/*! 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 26 Compute the directional derivati... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 26

Compute the directional derivative of the following functions at the given point \(P\) in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$f(x, y)=x /(x-y) ; P(4,1) ;\langle-1,2\rangle$$

Short Answer

Expert verified
Answer: The directional derivative of the function at point \(P(4,1)\) in the direction of the vector \(\langle -1, 2 \rangle\) is \(\frac{-1}{\sqrt{5}}\).

Step by step solution

01

Compute the gradient vector of the function

To find the gradient vector, calculate the partial derivatives with respect to x and y. For the function \(f(x,y) = \frac{x}{x-y}\), we compute the partial derivatives as follows: $$\frac{\partial f}{\partial x} = \frac{y}{(x-y)^2}$$ $$\frac{\partial f}{\partial y} = -\frac{x}{(x-y)^2}$$ So the gradient vector, denoted by \(\nabla f\), is: $$\nabla f = \left\langle \frac{y}{(x-y)^2}, -\frac{x}{(x-y)^2}\right\rangle$$
02

Convert the given direction vector to a unit direction vector

The given direction vector is \(\langle -1, 2 \rangle\). We need to convert it into a unit vector. First, let's compute its magnitude: $$||\mathbf{v}|| = \sqrt{(-1)^2 + 2^2} = \sqrt{5}$$ Now, divide the direction vector by its magnitude to find the unit direction vector, denoted by \(\mathbf{\hat{v}}\): $$\mathbf{\hat{v}} = \frac{\langle -1, 2 \rangle}{\sqrt{5}} = \left\langle -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle$$
03

Compute the directional derivative

The directional derivative of the function at point \(P(4,1)\) in the direction of the unit vector \(\mathbf{\hat{v}}\) is given by the dot product of the gradient vector with the unit direction vector at point P: $$\nabla{f}(4,1) \cdot \mathbf{\hat{v}} = \left\langle \frac{1}{(4-1)^2} , -\frac{4}{(4-1)^2}\right\rangle \cdot \left\langle -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle$$ Now, compute the dot product: $$\left\langle \frac{1}{3^2}, -\frac{4}{3^2}\right\rangle \cdot \left\langle -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle = \left(\frac{1}{9}\right)\left(-\frac{1}{\sqrt{5}}\right) + \left(-\frac{4}{9}\right)\left(\frac{2}{\sqrt{5}}\right)$$ This gives us: $$D_{\mathbf{\hat{v}}}f(4,1) = -\frac{1}{9\sqrt{5}}-\frac{8}{9\sqrt{5}} = -\frac{9}{9\sqrt{5}} = \frac{-1}{\sqrt{5}}$$ So, the directional derivative of the function \(f(x, y)=\frac{x}{x-y}\) at point \(P(4,1)\) in the direction of the vector \(\langle -1,2 \rangle\) is: $$D_{\mathbf{\hat{v}}}f(4,1) = \frac{-1}{\sqrt{5}}$$

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 91影视!

Key Concepts

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

Gradient Vector
Understanding the gradient vector is pivotal in the field of multivariable calculus. It represents the direction and rate of the steepest ascent of a function at any given point. When we compute the gradient vector, we're actually stringing together all the partial derivatives of a function into a neatly packaged vector.

In the equation abla f = \( \langle \frac{y}{(x-y)^2}, -\frac{x}{(x-y)^2}\rangle \), each component of this vector stems from the rate of change of the function with respect to each variable, x and y respectively. The gradient vector points in the direction where the function increases the most rapidly from the given point and its magnitude tells us the rate of that increase.
Partial Derivatives
Partial derivatives play a critical role when dissecting the behavior of multivariable functions. They measure the rate at which the function changes as you nudge one variable, keeping all others constant.

Delving into our specific function, f(x, y) = \( \frac{x}{x-y} \), finding the partial derivative with respect to x and then y yields \( \frac{\partial f}{\partial x} = \frac{y}{(x-y)^2} \) and \( \frac{\partial f}{\partial y} = -\frac{x}{(x-y)^2} \), respectively. These computations reveal not only how the function slopes in the x-direction but also in the y-direction, pertinent facts for grasping the terrain of the function's graph.
Unit Vector
The notion behind a unit vector is that it serves as the standardized, pure direction element in calculus and physics. It has no other component but direction; its magnitude is exactly 1. This becomes pivotal when we want to isolate the directional influence in operations like the directional derivative.

For the direction vector \( \langle -1, 2 \rangle \), we normalize it by dividing by its magnitude, resulting in the unit vector \( \mathbf{\hat{v}} = \langle -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \), which points in the same direction as our original vector but stretched or shrunk to a standardized length, making further calculations uniform and comparable.
Dot Product
The dot product, an essential concept in vector calculus, bridges the algebraic and geometric by multiplying vectors in a way that extracts information like the angle between them and projections. In the calculation of the directional derivative, the dot product is crucial because it combines the gradient vector, which encodes the function's steepest ascent, with a unit vector pointing in our direction of interest.

The resulting product, \( abla{f}(4,1) \cdot \mathbf{\hat{v}} \), gives a scalar value. That scalar is the directional derivative, describing how fast the function increases in the direction of the unit vector at the point (4,1). The result, \( D_{\mathbf{\hat{v}}}f(4,1) = \frac{-1}{\sqrt{5}} \), tells us that moving in the direction of \( \langle -1, 2 \rangle \) leads to a decrease in the function's value, at that steepness, from the point in question.

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

Flow in a cylinder Poiseuille's Law is a fundamental law of fluid dynamics that describes the flow velocity of a viscous incompressible fluid in a cylinder (it is used to model blood flow through veins and arteries). It says that in a cylinder of radius \(R\) and length \(L,\) the velocity of the fluid \(r \leq R\) units from the center-line of the cylinder is \(V=\frac{P}{4 L \nu}\left(R^{2}-r^{2}\right),\) where \(P\) is the difference in the pressure between the ends of the cylinder and \(\nu\) is the viscosity of the fluid (see figure). Assuming that \(P\) and \(\nu\) are constant, the velocity \(V\) along the center line of the cylinder \((r=0)\) is \(V=k R^{2} / L,\) where \(k\) is a constant that we will take to be \(k=1.\) a. Estimate the change in the centerline velocity \((r=0)\) if the radius of the flow cylinder increases from \(R=3 \mathrm{cm}\) to \(R=3.05 \mathrm{cm}\) and the length increases from \(L=50 \mathrm{cm}\) to \(L=50.5 \mathrm{cm}.\) b. Estimate the percent change in the centerline velocity if the radius of the flow cylinder \(R\) decreases by \(1 \%\) and the length \(L\) increases by \(2 \%.\) c. Complete the following sentence: If the radius of the cylinder increases by \(p \%,\) then the length of the cylinder must increase by approximately __________ \(\%\) in order for the velocity to remain constant.

An identity Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Determine whether the following statements are true and give an explanation or counterexample. a. Suppose you are standing at the center of a sphere looking at a point \(P\) on the surface of the sphere. Your line of sight to \(P\) is orthogonal to the plane tangent to the sphere at \(P\). b. At a point that maximizes \(f\) on the curve \(g(x, y)=0,\) the dot product \(\nabla f \cdot \nabla g\) is zero.

Given the production function \(P=f(K, L)=K^{a} L^{1-a}\) and the budget constraint \(p K+q L=B,\) where \(a, p, q,\) and \(B\) are given, show that \(P\) is maximized when \(K=a B / p\) and \(L=(1-a) B / q\).

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation
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.