/*! 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 20 The unit vector normal to the su... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 20

The unit vector normal to the surface \(z=f(x, y)\) is given by $$ \hat{\mathbf{n}}=\left(-\mathbf{i} \frac{\partial f}{\partial x}-\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k}\right) \mid \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}} $$ [see Equation (II-4)]. We have also established that \(\nabla F\) is a vector normal to the surface \(F(x, y, z)=\) const. (page 140 ) so that \(\nabla F /|\nabla F|\) is a unit vector normal to the surface \(F(x, y, z)=\) const. Show that these two expressions for the unit normal vector are identical if \(F(x, y, z)=\) const. and \(z=f(x, y)\) describe the same surface.

Short Answer

Expert verified
The two expressions for the unit normal vector are identical, which confirms they describe the same surface.

Step by step solution

01

Express the Gradient of the Surface

The surface described by \( F(x, y, z) = \text{const.} \) implies \( F(x, y, z) = z - f(x, y) = 0 \). Thus, \( F(x, y, z) = z - f(x, y) \). The gradient of this function, \( abla F \), which is a normal vector to the surface, is given by:\[ abla F = \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \].
02

Calculate the Magnitude of the Normal Vector

The magnitude of the gradient \( abla F \) is the square root of the sum of the squares of its components:\[ |abla F| = \sqrt{ \left( -\frac{\partial f}{\partial x} \right)^2 + \left( -\frac{\partial f}{\partial y} \right)^2 + 1^2 } = \sqrt{ 1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 } \].
03

Express the Unit Normal Vector

The unit normal vector is the normalized form of the gradient:\[ \frac{abla F}{|abla F|} = \left( \frac{-\frac{\partial f}{\partial x}}{|abla F|}, \frac{-\frac{\partial f}{\partial y}}{|abla F|}, \frac{1}{|abla F|} \right) \].
04

Compare with Given Unit Vector

The given unit vector normal to the surface \( z = f(x, y) \) is:\[ \hat{\mathbf{n}} = \left( -\mathbf{i} \frac{\partial f}{\partial x} - \mathbf{j} \frac{\partial f}{\partial y} + \mathbf{k} \right) \cdot \frac{1}{\sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2}} \].This matches the expression for \( \frac{abla F}{|abla F|} \), confirming they are the same.

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Key Concepts

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

Gradient Vector
The concept of a gradient vector is vital in understanding how a surface behaves in space. Think about a function of the form \( F(x, y, z) = z - f(x, y) \). Here, the gradient, often denoted as \( abla F \), is a vector that points in the direction of the steepest increase of the function. For our exercise, this function represents a surface, and the gradient vector is perpendicular to this surface.

The gradient vector \( abla F \) consists of partial derivatives \( \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \). Each component of this vector provides crucial insight:
  • The negative signs indicate the direction of decreasing x and y.
  • The partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) tell us how the function \( f \) changes as x or y changes.
  • The constant "1" shows that the z-component is only influenced by the constant term in our function, ensuring the vector points perpendicularly out of the plane.
Thus, the gradient vector helps us see how a surface element of \( F(x, y, z) \) is oriented in space.
Surface Normal
A surface normal is a vector that is perpendicular, or orthogonal, to a surface at a given point. This concept is essential in defining a unit normal vector. For a surface described by both \( z = f(x, y) \) and \( F(x, y, z) = \text{const.} \), our goal is to show that the normal vectors derived from these two expressions are identical. This involves finding the unit vector from the gradient.

For \( F(x, y, z) = z - f(x, y) \), we previously found the gradient vector \( abla F = \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \). Calculating the magnitude of this vector gives:\[|abla F| = \sqrt{ 1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 }\]This results in a normal vector \( abla F / |abla F| \), making it a unit normal vector. Thus, both expressions rely on the concept that for a surface, the gradient's direction gives a normal鈥攅ven more so when normalized. Ponder over how this insight helps in visualizing and calculating the behavior of surfaces.
Partial Derivatives
Partial derivatives are tools that examine how a function changes when one of its variables is varied while keeping the others constant. For our surface \( z = f(x, y) \), these partial derivatives are \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). They tell us how the height \( z \) changes concerning x or y, respectively. This change is represented in the components of the gradient vector.

Understanding partial derivatives is crucial as they define the rate of change of the surface's position:
  • \( \frac{\partial f}{\partial x} \): Indicates how \( z \) changes as x is varied, with y held constant.
  • \( \frac{\partial f}{\partial y} \): Describes the change in \( z \) when varying y, with x constant.
These derivatives show the tilt of the surface slope in different directions, helping us construct the vector \( abla F \). Thus, they are intrinsic in understanding the nature of surfaces and their orientations in space.

