Question

Find a unit vector in the direction in which \(f\) increases most rapidly at \(\mathbf{p} .\) What is the rate of change in this direction? \(f(x, y, z)=x^{2} y z ; \mathbf{p}=(1,-1,2)\)

Short answer

Unit vector: \((-\frac{4}{\sqrt{21}}, \frac{2}{\sqrt{21}}, -\frac{1}{\sqrt{21}})\); Rate of change: \(\sqrt{21}\).

Step-by-step solution

Find the Gradient of the Function

The gradient of a function indicates the direction of the steepest ascent. For the function \( f(x, y, z) = x^2 y z \), we calculate the gradient \( abla f(x, y, z) \).The partial derivatives are:- \( \frac{\partial f}{\partial x} = 2xy z \)- \( \frac{\partial f}{\partial y} = x^2 z \)- \( \frac{\partial f}{\partial z} = x^2 y \)Thus, the gradient is:\[ abla f(x, y, z) = (2xyz, x^2z, x^2y) \]

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

Now, we place the coordinates of point \( \mathbf{p} = (1, -1, 2) \) into the gradient:- \( 2xyz = 2 \cdot 1 \cdot (-1) \cdot 2 = -4 \)- \( x^2z = 1^2 \cdot 2 = 2 \)- \( x^2y = 1^2 \cdot (-1) = -1 \)So, \( abla f(1, -1, 2) = (-4, 2, -1) \).

Find the Unit Vector in the Direction of the Gradient

To obtain a unit vector, divide the gradient vector by its magnitude.First, calculate the magnitude of \( abla f(1, -1, 2) \):\[\| abla f(1, -1, 2) \| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}.\]The unit vector is:\[\mathbf{u} = \frac{1}{\sqrt{21}}(-4, 2, -1).\]

Calculate the Rate of Change

The rate of change of \( f \) in the direction of the gradient is given by the magnitude of the gradient vector. Thus, the rate of change at \( \mathbf{p} \) is:\[\| abla f(1, -1, 2) \| = \sqrt{21}.\]

Key concepts

Unit Vector

A unit vector is a vector with a length, or magnitude, of exactly 1. It is often used to specify a direction without regard to any specific scale or magnitude. To find a unit vector in a particular direction, you take the original vector that indicates the direction and divide each of its components by its magnitude. This process normalizes the vector, converting it from any length to a unit length of 1, while maintaining its original direction.
In our example, the vector derived from the gradient of the function at the point \( \mathbf{p} = (1, -1, 2) \) is \((-4, 2, -1)\). The magnitude of this gradient vector is calculated as \( \sqrt{21} \).

• First, calculate each component magnitude: \((-4)^2 = 16\), \(2^2 = 4\), \((-1)^2 = 1\).
• Sum these values to get \(16 + 4 + 1 = 21\).
• The square root of 21 gives us the magnitude \( \sqrt{21} \).

To find the unit vector, each component of the gradient vector is divided by \( \sqrt{21} \), giving us the unit vector \( \mathbf{u} = \frac{1}{\sqrt{21}}(-4, 2, -1) \). This vector maintains the direction of the original gradient but with a magnitude of 1.

Partial Derivatives

Partial derivatives measure how a function changes as its variables change, but with all other variables held constant. They are crucial in multivariable calculus and are used to calculate the gradient vector, which points in the direction of the steepest ascent of a function.
For the function \( f(x, y, z) = x^2 y z \), we find partial derivatives with respect to each variable:

• \( \frac{\partial f}{\partial x} = 2xyz \) — This derivative tells us how \( f \) changes as \( x \) changes, keeping \( y \) and \( z \) constant.
• \( \frac{\partial f}{\partial y} = x^2z \) — This shows the rate of change of \( f \) with respect to \( y \), with \( x \) and \( z \) fixed.
• \( \frac{\partial f}{\partial z} = x^2y \) — This describes how \( f \) changes as \( z \) changes, while \( x \) and \( y \) are constant.

These partial derivatives are combined to form the gradient vector \( abla f = (2xyz, x^2z, x^2y) \). At point \( \mathbf{p} = (1, -1, 2) \), the gradient becomes \( (-4, 2, -1) \), indicating both magnitude and direction for the steepest ascent.

Rate of Change

The rate of change of a function at a certain point in a specific direction is a fundamental concept in calculus, particularly in the study of multivariable functions. When we discuss the rate of change in the context of the gradient, we're specifically looking at how quickly the function's value is increasing or decreasing in the direction of the gradient.
The gradient vector itself points in the direction of the steepest increase of the function, and its magnitude tells us just how quickly the function is changing in that direction.

• For the function \( f(x, y, z) = x^2 y z \) at the point \( \mathbf{p} = (1, -1, 2) \), the gradient vector was found to be \( (-4, 2, -1) \) with a magnitude of \( \sqrt{21} \).
• This magnitude, \( \sqrt{21} \), represents the instantaneous rate of change of the function \( f \) at point \( \mathbf{p} \) in the direction of the gradient.

Therefore, the rate of change at \( \mathbf{p} \) in the direction where \( f \) increases most rapidly is exactly \( \sqrt{21} \). This means that if you move in the direction of the gradient, the value of the function \( f \) will increase by \( \sqrt{21} \) units per unit step.