Problem 11 - 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? \(f(x, y, z)=x^{2} y z ; \mathbf{p}=(1,-1,2)\)

Step-by-Step Solution

Answer

The unit vector is \( \frac{1}{\sqrt{21}} (-4, 2, -1) \) and the rate of change is \( \sqrt{21} \).

Step 1: Find the Gradient of f

The gradient of a function \( f(x, y, z) \) is a vector of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). For our function \( f(x,y,z) = x^2yz \), calculate the partial derivatives. \( \frac{\partial f}{\partial x} = 2xyz \), \( \frac{\partial f}{\partial y} = x^2z \), \( \frac{\partial f}{\partial z} = x^2y \). Thus, \( abla f = (2xyz, x^2z, x^2y) \).

Step 2: Evaluate the Gradient at Point p

Substitute the point \( \mathbf{p} = (1, -1, 2) \) into the gradient. The gradient becomes \( abla f(1, -1, 2) = (2(1)(-1)(2), (1)^2(2), (1)^2(-1)) \). Simplifying the components, we get \( abla f(1, -1, 2) = (-4, 2, -1) \).

Step 3: Find the Magnitude of the Gradient

Calculate the magnitude of the gradient vector, which is \( \|abla f(1, -1, 2)\| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \).

Step 4: Determine the Unit Vector in the Direction of the Gradient

The unit vector in the direction of the gradient is found by dividing the gradient vector by its magnitude. Thus, the unit vector is \( \frac{1}{\sqrt{21}} (-4, 2, -1) \).

Step 5: State the Rate of Change in the Direction of the Gradient

The rate of change in the direction of the gradient is the magnitude of the gradient. Thus, the rate of change is \( \sqrt{21} \).

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in understanding how multivariable functions change. In single variable calculus, derivatives tell us the rate at which a function changes as its input changes. For functions of multiple variables, we need to examine how the function behaves with respect to each variable independently.

This is where partial derivatives come into play. Consider a function like \( f(x, y, z) = x^2yz \), which depends on three variables \( x, y, z \). The partial derivative with respect to \( x \) would be calculated as if \( y \) and \( z \) were constants. It's expressed as \( \frac{\partial f}{\partial x} = 2xyz \). Similarly, the partial derivatives with respect to \( y \) and \( z \) are \( \frac{\partial f}{\partial y} = x^2z \) and \( \frac{\partial f}{\partial z} = x^2y \).

These partial derivatives are components of the gradient vector \( abla f \). This vector combines all the partial derivatives and provides crucial information about the direction in which the function increases most rapidly. It's a navigational tool in the multidimensional landscape of the function.

Rate of Change

The rate of change is a powerful concept, particularly when dealing with functions of several variables. It indicates how quickly a function's value is changing at a particular point. In the context of gradients, it refers to the maximum rate at which the function increases in a specific direction.

For the function \( f(x, y, z) = x^2yz \) at the point \( \mathbf{p} = (1, -1, 2) \), we first find the gradient vector: \( abla f = (2xyz, x^2z, x^2y) \). Substituting \( \mathbf{p} \) gives us \( (-4, 2, -1) \). This vector points in the direction where \( f \) increases most rapidly.

The magnitude of this gradient vector, calculated as \( \sqrt{21} \), represents the maximum rate of change of the function at that point. Essentially, it's the steepest slope of the function as you move through its landscape from that specific location. Hence, the rate of change in the direction of the gradient provides insight into how sharply the function is increasing.

Unit Vector

A unit vector is a vector with a length of one. It doesn't change the direction of the vector it's derived from, but it does give a sense of direction alone, stripped of any notion of length or size. When we talk about a unit vector in the direction of the gradient, we're essentially pointing out the direction in which the function increases most rapidly, without confounding it with how quickly it increases.

To find a unit vector, we divide a given vector by its magnitude, which normalizes the vector to length one. For example, from the gradient vector \( (-4, 2, -1) \), we calculate its magnitude \( \| (-4, 2, -1) \| = \sqrt{21} \). This gives the unit vector \( \frac{1}{\sqrt{21}} (-4, 2, -1) \).

Thus, unit vectors are vital in clearly indicating directions in multidimensional space without being skewed by magnitude, which is crucial for understanding vector directionality in calculus and physics.