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Find the gradient field \(\mathbf{F}=\nabla \varphi\) for the follow. ing potential functions \(\varphi\). $$\varphi(x, y)=\tan ^{-1}(x / y)$$

Short Answer

Expert verified
Answer: The gradient field of the potential function \(\varphi(x, y) = \tan^{-1}(x / y)\) is \(\mathbf{F} = \left(\frac{y}{y^2 + x^2}, -\frac{x}{y^2 + x^2}\right)\).

Step by step solution

01

Calculate the partial derivative of \(\varphi\) with respect to \(x\)

To compute the partial derivative of \(\varphi(x, y)=\tan^{-1}(x/y)\) with respect to \(x\), we need to differentiate it partially, keeping \(y\) as a constant. Using the chain rule, we get: $$\frac{\partial \varphi}{\partial x} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y}{y^2 + x^2}$$
02

Calculate the partial derivative of \(\varphi\) with respect to \(y\)

Similarly, to compute the partial derivative of \(\varphi(x, y)=\tan^{-1}(x/y)\) with respect to \(y\), we need to differentiate it partially, keeping \(x\) as a constant. Using the chain rule, we get: $$\frac{\partial \varphi}{\partial y} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \frac{d}{dy}\left(\frac{x}{y}\right) = -\frac{x}{y^2 + x^2}$$
03

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

Now that we have the partial derivatives, we can obtain the gradient field \(\mathbf{F}=\nabla \varphi\) by setting it as a vector: $$\mathbf{F} = \left(\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}\right) = \left(\frac{y}{y^2 + x^2}, -\frac{x}{y^2 + x^2}\right)$$ , So, the gradient field of the given potential function \(\varphi(x,y) = \tan^{-1}(x / y)\) is $$\mathbf{F} = \left(\frac{y}{y^2 + x^2}, -\frac{x}{y^2 + x^2}\right)$$

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables, like our potential function \(\varphi(x, y) = \tan^{-1}(x / y)\). In simple terms, a partial derivative measures how a function changes as one of its variables changes, while keeping all other variables constant. For example, when we find \(\frac{\partial \varphi}{\partial x}\), we are looking at how \(\varphi\) changes as \(x\) changes, keeping \(y\) constant.
To calculate a partial derivative, you treat all the other variables as constants and differentiate with respect to the variable of interest. So, when calculating the partial derivative of \(\varphi(x, y)\) with respect to \(x\), it's as if \(y\) is just a known value, allowing us to focus solely on how changes in \(x\) influence \(\varphi\). Similarly, to find \(\frac{\partial \varphi}{\partial y}\), \(x\) is treated as a constant.
  • Partial derivatives help us understand the rate of change along different directions in a multi-dimensional space.
  • They are essential in forming the gradient field.
  • They simplify complex problems into easier, single-variable derivations.
Potential Function
A potential function, like the one we have here \(\varphi(x, y) = \tan^{-1}(x / y)\), is a scalar field whose gradient forms a vector field. In physics and mathematics, potential functions are often used to describe fields like electrostatic or gravitational fields. They are instrumental in simplifying problems where we deal with forces or flow of any kind.
The interesting part of a potential function is its relation to gradient fields. The gradient of a potential function gives us a vector field, which points in the direction of the greatest rate of increase of the function. This means if you walk in the direction of the gradient vector at a point, you ascend most steeply there.
  • Potential functions make it easier to find and visualize forces in a field.
  • They are pivotal for constructing gradient fields used to analyze integrals and flow problems.
  • The gradient of the potential function \(\varphi\) becomes the vector field \(\mathbf{F}\).
Chain Rule
The chain rule is a vital differentiation tool used when we deal with composite functions, and it's crucial for correctly computing partial derivatives in our example. Consider the potential function \(\varphi(x, y) = \tan^{-1}(x / y)\). Here, \(x / y\) is a function within the arctan function, making \(\varphi\) a composite function.
When applying the chain rule, we differentiate the outer function and multiply it by the derivative of the inner function. For instance, to find \(\frac{\partial \varphi}{\partial x}\), we first differentiate \(\tan^{-1}(u)\), where \(u = x / y\), with respect to \(u\), which gives \(\frac{1}{1 + u^2}\), and then multiply it by \(\frac{du}{dx}\), which is \(\frac{1}{y}\).
  • The chain rule simplifies the differentiation of nested functions.
  • It is essential for obtaining accurate partial derivatives when a function depends on multiple variables.
  • Understanding the chain rule enables one to navigate through complex derivatives systematically.

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Most popular questions from this chapter

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) $$\begin{aligned} &\text { -3. } T(x, y, z)=-\ln \left(x^{2}+y^{2}+z^{2}\right) ; S \text { is the sphere }\\\ &x^{2}+y^{2}+z^{2}=a^{2} \end{aligned}$$

Let \(f\) be differentiable and positive on the interval \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. Use Theorem 17.14 to show that the area of \(S\) (as given in Section 6.6 ) is $$ \int_{a}^{b} 2 \pi f(x) \sqrt{1+f^{\prime}(x)^{2}} d x $$

The Navier-Stokes equation is the fundamental equation of fluid dynamics that models the flow in everything from bathtubs to oceans. In one of its many forms (incompressible, viscous flow), the equation is $$\rho\left(\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V} \cdot \nabla) \mathbf{v}\right)=-\nabla p+\mu(\nabla \cdot \nabla) \mathbf{V}$$ In this notation, \(\mathbf{V}=\langle u, v, w\rangle\) is the three-dimensional velocity field, \(p\) is the (scalar) pressure, \(\rho\) is the constant density of the fluid, and \(\mu\) is the constant viscosity. Write out the three component equations of this vector equation. (See Exercise 40 for an interpretation of the operations.)

Stokes' Theorem on closed surfaces Prove that if \(\mathbf{F}\) satisfies the conditions of Stokes' Theorem, then \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S=0\) where \(S\) is a smooth surface that encloses a region.

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) \(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; D\) is the sphere of radius \(a\) centered at the origin.

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