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Most popular questions from this chapter

Verify the following identities in which \(f\) and \(g\) are arbitrary differentiable scalar functions of position, and \(\mathbf{F}\) and \(\mathbf{G}\) are arbitrary differentiable vector functions of position. (a) \(\nabla(f g)=f \nabla g+g \nabla f\). (b) \(\nabla(\mathbf{F} \cdot \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}+(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F} \times(\nabla \times \mathbf{G})+\mathbf{G} \times\) \((\nabla \times \mathbf{F})\) (c) \(\nabla \cdot(f \mathbf{F})=f \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla f\) (d) \(\nabla \cdot(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G})\) (e) \(\nabla \times(f \mathbf{F})=f \nabla \times \mathbf{F}+(\nabla f) \times \mathbf{F}\) (f) \(\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})-\mathbf{G}(\nabla \cdot \mathbf{F})\) (g) \(\nabla \times(\nabla \times \mathbf{F})=\nabla(\nabla \cdot \mathbf{F})-\nabla^{2} \mathbf{F}\)

Suppose you find a solution of Laplace's equation that satisfies certain boundary conditions. Is this solution unique or are there others? This problem will answer that question in certain simple cases. Consider the region of space completely enclosed by a surface \(S_{0}\) and containing in its interior objects \(1,2,3, \ldots\) (two of which are pictured in the diagram). Suppose that \(S_{0}\) is maintained at a constant potential \(\Phi_{0}\), object no. 1 at \(\Phi_{1}\), object no. 2 at \(\Phi_{2}\), and so on. Then in the charge-free region \(R\) enclosed by \(S_{0}\) and between the objects, the potential must satisfy Laplace's equation $$ \nabla^{2} \Phi=0 $$ and the boundary conditions $$ \Phi=\left\\{\begin{array}{c} \Phi_{0} \text { on } S_{0} \\ \Phi_{1} \text { on } S_{1} \\ \Phi_{2} \text { on } S_{2} \\ \vdots \end{array}\right. $$ The following steps will guide you through a proof that \(\Phi\) is unique. (a) Assume that there are two potentials \(u\) and \(v\), both of which satisfy Laplace's equation and the boundary conditions listed earlier. Form their difference \(w=u-v\). Show that \(\nabla^{2} w=0\) in \(R\). (b) What are the boundary conditions satisfied by \(w\) ? (c) Apply the divergence theorem to $$ \iint_{S} \hat{\mathbf{n}} \cdot(w \nabla w) d S $$ where the integration is carried out over the surface \(S_{0}+S_{1}+S_{2}\) \(+\cdots\), and show thereby that $$ \iiint_{V}|\nabla w|^{2} d V=0 $$ where \(V\) is the volume of the region \(R\). (d) From the result of (c) argue that \(\nabla w=0\) and that this, in turn, means \(w\) is a constant. (e) If \(w\) is a constant, what is its value? (Use the boundary conditions on \(w\) to answer this.) What does this say about \(u\) and \(v ?\) (f) The uniqueness proof outlined in (a) to (e) involves specifying the value of the potential on various surfaces. Might we have specified a different kind of boundary condition and still proved uniqueness? If so, in what way or ways would the proof and the result differ from those given above?

Show that \(\nabla \times \nabla f=0\) where \(f(x, y, z)\) is an arbitrary differentiable scalar function. Assume that mixed second-order partial derivatives are independent of the order of differentiation. For example, \(\partial^{2} f / \partial x \partial z\) \(=\partial^{2} f / \partial z \partial x\)

The heat \(Q\) in a body of volume \(V\) is given by $$ Q=c \iiint_{V} T \rho d V $$ where \(c\) is a constant called the specific heat of the body, and \(T(x, y, z, t)\) and \(\rho(x, y, z)\) are, respectively, the temperature and density of the body. (Note that we are assuming the density to be independent of time.) The rate at which heat flows through \(S\), the bounding surface of the body, is given by $$ \frac{d Q}{d t}=k \iint_{S} \hat{\mathbf{n}} \cdot \nabla T d S $$ where \(k\) (assumed constant) is the thermal conductivity of the body, and the integral is taken over the surface \(S\) bounding the body. Use these facts to derive the heat flow equation $$ \nabla^{2} T=\alpha \frac{\partial T}{\partial t} $$ where \(\alpha=c \rho / k\)

Here is a "proof" that there is no such thing as magnetism. One of Maxwell's equations tells us that $$ \nabla \cdot \mathbf{B}=0 $$ where \(\mathbf{B}\) is any magnetic field. Then using the divergence theorem, we find $$ \iint_{S} \mathbf{B} \cdot \hat{\mathbf{n}} d S=\iiint_{V} \nabla \cdot \mathbf{B} d V=0 $$ Because B has zero divergence, we know (see Problem III-24) there exists a vector function, call it \(\mathbf{A}\), such that $$ \mathbf{B}=\nabla \times \mathbf{A} $$ Combining these last two equations, we get $$ \iint_{s} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Next we apply Stokes' theorem and the preceding result to find $$ \oint_{C} \mathbf{A} \cdot \hat{\mathbf{t}} d s=\iint_{S} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Thus we have shown that the circulation of \(\mathbf{A}\) is path independent. It follows that we can write \(\mathbf{A}=\nabla \psi\), where \(\psi\) is some scalar function. Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that $$ \mathbf{B}=\nabla \times \nabla \psi=0 $$ that is, all magnetic fields are zero! Where did we go wrong? [Taken from G. Arfken, Amer. J. Phys., 27, 526 (1959).]
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