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For the potential function \(\varphi\) and points \(A, B, C,\) and D on the level curve \(\varphi(x, y)=0,\) complete the following steps. a. Find the gradient field \(\mathbf{F}=\nabla \varphi.\) b. Evaluate \(\mathbf{F}\) at the points \(A, B, C,\) and \(D.\) c. Plot the level curve \(\varphi(x, y)=0\) and the vectors \(\mathbf{F}\) at the points \(A\) \(B, C,\) and \(D.\) $$\begin{aligned} &\varphi(x, y)=-y+\sin x ; A(\pi / 2,1), B(\pi, 0), C(3 \pi / 2,-1)\text { and } D(2 \pi, 0) \end{aligned}$$

Short Answer

Expert verified
The gradient field is \(\mathbf{F} = (\cos x, -1)\). The evaluations of the gradient field at points A, B, C, and D are: 1. A\((\pi/2,1)\) has a vector of \((0,-1)\). 2. B\((\pi,0)\) has a vector of \((-1,-1)\). 3. C\((3\pi/2,-1)\) has a vector of \((0,-1)\). 4. D\((2\pi,0)\) has a vector of \((1,-1)\). The plot consists of the sine curve \(y = \sin x\) with gradient field vectors at each of the specified points A, B, C, and D, showing the gradient field direction at each of these points.

Step by step solution

01

Finding the gradient field \(\mathbf{F}=\nabla \varphi\)

First, we should compute the gradient field \(\mathbf{F}=\nabla \varphi\). The gradient field is defined as \(\mathbf{F}=(\frac{\partial \varphi}{\partial x},\frac{\partial \varphi}{\partial y})\). Let's find the partial derivatives: $$\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x}(-y+\sin x) = 0+\cos x = \cos x$$ $$\frac{\partial \varphi}{\partial y} = \frac{\partial}{\partial y}(-y+\sin x) = -1+0=-1$$ Thus, the gradient field is \(\mathbf{F}=\nabla\varphi=(\cos x, -1)\).
02

Evaluating the gradient field at points A, B, C, and D

Now, we should evaluate the gradient field \(\mathbf{F}\) at each point A, B, C, and D as follows: A\((\pi/2,1)\): \(\mathbf{F}(\pi/2,1) = (\cos(\pi/2), -1) = (0, -1)\) B\((\pi,0)\): \(\mathbf{F}(\pi,0)=(\cos(\pi), -1) = (-1, -1)\) C\((3\pi/2,-1)\): \(\mathbf{F}(3\pi/2,-1) = (\cos(3\pi/2), -1) = (0, -1)\) D\((2\pi,0)\): \(\mathbf{F}(2\pi,0) = (\cos(2\pi), -1) = (1,-1)\)
03

Plotting the level curve and the gradient field at points A, B, C, and D

First, solve for \(\varphi(x, y) = 0\) to get the level curve: $$-y+\sin x = 0$$ $$y=\sin x$$ Now, plot the sine curve \(y=\sin x\) and add the gradient field vectors at the points A, B, C, and D: 1. A\((\pi/2,1)\) has a vector of \((0,-1)\). 2. B\((\pi,0)\) has a vector of \((-1,-1)\). 3. C\((3\pi/2,-1)\) has a vector of \((0,-1)\). 4. D\((2\pi,0)\) has a vector of \((1,-1)\). The plot will have a sine curve with vectors at the specified points A, B, C, and D showing the gradient field direction at each of these points.

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Key Concepts

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

Level Curve
A level curve is a set of points where a function of two variables remains constant. Here, we're dealing with the function \(\varphi(x, y) = -y + \sin x\). When the level curve is given as \(\varphi(x, y) = 0\), it means we're looking for points where the function's value remains zero. By setting \(-y + \sin x = 0\), we can solve for \(y\) in terms of \(x\) to get \(y = \sin x\).

This equation, \(y = \sin x\), forms a curve that is equal to the sine wave along the plane at the height where the original potential function becomes zero. Understanding level curves helps us visualize how functions behave across those specified two-dimensional spaces.
Potential Function
The potential function \(\varphi(x, y)\) can be thought of as a surface in three-dimensional space, giving a height for each point \((x, y)\). A key role of the potential function in physics and mathematics is to represent fields of force. The vector field, called a gradient field \(\mathbf{F} = abla \varphi\), is derived from this potential function.

The potential function allows us to find this gradient through its partial derivatives. When these functions equal zero on a level curve, they form pathways or curves on the plane where the potential is constant. The simplicity of potential functions till now, like \(-y + \sin x\), ensures clear insights into more complex dynamics.
Partial Derivatives
Partial derivatives help us understand how a function changes as each variable changes, separately. For the potential function \(\varphi(x, y) = -y + \sin x\), the gradient field \(\mathbf{F}\) involves taking partial derivatives with respect to \(x\) and \(y\).

- The partial derivative with respect to \(x\), \(\frac{\partial \varphi}{\partial x}\), evaluates changes along the \(x\)-axis. For \(-y + \sin x\), it equals \(\cos x\).
- The partial derivative with respect to \(y\), \(\frac{\partial \varphi}{\partial y}\), determines changes along the \(y\)-axis. In this case, it's \(-1\).

By combining these, we get the gradient field \(\mathbf{F} = (\cos x, -1)\). This vector shows the direction and rate of change of the potential function around any point.
Vector Plotting
Vector plotting allows us to graphically represent the gradient fields. After evaluating the gradient field \(\mathbf{F} = abla \varphi = (\cos x, -1)\) at specific points \(A, B, C\), and \(D\), we obtained vectors for these points. Each vector signifies both direction and magnitude of change in the potential function.

For this scenario:
  • At point \(A(\pi/2, 1)\), the vector (0, -1) shows no movement in \(x\) but a downward pull in \(y\).
  • At \(B(\pi, 0)\), the vector (-1, -1) indicates left and downward movement.
  • \(C(3\pi/2, -1)\) yields (0, -1), similar to \(A\).
  • While \(D(2\pi, 0)\) results in (1, -1), indicating right and downward motion.
By plotting these vectors alongside the level curve, we provide a detailed way to understand how the field behaves and changes around these specific points.

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Most popular questions from this chapter

Conservation of energy 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 \(R\).

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. $$\begin{aligned} &\iint_{S} \frac{\langle x, 0, z\rangle}{\sqrt{x^{2}+z^{2}}} \cdot \mathbf{n} d S, \text { where } S \text { is the cylinder } x^{2}+z^{2}=a^{2}\\\ &|y| \leq 2 \end{aligned}$$

What's wrong? Consider the radial field \(\mathbf{F}=\frac{\langle x, y\rangle}{x^{2}+y^{2}}\) a. Verify that the divergence of \(\mathbf{F}\) is zero, which suggests that the double integral in the flux form of Green's Theorem is zero. b. Use a line integral to verify that the outward flux across the unit circle of the vector field is \(2 \pi\) c. Explain why the results of parts (a) and (b) do not agree.

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

Cone and sphere The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\) b. Find the surface area of the sphere above \(C\) c. Find the surface area of the cone below \(C\), for \(z \geq 0\)

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