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Chapter 12: Problem 59

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 directional derivative of the function \(f\) at point \(P\) in the given direction is \(-1\).

Step by step solution

01

Calculate Partial Derivatives

Compute the partial derivatives of the function \(f(x, y, z) = 1 + \sin(x + 2y - z)\) with respect to \(x\), \(y\), and \(z\). $$\frac{\partial f}{\partial x} = \cos(x + 2y - z)$$ $$\frac{\partial f}{\partial y} = 2\cos(x + 2y - z)$$ $$\frac{\partial f}{\partial z} = -\cos(x + 2y - z)$$
02

Gradient Vector

Compute the gradient vector using the partial derivatives found in step 1. $$\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle = \left\langle\cos(x + 2y - z), 2\cos(x + 2y - z), -\cos(x + 2y - z)\right\rangle$$
03

Evaluate Gradient Vector at Point P

Substitute the coordinates of point \(P\left(\frac{\pi}{6}, \frac{\pi}{6}, -\frac{\pi}{6}\right)\) into the gradient vector. $$\nabla f|_P = \left\langle\cos\left(\frac{\pi}{6} + 2 \cdot \frac{\pi}{6} + \frac{\pi}{6}\right), 2\cos\left(\frac{\pi}{6} + 2 \cdot \frac{\pi}{6} + \frac{\pi}{6}\right), -\cos\left(\frac{\pi}{6} + 2 \cdot \frac{\pi}{6} + \frac{\pi}{6}\right)\right\rangle = \left\langle\cos\left(\pi\right), 2\cos\left(\pi\right), -\cos\left(\pi\right)\right\rangle = \left\langle-1, -2, 1\right\rangle$$
04

Unit Vector of Maximum Increase

Normalized gradient vector at point \(P\) is the unit vector of maximum increase. Since the gradient vector at point \(P\) is \(\left\langle-1, -2, 1\right\rangle\), its magnitude is: $$\|\nabla f|_P\| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{6}$$ Then, normalize the gradient vector: $$\text{Unit Vector} = \frac{\nabla f|_P}{\|\nabla f|_P\|} = \frac{1}{\sqrt{6}} \left\langle-1, -2, 1\right\rangle$$
05

Maximum Rate of Change

The rate of change of \(f\) in the direction of maximum increase is the magnitude of the gradient vector at point \(P\). $$\text{Max Rate of Change} = \|\nabla f|_P\| = \sqrt{6}$$
06

Directional Derivative in the Given Direction

Compute the directional derivative at point \(P\) in the direction of the given unit vector \(\left\langle\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right\rangle\). The directional derivative is the dot product of the gradient vector and the unit vector: $$D_\mathbf{u}f|_P = \nabla f|_P \cdot \mathbf{u} = \left\langle-1, -2, 1\right\rangle \cdot \left\langle\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right\rangle = -\frac{1}{3} -\frac{4}{3} +\frac{2}{3} = - \frac{3}{3}$$ Therefore, the directional derivative of \(f\) at point \(P\) in the given direction is \(-1\).

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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 an essential concept in calculus when dealing with functions of multiple variables. They provide the rate at which the function changes as we vary one variable while keeping others constant. The partial derivative with respect to a variable is represented by \( \frac{\partial f}{\partial x} \), where \( f \) is the function and \( x \) is the variable of interest. In our function \( f(x, y, z) = 1 + \sin(x + 2y - z) \), the partial derivatives are calculated as follows:
  • \( \frac{\partial f}{\partial x} = \cos(x + 2y - z) \) which represents the rate of change of \( f \) as \( x \) changes.
  • \( \frac{\partial f}{\partial y} = 2\cos(x + 2y - z) \), providing insight into how \( y \) affects \( f \).
  • \( \frac{\partial f}{\partial z} = -\cos(x + 2y - z) \), showing how changes in \( z \) impact \( f \).
These derivatives form the components of the gradient vector, which will lead us to understand directions of change in the function.
Directional Derivative
The directional derivative extends the concept of partial derivatives. It measures the rate of change of a function in any specified direction, not just along the axes. To find it, you need the gradient vector of the function and the direction vector. The directional derivative is calculated as the dot product of these two vectors.Let's break it down step-by-step:
  • The directional derivative of \( f \) at point \( P \) in the direction of a vector \( \mathbf{u} \) is denoted \( D_\mathbf{u}f|_P \).
  • Using the gradient vector \( abla f = \langle-1, -2, 1\rangle \) at \( P \) and the unit direction vector \( \mathbf{u} = \langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \rangle \), the directional derivative is calculated as:\[D_\mathbf{u}f|_P = abla f|_P \cdot \mathbf{u}\]
  • Performing the dot product, we get \(-\frac{1}{3} -\frac{4}{3} +\frac{2}{3} = -1\).
Thus, the directional derivative tells us how much \( f \) is increasing or decreasing at \( P \) in the specified direction of \( \mathbf{u} \). A negative value of -1 indicates a decrease in that direction.
Rate of Change
The rate of change can take many forms in calculus, but when dealing with multiple variables, it's often expressed using the gradient vector. This vector is formed by the partial derivatives and indicates how the function changes at a point in space. In the context of maximum increase, the gradient vector \( abla f|_P \) points in the direction where the function increases most rapidly.To find the rate of change in this maximum direction:
  • Calculate the magnitude of the gradient vector, \( \|abla f|_P\| \), which gives the rate of change in that direction.
  • In our example, \( abla f|_P = \langle -1, -2, 1 \rangle \), so the magnitude is:\[\|abla f|_P\| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{6}\]
Therefore, the rate of change at point \( P \) in the direction of maximum increase is \( \sqrt{6} \). This magnitude gives a numeric value for how rapidly \( f \) increases in the steepest ascent direction.

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