrix}$$
03

Compute the Laplacian of the temperature distribution function

Now, we need to find the Laplacian of the given temperature distribution function, which is the sum of second partial derivatives with respect to \(x, y, z\). So, we will compute \(\nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}\). We have calculated the first partial derivatives in step 1, now we will find the second partial derivatives: - \(\frac{\partial^2 T}{\partial x^2} = -400 e^{-x^2 + y^2 + z^2} + 800x^2 e^{-x^2 + y^2 + z^2}\) - \(\frac{\partial^2 T}{\partial y^2} = 400 e^{-x^2 + y^2 + z^2} - 800y^2 e^{-x^2 + y^2 + z^2}\) - \(\frac{\partial^2 T}{\partial z^2} = -400 e^{-x^2 + y^2 + z^2} + 800z^2 e^{-x^2 + y^2 + z^2}\) Now, we sum up the second partial derivatives: $$\nabla^2 T = (-400 + 800x^2)e^{-x^2 + y^2 + z^2} + (400 - 800y^2)e^{-x^2 + y^2 + z^2} + (-400 + 800z^2)e^{-x^2 + y^2 + z^2}$$
04

Compute the divergence of the heat flow vector field

Now that we have the Laplacian of \(T\), we can calculate the divergence of the heat flow vector field \(\nabla \cdot \mathbf{F} = -k \nabla^2 T\). We are given: $$\nabla \cdot \mathbf{F} = -k \nabla^2 T = -k((-400 + 800x^2)e^{-x^2 + y^2 + z^2} + (400 - 800y^2)e^{-x^2 + y^2 + z^2} + (-400 + 800z^2)e^{-x^2 + y^2 + z^2})$$ Thus, the heat flow vector field is: $$\mathbf{F} = \begin{bmatrix} 200 kx e^{-x^2 + y^2 + z^2} \\ -200 k y e^{-x^2 + y^2 + z^2} \\ 200 kz e^{-x^2 + y^2 + z^2} \end{bmatrix}$$ And the divergence of the heat flow vector field is: $$\nabla \cdot \mathbf{F} = -k((-400 + 800x^2)e^{-x^2 + y^2 + z^2} + (400 - 800y^2)e^{-x^2 + y^2 + z^2} + (-400 + 800z^2)e^{-x^2 + y^2 + z^2})$$

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Key Concepts

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

Temperature Distribution
In the context of heat flow, the **temperature distribution** describes how the temperature varies at different points within a solid object. For a function defined as \( T(x, y, z) \), each coordinate \( (x, y, z) \) corresponds to a specific location in the object, providing a temperature value at that point. The given example function, \( T(x, y, z) = 100 e^{-x^2 + y^2 + z^2} \), implies an exponential decay, suggesting that the temperature changes at varying speeds depending on the coordinates.

This function essentially describes how hot or cold a particular point is within the object. The rate and direction at which these temperature changes occur lead us to concepts like the gradient, as they tell us how quickly and in which direction the temperature is increasing or decreasing. Applying mathematical functions to these temperature distributions can help predict and manage heat flow within various structures.
Gradient
The **gradient** of a temperature field is a vector that points in the direction of the greatest rate of increase of the temperature. Mathematically, it is denoted as \( abla T \) and comprises the partial derivatives of the temperature function with respect to each spatial coordinate. For the temperature distribution given, \( T(x, y, z) = 100 e^{-x^2 + y^2 + z^2} \), the gradient vector is calculated by finding the partial derivatives of \( T \) with respect to \( x \), \( y \), and \( z \).

  • The partial derivative with respect to \( x \) is \( \frac{\partial T}{\partial x} = -200 x e^{-x^2 + y^2 + z^2} \).
  • The partial derivative with respect to \( y \) is \( \frac{\partial T}{\partial y} = 200 y e^{-x^2 + y^2 + z^2} \).
  • The partial derivative with respect to \( z \) is \( \frac{\partial T}{\partial z} = -200 z e^{-x^2 + y^2 + z^2} \).
These derivatives are then combined into the gradient vector \( abla T = \begin{bmatrix} -200 x e^{-x^2 + y^2 + z^2} \ 200 y e^{-x^2 + y^2 + z^2} \ -200 z e^{-x^2 + y^2 + z^2} \end{bmatrix} \). This gradient indicates the rate and direction in which the temperature changes most rapidly in the object, crucial for understanding how heat might naturally flow through it.
Divergence
**Divergence** measures the magnitude of a source or sink at a given point within a vector field, essentially quantifying how much a vector field spreads out from a point. In the context of heat flow, the divergence of a vector field like \( \mathbf{F} = -k abla T \) gives insights into whether the field represents a converging or diverging flow at any point.

