Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-12, find a unit vector in the direction in which \(f\) increases most rapidly at p. What is the rate of change in this direction? \(f(x, y, z)=x e^{y z} ; \mathbf{p}=(2,0,-4)\)

Step-by-Step Solution

Answer

The unit vector is \( \left( \frac{1}{\sqrt{65}}, \frac{-8}{\sqrt{65}}, 0 \right) \) and the rate of change is \( \sqrt{65} \).

Step 1: Find the Gradient Vector

The direction in which a function increases most rapidly is given by its gradient. The gradient of the function \( f(x, y, z) = x e^{yz} \) is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Calculate each partial derivative:- \( \frac{\partial f}{\partial x} = e^{yz} \)- \( \frac{\partial f}{\partial y} = xz e^{yz} \)- \( \frac{\partial f}{\partial z} = xy e^{yz} \)So, \( abla f = \left( e^{yz}, xz e^{yz}, xy e^{yz} \right) \).

Step 2: Evaluate the Gradient at Point \( \mathbf{p} \)

To find the gradient at point \( \mathbf{p} = (2, 0, -4) \), substitute \( x = 2 \), \( y = 0 \), and \( z = -4 \) into the gradient:\[ abla f(2, 0, -4) = \left( e^{0 \cdot (-4)}, 2 \cdot (-4) \cdot e^{0 \cdot (-4)}, 2 \cdot 0 \cdot e^{0 \cdot (-4)} \right) \]Simplify:\[ abla f(2, 0, -4) = (1, -8, 0) \]

Step 3: Compute the Magnitude of the Gradient Vector

Calculate the magnitude of the gradient vector \( abla f(2, 0, -4) = (1, -8, 0) \) using the formula:\[ \| abla f \| = \sqrt{1^2 + (-8)^2 + 0^2} = \sqrt{1 + 64} = \sqrt{65} \]

Step 4: Determine the Unit Vector

The unit vector in the direction of the gradient is obtained by dividing the gradient by its magnitude:\[ \mathbf{u} = \frac{1}{\sqrt{65}}(1, -8, 0) = \left( \frac{1}{\sqrt{65}}, \frac{-8}{\sqrt{65}}, 0 \right) \]

Step 5: Find the Rate of Change in This Direction

The rate of change of the function in the direction of the gradient is given by the magnitude of the gradient vector \( \| abla f \| \):\[ \text{Rate of Change} = \sqrt{65} \]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus that deals with functions of multiple variables. The idea is to take the derivative of the function with respect to one variable, while keeping others constant.This allows us to understand how the function changes as each variable changes individually.
For a function like \(f(x, y, z) = x e^{yz}\), calculating partial derivatives involves differentiating with respect to each variable:

• The partial derivative \(\frac{\partial f}{\partial x}\) is\( e^{yz} \). Here, \(y\) and \(z\) are treated as constants.
• The partial derivative \(\frac{\partial f}{\partial y}\) is \( xz e^{yz} \). The variable \(x\) is treated as a constant when differentiating with respect to \(y\).
• The partial derivative \(\frac{\partial f}{\partial z}\) is \( xy e^{yz} \). Here, \(x\) and \(y\) are constants when differentiating with respect to \(z\).

These derivatives help to form the gradient vector, which points in the direction of the steepest slope.

Magnitude of Vector

The magnitude of a vector, sometimes called its length or norm, captures how "long" or "large" the vector is.This concept applies to both 2-dimensional and 3-dimensional vectors, and is crucial in physics and engineering.To find the magnitude, we use the formula:

• For a vector \( \mathbf{v} = (a, b, c) \), the magnitude \( \| \mathbf{v} \| \) is given by \( \sqrt{a^2 + b^2 + c^2} \).

In our exercise, the gradient vector at point \( \mathbf{p} = (2, 0, -4) \) is \( (1, -8, 0) \).Thus, its magnitude is calculated as:

• \( \| abla f \| = \sqrt{1^2 + (-8)^2 + 0^2} = \sqrt{1 + 64} = \sqrt{65} \).

This magnitude tells us about the steepness of the slope and how quickly the function increases.

Unit Vector

A unit vector is a vector that has a magnitude of 1. Unit vectors are often used to indicate direction without implying any change in magnitude.To convert any non-zero vector into a unit vector, you divide each component of the vector by the magnitude of the vector.
Here's how it works:

• Given a vector \( \mathbf{v} = (a, b, c) \) with magnitude \( \| \mathbf{v} \| \), the unit vector \( \mathbf{u} \) is \( \frac{1}{\| \mathbf{v} \|}(a, b, c) \).

In our example, the gradient vector at point \( \mathbf{p} \) is \( (1, -8, 0) \) with a magnitude of \( \sqrt{65} \).So, the unit vector \( \mathbf{u} \) in this direction is:

• \( \mathbf{u} = \left( \frac{1}{\sqrt{65}}, \frac{-8}{\sqrt{65}}, 0 \right) \).

This unit vector points in the direction of the gradient, showing where the function \( f \) increases most steeply.

Rate of Change

The rate of change indicates how a function's value changes with respect to a change in its input. When talking about the rate of change in a specific direction, it often refers to the directional derivative.In mathematics, the rate of change in the direction of maximum increase is directly linked with the gradient vector's magnitude.
For a gradient vector \( abla f \), the rate of change \(\| abla f \|\) tells us how fast the value of the function \( f \) changes as one moves in the direction of the gradient vector.In our exercise:

• The rate of change at point \( \mathbf{p} = (2, 0, -4) \) is \( \sqrt{65} \).

This means if you move from point \( \mathbf{p} \) in the direction where \( f \) increases most rapidly, the function will increase at this rate.This rate provides insights into the steepness and pace of change at any point on the function.