/*! 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 21 \(21-26\) Find the maximum rate ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 21

\(21-26\) Find the maximum rate of change of \(f\) at the given point and the direction in which it occurs. $$f(x, y)=y^{2} / x, \quad(2,4)$$

Short Answer

Expert verified
The maximum rate of change is \(4\sqrt{2}\) in the direction of \((-1/\sqrt{2}, 1/\sqrt{2})\).

Step by step solution

01

Understanding the Concept of Gradient

The maximum rate of change of a function at a given point occurs in the direction of its gradient. The gradient of a function \(f(x, y)\) is given by the vector \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).
02

Compute the Partial Derivative with Respect to x

The function given is \(f(x, y) = \frac{y^{2}}{x}\). The partial derivative with respect to \(x\) is found by keeping \(y\) constant: \[ \frac{\partial f}{\partial x} = -\frac{y^2}{x^2} \]
03

Compute the Partial Derivative with Respect to y

To find the partial derivative with respect to \(y\), keep \(x\) constant:\[ \frac{\partial f}{\partial y} = \frac{2y}{x} \]
04

Evaluate the Gradient at the Point

Now, evaluate the gradient at the point \( (2, 4) \):\[ abla f (2, 4) = \left( -\frac{4^2}{2^2}, \frac{2(4)}{2} \right) = (-4, 4) \]
05

Determine the Maximum Rate of Change

The magnitude of the gradient gives the maximum rate of change. Calculate the magnitude:\[ \|abla f (2, 4)\| = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]
06

Determine the Direction of Maximum Rate of Change

The direction of the maximum rate of change is in the direction of the gradient vector \((-4, 4)\). The direction vector is \((-1, 1)\) after normalization \[ \text{Direction: } (-1/\sqrt{2}, 1/\sqrt{2}) \].

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.

Partial Derivatives
In calculus, a partial derivative represents how a function changes as one of its independent variables is varied. To find the partial derivative of a function with respect to
  • a variable, treat all other variables as constants.
  • For example, consider the function \( f(x, y) = \frac{y^2}{x} \).
  • When calculating the partial derivative with respect to \( x \), the \( y \) term is treated as a constant.
  • This results in the partial derivative \( \frac{\partial f}{\partial x} = -\frac{y^2}{x^2} \).
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = \frac{2y}{x} \), considering \( x \) constant.
Partial derivatives are crucial for exploring how each variable individually affects the function's overall change.
Each derivative provides insight into the function's behavior along a single direction.
Maximum Rate of Change
The maximum rate of change of a function at a given point is directed along its gradient. Imagine standing on a hilly landscape:
  • The steepest ascent path from where you are is represented by the maximum rate of change.
  • It always points in the direction of the gradient vector.
  • For the function \( f(x, y) = \frac{y^2}{x} \), at the point \((2, 4)\), the gradient vector is \((-4, 4)\).
To find how fast the change occurs, you calculate the magnitude of the gradient. This tells you how steep or swift the change in the function is along that direction. At \((2, 4)\), the maximum rate of change is determined by the magnitude \( 4\sqrt{2} \).
Gradient Vector
The gradient vector is a powerful tool in multivariable calculus. It indicates both the direction and the rate of the steepest ascent for a function. If you visualize
  • a mountain range, the gradient vector tells you which way is uphill and just how steep it is.
  • Formally, it combines the function's partial derivatives into a vector.
  • For example, \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
  • At the point \((2, 4)\), for our function, the gradient vector is calculated as \((-4, 4)\).
The crux of using the gradient is understanding that it immensely helps in determining optimization and is pivotal for real-life applications such as navigation and design.
Magnitude of Gradient
The magnitude of the gradient vector represents the maximum rate of change of a function at a specific point. Think of it as not just identifying the direction of change,
  • but also evaluating how intense the change is.
  • For any function \( f \), the gradient's magnitude at a given point is computed using:
  • \[ \|abla f(x, y)\| = \sqrt{ \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 } \].
  • It helps quantify how rapidly the function value increases or decreases in the gradient's direction.
In our case, at the point \((2, 4)\), the magnitude is \( 4\sqrt{2} \). This gives a clear understanding that the function changes quite significantly in that particular direction.

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

The ellipsoid \(4 x^{2}+2 y^{2}+z^{2}=16\) intersects the plane \(y=2\) in an ellipse. Find parametric equations for the tangent line to this ellipse at the point \((1,2,2)\) .

Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely. $$\begin{array}{l}{f(x, y)=\sin x+\sin y+\sin (x+y)} \\ {0 \leqslant x \leqslant 2 \pi, 0 \leqslant y \leqslant 2 \pi}\end{array}$$

Describe how the graph of \(g\) is obtained from the graph of \(f .\) (a) \(g(x, y)=f(x-2, y) \quad\) (b) \(g(x, y)=f(x, y+2)\) (c)\(g(x, y)=f(x+3, y-4)\)

The wind-chill index is modeled by the function $$W=13.12+0.6215 T-11.37 v^{0.16}+0.3965 T v^{0.16}$$ where \(T\) is the temperature \(\left(^{\circ} \mathrm{C}\right)\) and \(v\) is the wind speed \((\mathrm{km} / \mathrm{h}) .\) When \(T=-15^{\circ} \mathrm{C}\) and \(v=30 \mathrm{km} / \mathrm{h}\) , by how much would you expect the apparent temperature \(W\) to drop if the actual temperature decreases by \(1^{\circ} \mathrm{C} ?\) What if the wind speed increases by 1 \(\mathrm{km} / \mathrm{h}\) ?

The total resistance \(R\) produced by three conductors with resistances \(R_{1}, R_{2}, R_{3}\) connected in a parallel electrical circuit is given by the formula $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$$ Find \(\partial R / \partial R_{1}\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

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.