/*! 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 14 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 \(P\). $$f(x, y)=4 x^{2}-2 x y+y^{2} ; P(-1,-5)$$

Short Answer

Expert verified
To compute the gradient of the function \(f(x,y) = 4x^2 - 2xy + y^2\) at the point \(P(-1, -5)\), first, we find the partial derivatives with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x = 8x - 2y\), and with respect to \(y\) is \(f_y = -2x + 2y\). By substituting the point \(P(-1,-5)\) into the partial derivatives, we get \(f_x(-1, -5) = 2\) and \(f_y(-1, -5) = -8\). Therefore, the gradient of the function at the point \(P(-1, -5)\) is \(\langle 2, -8 \rangle\).

Step by step solution

01

Find the partial derivatives of the function

To find the gradient of the function, we first need to find the partial derivatives with respect to each variable. Let's compute the partial derivative of the function \(f(x,y)\) with respect to \(x\) and \(y\): Partial derivative with respect to \(x\), denoted as \(f_x\): $$f_x = \frac{\partial}{\partial x} (4x^2 - 2xy + y^2)$$ Partial derivative with respect to \(y\), denoted as \(f_y\): $$f_y = \frac{\partial}{\partial y} (4x^2 - 2xy + y^2)$$
02

Compute the partial derivatives

Now we will compute the partial derivatives: For \(f_x\): $$f_x = \frac{\partial}{\partial x} (4x^2 - 2xy + y^2) =(8x - 2y)$$ For \(f_y\): $$f_y = \frac{\partial}{\partial y} (4x^2 - 2xy + y^2) =(-2x + 2y)$$
03

Evaluate the partial derivatives at the given point

We are given the point \(P(-1, -5)\). To find the gradient at this point, we will substitute these values into the partial derivatives: $$f_x(-1, -5) = (8(-1) - 2(-5)) = -8 + 10 = 2$$ $$f_y(-1, -5) = (-2(-1) + 2(-5)) = 2 -10 = -8$$
04

Compute the gradient vector

The gradient of the function \(f(x,y)\) is a vector containing both partial derivatives, written as: Gradient\((f(x,y)) = \langle f_x(x,y), f_y(x,y) \rangle\) We've already found the partial derivatives and evaluated them at the given point, so we can now write the gradient vector: Gradient\((f(x,y))|_{(-1,-5)} = \langle 2, -8 \rangle\) Thus, the gradient of the function \(f(x, y) = 4x^2 - 2xy + y^2\) at the point \(P(-1, -5)\) is \(\langle 2, -8 \rangle\).

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!

Key Concepts

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

Partial Derivatives
The concept of partial derivatives is a foundational element in multivariable calculus, which deals with functions of more than one variable. Unlike a regular derivative that measures how a function changes as its single input variable changes, a partial derivative measures how a function changes as one of its multiple input variables changes while keeping the other variables held constant.

In the context of our exercise, we are examining a function, namely, \(f(x, y) = 4x^2 - 2xy + y^2\), which depends on two variables: \(x\) and \(y\). To understand how \(f\) changes in response to changes in just one of these variables at a time, we compute the partial derivatives \(f_x\) and \(f_y\). These derivatives are represented as \(\frac{\partial}{\partial x}\) and \(\frac{\partial}{\partial y}\), respectively.

It is crucial for students to remember that when calculating \(f_x\), you treat \(y\) as a constant, and similarly, when calculating \(f_y\), you treat \(x\) as a constant. In this manner, partial derivatives provide a way to analyze the slope of the function in each dimension independently. Understandably, they play an indispensable role when it comes to understanding functions in higher dimensions and are a stepping stone to more complex concepts like the gradient vector.
Gradient Vector
The gradient vector is a central concept in vector calculus, particularly when dealing with functions of multiple variables. It is a vector that points in the direction of the greatest rate of increase of the function and whose magnitude is the rate of increase in that direction. The components of the gradient vector are exactly the partial derivatives found for each independent variable.

For instance, with our given function \(f(x, y)\), the gradient vector at any point \((x, y)\) would be denoted as \(abla f(x, y)\) and is calculated as \(\langle f_x(x, y), f_y(x, y) \rangle\), where \(f_x\) and \(f_y\) are the partial derivatives with respect to \(x\) and \(y\), respectively.

For our specific example, upon computing the partial derivatives, the gradient at point \(P(-1, -5)\) is \(\langle 2, -8 \rangle\). This means that at point \(P\), the function increases most rapidly in the direction of the vector \(\langle 2, -8 \rangle\). The concept of the gradient is hugely important, as it is used in optimization problems to find local maxima and minima of functions, which has practical applications in areas such as physics, economics, and engineering.
Multivariable Calculus
Finally, multivariable calculus is the extension of calculus to functions of several variables. It goes beyond the one-dimensional cases to embrace the full spectrum of problems involving multiple dimensions. This field is not just about evaluating multiple integrals or derivatives; it provides the tools for understanding and predicting the behavior of complex systems where several factors are at play.

Concepts such as partial derivatives and the gradient vector, which we have explored in relation to our exercise, are just the tip of the iceberg. Multivariable calculus allows us to handle curves and surfaces in three dimensions or more, analyze vector fields, and even apply calculus to shapes of differing dimensions, like curves, surfaces, and solids.

The theory and application of multivariable calculus are critical in various scientific disciplines, such as physics, engineering, economics, and statistics. By learning how to compute gradients and understanding their meaning, students gain a powerful toolset for analyzing real-world problems where change occurs in more than one direction or dimension.

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

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

A snapshot (frozen in time) of a set of water waves is described by the function \(z=1+\sin (x-y),\) where \(z\) gives the height of the waves and \((x, y)\) are coordinates in the horizontal plane \(z=0\) a. Use a graphing utility to graph \(z=1+\sin (x-y)\) b. The crests and the troughs of the waves are aligned in the direction in which the height function has zero change. Find the direction in which the crests and troughs are aligned. c. If you were surfing on one of these waves and wanted the steepest descent from the crest to the trough, in which direction would you point your surfboard (given in terms of a unit vector in the \(x y\) -plane)? d. Check that your answers to parts (b) and (c) are consistent with the graph of part (a).

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) In two dimensions, the motion of an ideal fluid (an incompressible and irrotational fluid) is governed by a velocity potential \(\varphi .\) The velocity components of the fluid, \(u\) in the \(x\) -direction and \(v\) in the \(y\) -direction, are given by \(\langle u, v\rangle=\nabla \varphi .\) Find the velocity components associated with the velocity potential \(\varphi(x, y)=\sin \pi x \sin 2 \pi y\)

Many gases can be modeled by the Ideal Gas Law, \(P V=n R T\), which relates the temperature \((T,\) measured in kelvins ( \(\mathrm{K}\) )), pressure ( \(P\), measured in pascals (Pa)), and volume ( \(V\), measured in \(\mathrm{m}^{3}\) ) of a gas. Assume the quantity of gas in question is \(n=1\) mole (mol). The gas constant has a value of \(R=8.3 \mathrm{m}^{3} \mathrm{Pa} / \mathrm{mol}-\mathrm{K}\) a. Consider \(T\) to be the dependent variable, and plot several level curves (called isotherms) of the temperature surface in the region \(0 \leq P \leq 100,000\) and \(0 \leq V \leq 0.5\). b. Consider \(P\) to be the dependent variable, and plot several level curves (called isobars) of the pressure surface in the region \(0 \leq T \leq 900\) and \(0

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$G(x, y)=-\sqrt{1+x^{2}+y^{2}}$$

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.