/*! 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 58 Suppose a solid object in \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T\) (the Laplacian of \(T\)). Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100(1+\sqrt{x^{2}+y^{2}+z^{2}})$$

Short Answer

Expert verified
Question: Calculate the divergence of the heat flow vector field for the temperature distribution \(T(x, y, z) = 100(1+\sqrt{x^{2}+y^{2}+z^{2}})\), given that the heat flow vector field is \(\mathbf{F} = -k \nabla T\), where k > 0. Answer: The divergence of the heat flow vector field is given by \(\nabla \cdot \mathbf{F} = 100k(x^2+y^2+z^2)^{-1/2}\).

Step by step solution

01

Find the gradient of T(x, y, z)

To find the gradient of the temperature distribution \(T(x, y, z) = 100(1+\sqrt{x^{2}+y^{2}+z^{2}})\), we need to find the partial derivatives with respect to each variable and arrange them as a vector: $$\nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right)$$ Compute the partial derivatives: $$\frac{\partial T}{\partial x} = 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}$$ $$\frac{\partial T}{\partial y} = 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}$$ $$\frac{\partial T}{\partial z} = 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}}$$ Combining these partial derivatives, we have the gradient of \(T\): $$\nabla T = \left( 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}} \right)$$
02

Compute the heat flow vector field F(x, y, z)

We are given that \(\mathbf{F} = -k \nabla T\), where \(k > 0\) is the conductivity. Substitute the gradient of \(T\) found in Step 1: $$\mathbf{F} = -k \cdot \left( 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}} \right)$$
03

Compute the divergence of the heat flow vector field F(x, y, z)

From \(\nabla \cdot \mathbf{F} = -k \nabla^{2} T\), we want to compute \(\nabla^{2} T\). The Laplacian of \(T\) is given by: $$\nabla^{2} T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}$$ Compute the second partial derivatives and sum them up: $$\frac{\partial^2 T}{\partial x^2} = -100 \cdot \frac{x^2}{(x^2+y^2+z^2)^{3/2}}$$ $$\frac{\partial^2 T}{\partial y^2} = -100 \cdot \frac{y^2}{(x^2+y^2+z^2)^{3/2}}$$ $$\frac{\partial^2 T}{\partial z^2} = -100 \cdot \frac{z^2}{(x^2+y^2+z^2)^{3/2}}$$ Summing them up, we have: $$\nabla^{2} T = -100 \cdot \frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^{3/2}} = -100(x^2+y^2+z^2)^{-1/2}$$ Finally, calculate the divergence of the heat flow vector field as \(\nabla \cdot \mathbf{F} = -k \nabla^{2} T\): $$\nabla \cdot \mathbf{F} = -k(-100(x^2+y^2+z^2)^{-1/2}) = 100k(x^2+y^2+z^2)^{-1/2}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Temperature Distribution
In the study of heat flow, understanding the temperature distribution within a solid object is crucial. This distribution reflects how temperature varies at different points in the object. For example, consider the function \(T(x, y, z) = 100(1+\sqrt{x^2+y^2+z^2})\). Here, \(T\) denotes the temperature at any given point \((x, y, z)\) in the three-dimensional space \(\mathbb{R}^3\).

In this equation, the temperature depends on the distance from the origin, represented by \(\sqrt{x^2 + y^2 + z^2}\). This means that points further from the origin tend to have different temperatures.

Understanding temperature distribution helps us determine how heat will flow within the object. Areas with higher temperature will typically transfer heat to areas with lower temperature, influenced by the material's conductivity that dictates how well the heat is conducted.
Gradient of a Scalar Field
The gradient of a scalar field, like our temperature function \(T\), is a vector field that points in the direction of the greatest increase of the function. Mathematically, it is denoted by \(abla T\).

For our temperature function \(T(x, y, z) = 100(1+\sqrt{x^2+y^2+z^2})\), the gradient is calculated by finding the partial derivatives with respect to \(x\), \(y\), and \(z\):
  • \(\frac{\partial T}{\partial x} = 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}\)
  • \(\frac{\partial T}{\partial y} = 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}\)
  • \(\frac{\partial T}{\partial z} = 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}}\)
The gradient vector then becomes \(abla T = \left( 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}, 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}} \right)\).