For the heat flow vector field given by \( \mathbf{F} \), the divergence is determined as \( abla \cdot \mathbf{F} = -k abla^2 T \) where \( abla^2 T \) is the Laplacian of the temperature field \( T \). The calculation of divergence here involves utilizing the previously derived Laplacian and applying the thermal conductivity constant \( k \) effectively to determine its impact on heat flow dynamics. The divergence can indicate areas in the object where heat is accumulating or dissipating, key to solving complex thermal management problems.
Laplacian
The **Laplacian** is a scalar operator that is a critical concept in understanding the behavior of temperature distributions. It is defined as the divergence of the gradient of a field, denoted \( abla^2 T \) for a temperature function \( T(x, y, z) \). In practical terms, it provides a measure of the rate at which the average value of the function around a point compares to the value at the point itself. A positive Laplacian indicates a local minimum surrounded by higher values, whereas a negative one includes a local maximum.

For our example temperature function \( T(x, y, z) = 100 e^{-x^2 + y^2 + z^2} \), the Laplacian \( abla^2 T \) involves calculating the second partial derivatives with respect to each variable:
  • \( \frac{\partial^2 T}{\partial x^2} = -400 e^{-x^2 + y^2 + z^2} + 800x^2 e^{-x^2 + y^2 + z^2} \).
  • \( \frac{\partial^2 T}{\partial y^2} = 400 e^{-x^2 + y^2 + z^2} - 800y^2 e^{-x^2 + y^2 + z^2} \).
  • \( \frac{\partial^2 T}{\partial z^2} = -400 e^{-x^2 + y^2 + z^2} + 800z^2 e^{-x^2 + y^2 + z^2} \).
These are summed to find \( abla^2 T \), which participates directly in determining the divergence of the heat flow vector field. Through these calculations, the Laplacian offers insights into how temperature changes are distributed spatially within the object, a vital part of predicting heat flow behaviors.

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Most popular questions from this chapter

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.

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 \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\)

Surfaces of revolution Suppose \(y=f(x)\) is a continuous and positive function on \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. a. Show that \(S\) is described parametrically by \(\mathbf{r}(u, v)=\langle u, f(u) \cos v, f(u) \sin v\rangle,\) for \(a \leq u \leq b\) \(0 \leq v \leq 2 \pi\) b. Find an integral that gives the surface area of \(S\) c. Apply the result of part (b) to the surface generated with \(f(x)=x^{3},\) for \(1 \leq x \leq 2\) d. Apply the result of part (b) to the surface generated with \(f(x)=\left(25-x^{2}\right)^{1 / 2},\) for \(3 \leq x \leq 4\)

Rotated Green's Theorem Use Stokes' Theorem to write the circulation form of Green's Theorem in the \(y z\) -plane.

Alternative construction of potential functions in \(\mathbb{R}^{2}\) Assume the vector field \(\mathbf{F}\) is conservative on \(\mathbb{R}^{2}\), so that the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is independent of path. Use the following procedure to construct a potential function \(\varphi\) for the vector field \(\mathbf{F}=\langle f, g\rangle=\langle 2 x-y,-x+2 y\rangle\) a. Let \(A\) be (0,0) and let \(B\) be an arbitrary point \((x, y) .\) Define \(\varphi(x, y)\) to be the work required to move an object from \(A\) to \(B\) where \(\varphi(A)=0 .\) Let \(C_{1}\) be the path from \(A\) to \((x, 0)\) to \(B\), and let \(C_{2}\) be the path from \(A\) to \((0, y)\) to \(B\). Draw a picture. b. Evaluate \(\int_{C_{1}} \mathbf{F} \cdot d \mathbf{r}=\int_{C_{1}} f d x+g d y\) and conclude that \(\varphi(x, y)=x^{2}-x y+y^{2}\). c. Verify that the same potential function is obtained by evaluating the line integral over \(C_{2}\).

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