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

Short Answer

Expert verified
The force field \(\mathbf{F}=y \sin 2 x \mathbf{i}+\sin ^{2} x \mathbf{j}\) is conservative, and the scalar potential function is \(\phi(x,y) = \frac{y}{2} \cos 2x + C\).

Step by step solution

01

Check if the field is conservative

A vector field \(\mathbf{F} = P \mathbf{i} + Q \mathbf{j}\) is conservative if \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\). In this exercise, \(P = y \sin 2x\) and \(Q = \sin^2 x\). Compute \(\frac{\partial P}{\partial y}\) and \(\frac{\partial Q}{\partial x}\): \[\frac{\partial P}{\partial y} = \frac{\partial }{\partial y}(y \sin 2x) = \sin 2x\frac{\partial Q}{\partial x} = \frac{\partial }{\partial x}(\sin^2 x) = 2\sin x \cos x = \sin 2x\] Both partial derivatives are equal, so the field is conservative.
02

Find the scalar potential function \(\phi\)

Since \(\mathbf{F} = -abla \phi\), \(\mathbf{F} = (\frac{\partial \phi}{\partial x}) \mathbf{i} + (\frac{\partial \phi}{\partial y}) \mathbf{j}\) should hold. Therefore: \(\frac{\partial \phi}{\partial x} = - y \sin 2x\) and \(\frac{\partial \phi}{\partial y} = -\sin^2 x\).
03

Integrate to find \(\phi\) in terms of \(x\) and \(y\)

Integrate \(\frac{\partial \phi}{\partial x} = - y \sin 2x\) with respect to \(x\): \[\phi(x,y) = -y \int \sin 2x \, dx = -y \left( -\frac{1}{2} \cos 2x \right) + g(y) = \frac{y}{2} \cos 2x + g(y)\] Next, differentiate \(\phi(x,y)\) with respect to \(y\) to find \(g(y)\), and match it to \(\frac{\partial \phi}{\partial y}\): \[\frac{\partial \phi}{\partial y} = \frac{\partial }{\partial y} \left( \frac{y}{2} \cos 2x + g(y) \right) = \frac{1}{2} \cos 2x + g'(y)\ = -\sin^2 x\] Therefore, \(g'(y) = 0\) so \(g(y)\) is a constant. Let this constant be \(C\), then: \[\phi(x,y) = \frac{y}{2} \cos 2x + C\]
04

Write the final potential function \(\phi\)

The scalar potential function for the given force field is: \[ \phi(x,y) = \frac{y}{2} \cos 2x + C \]

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Key Concepts

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

Scalar Potential Function
A scalar potential function, often denoted as \(\phi\), is a mathematical tool used to describe a conservative force field. When a vector field \(\textbf{F}\) is derived from the gradient of some scalar function \(\textbf{F} = -\abla\phi\), it indicates the field is conservative. In our problem, we start by recognizing the given vector field: \(\textbf{F}=y \space sin \space 2x \space \textbf{i}+sin ^{2} x \space \textbf{j}\). Our task is to find the corresponding \(\phi\). The function \(\phi\) represents the stored energy or potential energy in a system where the conservative forces act.
To find \(\phi\):
  • We first identify components \(\textbf{F} = P \space \textbf{i} + Q \space \textbf{j}\), with \(\textbf{P} = y \space sin \space 2x\) and \(\textbf{Q} = sin^2 x\).
  • We then follow through calculations and integrate to express \(\phi\) in terms of the coordinates (x, y).

    Understanding the scalar potential function is key for interpreting physical phenomena like electric, gravitational, and other force fields.
Vector Field
A vector field assigns a vector to every point in space. These vectors typically represent some physical quantity such as force, velocity, or acceleration.
In our problem, the vector field is given by \(\textbf{F} = y \space sin \space 2x \textbf{i} + sin^2 x \textbf{j}\). Each component function defines the force in the respective direction (x or y).
  • Direction: \(\textbf{i}\) is the x-direction component, and \(\textbf{j}\) is the y-direction component.
  • P: The term y\sin2x corresponds to the x-component (P).
  • Q: The term sin^2x corresponds to the y-component (Q).

    Understanding vector fields is crucial in visualizing the direction and magnitude of forces over a region. In this case, recognizing how these components interact is essential to determining conservativeness and finding \(\phi\).
Partial Derivatives
Partial derivatives are used to understand how a function changes as we tweak one variable while keeping others constant. For checking the conservativeness of our vector field, computing partial derivatives helps verify this property.
We compute:
  • \(\frac{\text{{d}} P}{\text{{d}} y}\), the partial derivative of y\sin2x with respect to y
  • \(\frac{\text{{d}} Q}{\text{{d}} x}\), the partial derivative of sin^2x with respect to x
  • Conservativeness condition: If \(\frac{\text{{d}}\text{{P}}}{\text{{d}}\text{{y}}} = \frac{\text{{d}}\text{{Q}}}{\text{{d}}\text{{x}}}\), the field is conservative.

    In our problem, partial derivatives showcase that both are equal (\text{{sin 2x}}), confirming the field is conservative. Mastery of partial derivatives is vital in vector calculus for analyzing changes in multivariable contexts.
Integration
Integration allows us to find the original function given its derivative, helping us construct the scalar potential function \(\phi\).
To find \(\phi\) from \(\frac{\text{{d}} \phi}{\text{{d}} x} = - y \sin 2x\), we:
  • Integrate \(- y \sin 2x\) with respect to x:
    \(\text{{\phi(x,y) = -y \int sin 2x \, dx = \frac{y}{2} \cos 2x + g(y)}}\).
  • Differentiate \(\text{{\phi(x,y)}}\) with respect to y to determine \(\text{{g(y)}}\), a constant in our solution.
  • Recognize that \(\text{{g(y)}}\) simplifies to a constant C.

    These steps ensure we recover the potential function accurately. Integration is a vital tool in finding relationships between force fields and potential functions in vector calculus.

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Most popular questions from this chapter

Given the vector \(\mathbf{A}=\left(x^{2}-y^{2}\right) \mathbf{i}+2 x y \mathbf{j}\) (a) Find \(\nabla \times \mathbf{A}\) (b) Evaluate \(\iint(\nabla \times \mathbf{A}) \cdot d \sigma\) over a rectangle in the \((x, y)\) plane bounded by the lines \(x=0, x=a, y=0, y=b\) (c) Evaluate \(\oint \mathbf{A} \cdot d \mathbf{r}\) around the boundary of the rectangle and thus verify Stokes' theorem for this case.

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.

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The purpose in doing the following simple problems is to become familiar with the formulas we have discussed. So a good study method is to do them by hand and then check your results by computer. Compute the divergence and the curl of each of the following vector fields. $$\mathbf{V}=x \sin y \mathbf{i}+\cos y \mathbf{j}+x y \mathbf{k}$$

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