Problem 9 - Calculus with Concepts in Calculus

Question

Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y, z)=x^{3} y^{2} z ; P=(2,-1,2) ; \mathbf{a}=2 \mathbf{i}-\mathbf{j}-2 \mathbf{k} $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{64}{3}\).

Step 1: Calculate the Gradient of f

The gradient of a function \(f(x, y, z)\) is given by \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\). For \(f(x, y, z)=x^{3} y^{2} z\), we find:1. \(\frac{\partial f}{\partial x} = 3x^2 y^2 z\)2. \(\frac{\partial f}{\partial y} = 2x^3 y z\)3. \(\frac{\partial f}{\partial z} = x^3 y^2\)Thus, \(abla f = (3x^2 y^2 z, 2x^3 y z, x^3 y^2)\).

Step 2: Evaluate the Gradient at Point P

Substitute the coordinates of point \(P = (2, -1, 2)\) into the gradient obtained:\(abla f(2, -1, 2) = (3(2)^2(-1)^2(2), 2(2)^3(-1)(2), (2)^3(-1)^2)\)Which reduces to:\(abla f(2, -1, 2) = (24, -32, 8)\).

Step 3: Normalize the Direction Vector a

The direction vector \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} - 2 \mathbf{k} \). First, find its magnitude:\[ \|\mathbf{a}\| = \sqrt{(2)^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]Normalize \( \mathbf{a} \) by dividing each component by its magnitude:\( \mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \).

Step 4: Calculate the Dot Product

Calculate the dot product of the gradient \(abla f(2, -1, 2)\) and the unit vector \(\mathbf{u}\):\[(24, -32, 8) \cdot \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) = 24 \cdot \frac{2}{3} + (-32) \cdot \left(-\frac{1}{3}\right) + 8 \cdot \left(-\frac{2}{3}\right)\]Which simplifies to:\[16 + \frac{32}{3} - \frac{16}{3} = 16 + \frac{16}{3}\].

Step 5: Simplify and Find the Directional Derivative

Complete the simplification:\[16 + \frac{16}{3} = \frac{48}{3} + \frac{16}{3} = \frac{64}{3}\].Thus, the directional derivative of \(f\) at point \(P\) in the direction of \(\mathbf{a}\) is \(\frac{64}{3}\).

Key Concepts

Gradient of a Function

The gradient of a function is like a multi-dimensional derivative that tells us how a function changes in space. For a function with variables such as \( x, y, z \), the gradient, often denoted by \( abla f \), is a vector. It points in the direction of the greatest increase of the function. This vector consists of partial derivatives of the function with respect to each variable.

For example, given a function \( f(x, y, z) = x^3 y^2 z \), the gradient is calculated as:

• \( \frac{\partial f}{\partial x} = 3x^2 y^2 z \)
• \( \frac{\partial f}{\partial y} = 2x^3 y z \)
• \( \frac{\partial f}{\partial z} = x^3 y^2 \)

Thus, the gradient vector is \( abla f = (3x^2 y^2 z, 2x^3 y z, x^3 y^2) \). This vector is useful because it tells us how the function \( f \) is changing at any given point \( (x, y, z) \). By evaluating the gradient at a point, we know the rate and direction of change.

Unit Vector

A unit vector is essentially a direction marker. It has a length of one and is used to indicate direction without concern for magnitude. It's essential when defining directions in vector calculus because it standardizes directions.

Let's say you have a direction vector \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} - 2 \mathbf{k} \). To convert \( \mathbf{a} \) into a unit vector \( \mathbf{u} \), first calculate its magnitude:

• \( \|\mathbf{a}\| = \sqrt{(2)^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3 \)

Then, normalize each component by dividing them by the magnitude:

• \( \mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \)

This conversion helps us use \( \mathbf{u} \) in vector operations without altering the magnitude of outcomes.

Partial Derivative

Partial derivatives allow us to see how a function changes with respect to one of its variables, holding others constant. It's like observing the slope of the function in one particular direction, while not changing the direction of other variables.

For the function \( f(x, y, z) = x^3 y^2 z \), the partial derivatives are:

• With respect to \( x \): \( \frac{\partial f}{\partial x} = 3x^2 y^2 z \)
• With respect to \( y \): \( \frac{\partial f}{\partial y} = 2x^3 y z \)
• With respect to \( z \): \( \frac{\partial f}{\partial z} = x^3 y^2 \)

Each of these gives us insight into how the function changes as we move slightly in the direction of just one of the axes, holding others constant. Partial derivatives are fundamental in forming the gradient of a function.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar. It's valuable in determining how closely aligned two vectors are. If the vectors point in the same direction, their dot product is positive; if in opposite directions, it's negative.

Given the gradient vector \( abla f(2, -1, 2) = (24, -32, 8) \) and the unit vector \( \mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \), the dot product is calculated as:

• \( (24, -32, 8) \cdot \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) = 24 \cdot \frac{2}{3} + (-32) \cdot \left(-\frac{1}{3}\right) + 8 \cdot \left(-\frac{2}{3}\right) \)

This evaluates to \( 16 + \frac{32}{3} - \frac{16}{3} = \frac{64}{3} \). The result corresponds to the directional derivative, showing the rate of change of the function \( f \) at a point in the specific direction of \( \mathbf{u} \). This is crucial for understanding how the function behaves as you move through its domain.