Chapter 14: Problem 6
Why does a two-dimensional vector field with zero curl on a region have zero circulation on a closed curve that bounds the region?
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Chapter 14: Problem 6
Why does a two-dimensional vector field with zero curl on a region have zero circulation on a closed curve that bounds the region?
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Use the procedure in Exercise 57 to construct potential functions for the following fields. $$\mathbf{F}=\langle-y,-x\rangle$$
Prove that for a real number \(p\), with \(\mathbf{r}=\langle x, y, z\rangle, \nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{p(p-1)}{|\mathbf{r}|^{p+2}}.\)
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 \(R\)
Generalize Exercise 43 to show that the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r}\) circles the vector a in the counterclockwise direction looking along a from head to tail.
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 \(\mathbb{R}^{3}\). $$\nabla \cdot(\varphi \mathbf{F})=\nabla \varphi \cdot \mathbf{F}+\varphi \nabla \cdot \mathbf{F} \quad \text { (Product Rule) }$$
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