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Chapter 12: Problem 2

How do you compute the gradient of the functions \(f(x, y)\) and \(f(x, y, z) ?\)

Short Answer

Expert verified
Answer: To compute the gradient of a function with two variables, \(f(x, y)\), find the partial derivatives of the function with respect to each variable (\(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\)), then form the gradient vector: \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). For a function with three variables, \(f(x, y, z)\), find the partial derivatives with respect to each variable (\(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\)), then form the gradient vector: \(\nabla f(x, y, z) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)\).

Step by step solution

01

Explain the gradient of a function

The gradient of a scalar field (a scalar function of more than one variable) is a vector field that gives the direction of the greatest rate of change of the function, and its magnitude is the greatest rate of change in that direction.
02

Compute the gradient of a function with two variables, \(f(x, y)\)

To compute the gradient of a function of two variables, \(\nabla f(x, y)\), you need to find the partial derivatives of the function with respect to each variable: 1. \(\frac{\partial f}{\partial x}\): the partial derivative of \(f\) with respect to \(x\) 2. \(\frac{\partial f}{\partial y}\): the partial derivative of \(f\) with respect to \(y\) Once you find these partial derivatives, you can form the gradient vector of \(f\) as follows: $$\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$
03

Compute the gradient of a function with three variables, \(f(x, y, z)\)

To compute the gradient of a function of three variables, \(\nabla f(x, y, z)\), you need to find the partial derivatives of the function with respect to each variable: 1. \(\frac{\partial f}{\partial x}\): the partial derivative of \(f\) with respect to \(x\) 2. \(\frac{\partial f}{\partial y}\): the partial derivative of \(f\) with respect to \(y\) 3. \(\frac{\partial f}{\partial z}\): the partial derivative of \(f\) with respect to \(z\) Once you find these partial derivatives, you can form the gradient vector of \(f\) as follows: $$\nabla f(x, y, z) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$ In conclusion, to compute the gradient of a function \(f(x, y)\) or \(f(x, y, z)\), you need to find the partial derivatives of the function with respect to each of its variables and combine them into a vector.

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Most popular questions from this chapter

Use the Second Derivative Test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b)<0

Prove that for the plane described by \(f(x, y)=A x+B y,\) where \(A\) and \(B\) are nonzero constants, the gradient is constant (independent of \((x, y)\) ). Interpret this result.

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). $$u(x, t)=10 e^{-t} \sin x$$

Show that the plane \(a x+b y+c z=d\) and the line \(\mathbf{r}(t)=\mathbf{r}_{0}+\mathbf{v} t,\) not in the plane, have no points of intersection if and only if \(\mathbf{v} \cdot\langle a, b, c\rangle=0 .\) Give a geometric explanation of this result.

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=x\left(x^{2}-3 y^{2}\right)$$
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