/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Computing gradients Compute the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$g(x, y)=x^{2}-4 x^{2} y-8 x y^{2} ; P(-1,2)$$

Short Answer

Expert verified
Question: Compute the gradient of the function \(g(x, y)=x^{2}-4 x^{2} y-8 x y^{2}\) and evaluate it at point P(-1,2). Answer: The gradient of the function at point P(-1,2) is \(\begin{bmatrix}18\\ 28\end{bmatrix}\).

Step by step solution

01

Find the partial derivative with respect to x

Compute the partial derivative of the function \(g(x,y)\) with respect to x: $$\frac{\partial{g}}{\partial{x}} = \frac{\partial{(x^{2}-4 x^{2} y-8 x y^{2})}}{\partial{x}}$$ Take the derivative term by term: $$\frac{\partial{g}}{\partial{x}} = 2x - 8xy - 8y^2$$
02

Find the partial derivative with respect to y

Compute the partial derivative of the function \(g(x,y)\) with respect to y: $$\frac{\partial{g}}{\partial{y}} = \frac{\partial{(x^{2}-4 x^{2} y-8 x y^{2})}}{\partial{y}}$$ Take the derivative term by term: $$\frac{\partial{g}}{\partial{y}} = -4x^2 - 16xy$$
03

Form the gradient

Combine the results from step 1 and step 2 to form the gradient \(\nabla g(x,y)\): $$\nabla g(x,y) = \begin{bmatrix}2x - 8xy - 8y^2\\ - 4x^2 -16xy\end{bmatrix}$$
04

Evaluate the gradient at point P

Evaluate the gradient of \(g(x,y)\) at point \(P(-1, 2)\). Plug the coordinates of point P into the gradient vector: $$\nabla g(-1, 2) = \begin{bmatrix}2(-1) - 8(-1)(2) - 8(2)^2\\ - 4(-1)^2 - 16(-1)(2)\end{bmatrix}$$ Simplify and compute each component of the gradient: $$\nabla g(-1, 2) = \begin{bmatrix}18\\ 28\end{bmatrix}$$ The gradient of the function \(g(x, y)=x^{2}-4 x^{2} y-8 x y^{2}\) at point P(-1,2) is \(\begin{bmatrix}18\\ 28\end{bmatrix}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0,2)}(2 x y)^{x y}$$

Let \(x, y,\) and \(z\) be nonnegative numbers with \(x+y+z=200\). a. Find the values of \(x, y,\) and \(z\) that minimize \(x^{2}+y^{2}+z^{2}\). b. Find the values of \(x, y,\) and \(z\) that minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\). c. Find the values of \(x, y,\) and \(z\) that maximize \(x y z\). d. Find the values of \(x, y,\) and \(z\) that maximize \(x^{2} y^{2} z^{2}\).

Determine whether the following statements are true and give an explanation or counterexample. a. The plane passing through the point (1,1,1) with a normal vector \(\mathbf{n}=\langle 1,2,-3\rangle\) is the same as the plane passing through the point (3,0,1) with a normal vector \(\mathbf{n}=\langle-2,-4,6\rangle\) b. The equations \(x+y-z=1\) and \(-x-y+z=1\) describe the same plane. c. Given a plane \(Q\), there is exactly one plane orthogonal to \(Q\). d. Given a line \(\ell\) and a point \(P_{0}\) not on \(\ell\), there is exactly one plane that contains \(\ell\) and passes through \(P_{0}\) e. Given a plane \(R\) and a point \(P_{0},\) there is exactly one plane that is orthogonal to \(R\) and passes through \(P_{0}\) f. Any two distinct lines in \(\mathbb{R}^{3}\) determine a unique plane. g. If plane \(Q\) is orthogonal to plane \(R\) and plane \(R\) is orthogonal to plane \(S\), then plane \(Q\) is orthogonal to plane \(S\).

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y, z)=(x+y+z) e^{x y z}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.