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Chapter 6: Problem 14

Verify that each of the following force fields is conservative. Then find, for each, a scalar potential \(\phi\) such that \(\mathbf{F}=-\nabla \phi\). $$\mathbf{F}=\frac{y}{\sqrt{1-x^{2} y^{2}}} \mathbf{i}+\frac{x}{\sqrt{1-x^{2} y^{2}}} \mathbf{j}$$

Short Answer

Expert verified
Yes, \(\textbf{F}\) is conservative. \(\textbf{U}\)

Step by step solution

01

Identify Conservative Field Conditions

For a vector field \(\textbf{F} = P\textbf{i} + Q\textbf{j}\), it is conservative if \(\frac{\text{d}Q}{\text{d}x} = \frac{\text{d}P}{\text{d}y}\). Here, \(P = \frac{y}{\frac{1}{\frac{1}{x}}}\) and \(Q = \frac{x}{\frac{1}{\frac{1}{x}}}\).
02

Compute Partial Derivatives

Calculate \(\frac{\text{d}Q}{\text{d}x}\).\ set to zero. groovin']}
03

Compare Partial Derivatives

Verify if \(\frac{\text{d}P}{\text{d}y} = \frac{\text{d}Q}{\text{d}x}\). If the partial derivatives are equal, the field is conservative.
04

Find Scalar Potential Function

To find scalar potential \(\phi\), integrate \(P\) with respect to \(x\) and \(Q\) with respect to \(y\). Solve for \(\phi \) ensuring consistency in mixed partial derivatives.

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Key Concepts

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

Scalar Potential
A scalar potential function \( \phi \) is a single-valued function of space whose gradient is equal to a given vector field. If a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \) is conservative, then there exists a scalar potential \( \phi \) such that \( \mathbf{F} = -\abla \phi \). This means you can describe the force field using a single function whose rate of change in each direction matches the components of the force field. By finding \( \phi \), we simplify complex vector fields into more manageable scalar functions.
Partial Derivatives
Partial derivatives are derivatives of multivariable functions taken with respect to one variable while keeping other variables constant. For example, given \(P = \frac{y}{\frac{1}{\frac{1}{x}}}\textbf{i}\) and \(Q = \frac{x}{\frac{1}{\frac{1}{x}}}\textbf{j}\), calculating partial derivatives involves treating \(y\) as constant when differentiating with respect to \(x\), and vice versa. To determine if a force field is conservative, we compute \(\frac{\d Q}{\d x}\) and \(\frac{\d P}{\d y}\). If they are equal, the field is conservative.
Vector Calculus
Vector calculus deals with vector fields and operations on them. In our problem, we deal with identifying whether a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \) is conservative. A crucial theorem in vector calculus states that a two-dimensional vector field is conservative if the mixed partial derivatives of its components are equal, i.e., \(\frac{\d Q}{\d x} = \frac{\d P}{\d y}\). This property allows us to check the conservativeness of fields using partial derivatives, simplifying the otherwise complex vector analysis.
Integration
Integration is a fundamental concept for finding the scalar potential \( \phi \) from a conservative force field. By integrating one component of the vector field \(P\) with respect to \(x\) and the other component \(Q\) with respect to \(y\), we can identify \( \phi \). For example, we start by solving \(\frac{\partial \phi}{\partial x} = P\), then integrate \(P\) with respect to \(x\). This integration often yields a function \(\tilde{\phi}(x, y) = \phi(x, y) - g(y)\). We then use \(\frac{\partial \phi}{\partial y} = Q\) to determine \(g(y)\) by solving \(\frac{\d}{\d y} = g'(y)\). Both steps ensure we get a function \( \phi \) consistent across all partial derivatives.

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Most popular questions from this chapter

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iiint \nabla \cdot \mathbf{V} d \tau\) over the unit cube in the first octant, where $$\mathbf{V}=\left(x^{3}-x^{2}\right) y \mathbf{i}+\left(y^{3}-2 y^{2}+y\right) x \mathbf{j}+\left(z^{2}-1\right) \mathbf{k}$$.

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. If the temperature in the \((x, y)\) plane is given by \(T=x y-x,\) sketch a few isothermal curves, say for \(T=0,1,2,-1,-2 .\) Find the direction in which the temperature changes most rapidly with distance from the point \((1,1),\) and the maximum rate of change. Find the directional derivative of \(T\) at (1,1) in the direction of the vector \(3 \mathbf{i}-4 \mathbf{j} .\) Heat flows in the direction \(-\nabla T\) (perpendicular to the isothermals). Sketch a few curves along which heat would flow.

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iint \mathbf{r} \cdot \mathbf{n} d \sigma\) over the whole surface of the cylinder bounded by \(x^{2}+y^{2}=1, z=0\), and \(z=3 ; \mathbf{r}\) means \(\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\).

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the closed surface of the tin can bounded by \(x^{2}+y^{2}=9, z=0\) \(z=5,\) if $$\mathbf{V}=2 x y \mathbf{i}-y^{2} \mathbf{j}+(z+x y) \mathbf{k}$$

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. (a) Given \(\phi=x^{2}-y^{2},\) sketch on one graph the curves \(\phi=4, \phi=1, \phi=0\), \(\phi=-1, \phi=-4 .\) If \(\phi\) is the electrostatic potential, the curves \(\phi=\) const. are equipotentials, and the electric field is given by \(\mathbf{E}=-\nabla \phi\). If \(\phi\) is temperature, the curves \(\phi=\) const. are isothermals and \(\nabla \phi\) is the temperature gradient; heat flows in the direction \(-\nabla \phi\). (b) Find and draw on your sketch the vectors \(-\nabla \phi\) at the points \((x, y)=(\pm 1,\pm 1)\), \((0,\pm 2),(\pm 2,0) .\) Then, remembering that \(\nabla \phi\) is perpendicular to \(\phi=\) const., sketch, without computation, several curves along which heat would flow [see(a)].
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