/*! 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 63 Consider the following functions... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following functions \(f,\) points \(P,\) and unit vectors \(\mathbf{u}\). a. Compute the gradient of \(f\) and evaluate it at \(P\). b. Find the unit vector in the direction of maximum increase of \(f\) at \(P\). c. Find the rate of change of the function in the direction of maximum increase at \(P\) d. Find the directional derivative at \(P\) in the direction of the given vector. $$f(x, y, z)=1+\sin (x+2 y-z) ; P\left(\frac{\pi}{6}, \frac{\pi}{6},-\frac{\pi}{6}\right) ;\left\langle\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right\rangle$$

Short Answer

Expert verified
Answer: The rate of change of the function in the direction of maximum increase at the given point \(P\) is 0, as there is no direction of maximum increase since the gradient of f at P is the zero vector.

Step by step solution

01

Find the partial derivatives of f

To compute the gradient, we first need to find the partial derivatives \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\): Partial derivatives of f: $$\frac{\partial f}{\partial x} = \cos(x + 2y - z)$$ $$\frac{\partial f}{\partial y} = 2\cos(x + 2y - z)$$ $$\frac{\partial f}{\partial z} = -\cos(x + 2y - z)$$
02

Evaluate the gradient of f at P

Now we will evaluate the gradient at the given point \(P\left(\frac{\pi}{6},\frac{\pi}{6},-\frac{\pi}{6}\right)\): Gradient of f at P: $$\nabla f_P = \left\langle \cos\left(\frac{\pi}{6} + 2\frac{\pi}{6} + \frac{\pi}{6}\right), 2\cos\left(\frac{\pi}{6} + 2\frac{\pi}{6} + \frac{\pi}{6}\right), -\cos\left(\frac{\pi}{6} + 2\frac{\pi}{6} + \frac{\pi}{6}\right) \right\rangle = \left\langle 0, 0, 0 \right\rangle$$ #b. Find the unit vector in the direction of maximum increase of f at P.#
03

Find the unit vector of the gradient

Since the gradient of f at P is the zero vector, there is no direction of maximum increase at P. The function is flat at this point.
04

State the result

There is no direction of maximum increase at \(P\), as the function is flat at this point. #c. Find the rate of change of the function in the direction of maximum increase at P.#
05

State the result

Since there is no direction of maximum increase at \(P\), the rate of change of the function in the direction of maximum increase at \(P\) is 0. #d. Find the directional derivative at P in the direction of the given vector.#
06

Calculate the directional derivative

Given the unit vector \(\mathbf{u} = \left\langle\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right\rangle\), we can calculate the directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{u}\) by taking the dot product of the gradient of \(f\) at \(P\) and the unit vector \(\mathbf{u}\): Directional derivative of f at P in the direction of u: $$D_\mathbf{u}f_P = \nabla f_P \cdot \mathbf{u} = \left\langle 0, 0, 0 \right\rangle \cdot \left\langle \frac{1}{3}, \frac{2}{3},\frac{2}{3} \right\rangle = 0$$
07

State the result

The directional derivative of \(f\) at \(P\) in the direction of the given unit vector \(\mathbf{u}\) is 0.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental tool in understanding how functions behave in multivariable calculus. When dealing with a function of multiple variables, such as \( f(x, y, z) \), a partial derivative measures how the function changes as you tweak one variable, while keeping the others fixed. For example, the partial derivative \( \frac{\partial f}{\partial x} \) examines the rate at which \( f \) changes with respect to changes in \( x \), ignoring changes in \( y \) or \( z \).
To compute partial derivatives, differentiate the function with respect to the variable of interest, treating other variables as constants. This is akin to slicing through a graph at a constant value for the non-differentiated variables, giving a clear view of how one variable affects the overall function. It’s like considering the slope of a curve at a specific point on a larger surface.
Directional Derivative
The directional derivative extends the idea of partial derivatives by considering how a function changes as you move in any direction, not just along the axes. It is defined for a given direction represented by a vector and tells us the rate of change of the function in that specified direction.
To find the directional derivative of \( f \) at a point \( P \) in the direction of a vector \( \mathbf{u} \), we compute the dot product of the gradient of \( f \) at \( P \), denoted \( abla f_P \), and the unit vector \( \mathbf{u} \). The gradient is simply a vector comprising all the partial derivatives of \( f \), showing the general direction of steepest ascent.
  • If the result of the dot product is positive, the function increases in that direction.
  • If it's negative, the function decreases.
  • A zero result indicates that the function does not change in that direction.
Rate of Change
The rate of change of a function is crucial to understanding how sensitive the function is to changes in input variables. In multivariable calculus, the gradient vector at a point provides a concise representation of this rate. Each component of the gradient indicates the rate of change along the corresponding axis.
The magnitude of the gradient vector tells us the quickest rate at which the function value changes at the point. However, if the gradient is zero, as we found in our exercise at point \( P \), this indicates the function is flat at that point, meaning it exhibits no immediate change in any direction. This situation often arises at what’s called critical points, and further analysis is needed to classify these points as local maxima, minima, or saddle points.
Unit Vector
A unit vector is a vector of length one, and it is used to specify direction without regard to magnitude. When computing directional derivatives, unit vectors are essential because they strip away any impact of different vector lengths, allowing us to focus purely on the direction of interest.
To convert a general vector into a unit vector, it is divided by its magnitude. For instance, if a vector is \( \mathbf{v} = \langle x, y, z \rangle \), its magnitude is \( ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \). The corresponding unit vector is then:\ul>
  • \( \mathbf{u} = \langle \frac{x}{||\mathbf{v}||}, \frac{y}{||\mathbf{v}||}, \frac{z}{||\mathbf{v}||} \rangle \).
  • Unit vectors are powerful in a range of applications, making them essential in fields ranging from physics to computer graphics, ensuring calculations are consistent regardless of vector size.

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