This vector indicates the direction and rate of the greatest temperature increase at each point.
Divergence and Laplacian
The divergence of a vector field measures how much the field is spreading out from a point. For the heat flow vector field \(\mathbf{F} = -k abla T\), the divergence is \(abla \cdot \mathbf{F}\).

The Laplacian \(abla^2 T\) of the scalar field \(T\) is a measure of the field's curvature. It's critical in understanding how temperature changes across space.

To compute the Laplacian for \(T(x, y, z)\), we sum the second partial derivatives:
  • \(\frac{\partial^2 T}{\partial x^2} = -100 \cdot \frac{x^2}{(x^2+y^2+z^2)^{3/2}}\)
  • \(\frac{\partial^2 T}{\partial y^2} = -100 \cdot \frac{y^2}{(x^2+y^2+z^2)^{3/2}}\)
  • \(\frac{\partial^2 T}{\partial z^2} = -100 \cdot \frac{z^2}{(x^2+y^2+z^2)^{3/2}}\)
Adding these, the Laplacian is \(abla^2 T = -100(x^2+y^2+z^2)^{-1/2}\).

Finally, the divergence of \(\mathbf{F}\) is \(abla \cdot \mathbf{F} = 100k(x^2+y^2+z^2)^{-1/2}\), showing how heat flows away from or towards a point in the material.
Partial Derivatives
Partial derivatives are essential tools for analyzing functions of multiple variables, like our temperature function \(T(x, y, z)\).

They represent the rate of change of the function concerning one variable while keeping the others constant. In our example, to find \(\frac{\partial T}{\partial x}\), you treat \(y\) and \(z\) as constants and find how \(T\) changes as \(x\) changes.

Computing partial derivatives:
  • \(\frac{\partial T}{\partial x} = 100 \cdot \frac{x}{\sqrt{x^2+y^2+z^2}}\)
  • \(\frac{\partial T}{\partial y} = 100 \cdot \frac{y}{\sqrt{x^2+y^2+z^2}}\)
  • \(\frac{\partial T}{\partial z} = 100 \cdot \frac{z}{\sqrt{x^2+y^2+z^2}}\)
These derivatives combine to form the gradient \(abla T\), illustrating how the temperature changes in each direction.

Such calculations refine our understanding of the temperature distribution and the resulting heat flow within an object.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write Gauss' Formula of Exercise 48 in two dimensions- -that is, where \(\mathbf{F}=\langle f, g\rangle, D\) is a plane region \(R\) and \(C\) is the boundary of \(R .\) Show that the result is Green's Formula: $$\iint_{R} u\left(f_{x}+g_{y}\right) d A=\oint u(\mathbf{F} \cdot \mathbf{n}) d s-\iint_{R}\left(f u_{x}+g u_{y}\right) d A$$ Show that with \(u=1,\) one form of Green's Theorem appears. Which form of Green's Theorem is it?

Prove that the radial field \(\mathbf{F}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(p\) is a real number, is conservative on any region not containing the origin. For what values of \(p\) is \(\mathbf{F}\) conservative on a region that contains the origin?

The goal is to evaluate \(A=\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S,\) where \(\mathbf{F}=\langle y z,-x z, x y\rangle\) and \(S\) is the surface of the upper half of the ellipsoid \(x^{2}+y^{2}+8 z^{2}=1(z \geq 0)\) a. Evaluate a surface integral over a more convenient surface to find the value of \(A\) b. Evaluate \(A\) using a line integral.

Begin with the paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4,\) and slice it with the plane \(y=0\) Let \(S\) be the surface that remains for \(y \geq 0\) (including the planar surface in the \(x z\) -plane) (see figure). Let \(C\) be the semicircle and line segment that bound the cap of \(S\) in the plane \(z=4\) with counterclockwise orientation. Let \(\mathbf{F}=\langle 2 z+y, 2 x+z, 2 y+x\rangle\) a. Describe the direction of the vectors normal to the surface that are consistent with the orientation of \(C\). b. Evaluate \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) c. Evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) and check for agreement with part (b).

Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-\mathbf{G}(\nabla \cdot \mathbf{F})-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.