Problem 18 - Advanced Engineering Mathematics

Question

Find the directional derivative of the given function at the given point in the indicated direction. $$ F(x, y, z)=\frac{x^{2}-y^{2}}{z^{2}} ;(2,4,-1), \mathbf{i}-2 \mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{44}{\sqrt{6}} \).

Step 1: Find the Gradient of the Function

The gradient \( abla F(x, y, z) \) of the function \( F(x, y, z) = \frac{x^2 - y^2}{z^2} \) is calculated by finding the partial derivatives with respect to \( x \), \( y \), and \( z \). \[ abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \]Calculate each partial derivative:- \( \frac{\partial F}{\partial x} = \frac{2x}{z^2} \)- \( \frac{\partial F}{\partial y} = \frac{-2y}{z^2} \)- \( \frac{\partial F}{\partial z} = \frac{-2(x^2 - y^2)}{z^3} \)So, the gradient is: \[ abla F = \left( \frac{2x}{z^2}, \frac{-2y}{z^2}, \frac{-2(x^2 - y^2)}{z^3} \right) \]

Step 2: Calculate the Gradient at the Given Point

Substitute the point \((2, 4, -1)\) into the gradient expression:\[ abla F(2, 4, -1) = \left( \frac{2(2)}{(-1)^2}, \frac{-2(4)}{(-1)^2}, \frac{-2(2^2 - 4^2)}{(-1)^3} \right) \]Calculate each component:- First component: \( \frac{4}{1} = 4 \)- Second component: \( \frac{-8}{1} = -8 \)- Third component: \( \frac{-2(4 - 16)}{-1} = \frac{-2(-12)}{-1} = 24 \)Thus, \( abla F(2, 4, -1) = (4, -8, 24) \).

Step 3: Determine the Unit Vector in the Given Direction

The direction vector given is \( \mathbf{i} - 2\mathbf{j} + \mathbf{k} \), which can be represented as \( (1, -2, 1) \).The unit vector \( \mathbf{u} \) is obtained by normalizing this vector:\[ \mathbf{u} = \frac{1}{\sqrt{1^2 + (-2)^2 + 1^2}}(1, -2, 1) = \frac{1}{\sqrt{6}}(1, -2, 1) \].

Step 4: Compute the Directional Derivative

The directional derivative of \( F \) at point \( (2, 4, -1) \) in the direction of unit vector \( \mathbf{u} \) is given by the dot product:\[ D_{\mathbf{u}}F = abla F \cdot \mathbf{u} \]Substitute the known values:\[ D_{\mathbf{u}}F = (4, -8, 24) \cdot \left( \frac{1}{\sqrt{6}}(1, -2, 1) \right) \]Compute the dot product:\[ D_{\mathbf{u}}F = \frac{1}{\sqrt{6}} (4 \cdot 1 + (-8) \cdot (-2) + 24 \cdot 1) \]\[ = \frac{1}{\sqrt{6}} (4 + 16 + 24) = \frac{44}{\sqrt{6}} \].

Key Concepts

Gradient

The gradient of a function is a vector that contains all of its partial derivatives. It essentially provides the direction of maximum increase of the function at a given point. For a function of several variables like \( F(x, y, z) = \frac{x^2 - y^2}{z^2} \), the gradient is denoted by \( abla F \) and can be computed as:

• \( \frac{\partial F}{\partial x} \) - the partial derivative with respect to \( x \)
• \( \frac{\partial F}{\partial y} \) - the partial derivative with respect to \( y \)
• \( \frac{\partial F}{\partial z} \) - the partial derivative with respect to \( z \)

By calculating these derivatives, we can express the gradient as a vector. For our given function, the gradient is found to be \( abla F = \left( \frac{2x}{z^2}, \frac{-2y}{z^2}, \frac{-2(x^2 - y^2)}{z^3} \right) \).
This vector points in the direction where the function increases fastest, with its magnitude representing the rate of change. Substituting the given point \((2, 4, -1)\), gives \( abla F(2, 4, -1) = (4, -8, 24) \).
Thus, knowing the gradient is essential when determining the directional derivative, as the result heavily depends on it.

Partial Derivatives

Partial derivatives refer to the derivatives of a multivariable function with respect to one variable while keeping others constant. It helps us understand how the function changes as each variable changes individually.
In our exercise function \( F(x, y, z) = \frac{x^2 - y^2}{z^2} \), we compute the partial derivative with respect to each variable:

• \( \frac{\partial F}{\partial x} = \frac{2x}{z^2} \) illustrates how \( F \) changes with \( x \), keeping \( y \) and \( z \) constant.
• \( \frac{\partial F}{\partial y} = \frac{-2y}{z^2} \) illustrates how \( F \) changes with \( y \), keeping \( x \) and \( z \) constant.
• \( \frac{\partial F}{\partial z} = \frac{-2(x^2 - y^2)}{z^3} \) illustrates how \( F \) changes with \( z \), keeping \( x \) and \( y \) constant.

These derivatives are essential to construct the gradient vector, which plays a crucial role in understanding the function's behavior at specific points.

Unit Vector

A unit vector is a vector with a magnitude of 1 and is used to indicate direction. In the context of directional derivatives, it's vital to have the direction expressed as a unit vector, because the direction of change should not be influenced by the vector’s length.
Given a direction vector \( \mathbf{i} - 2\mathbf{j} + \mathbf{k} \), we need to convert it to a unit vector \( \mathbf{u} \). To do this, compute its magnitude, and divide each component by this magnitude. Here, the magnitude of the vector \((1, -2, 1)\) is \( \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \). Therefore, the unit vector is:

• \( \mathbf{u} = \frac{1}{\sqrt{6}}(1, -2, 1) \)

Using a unit vector ensures that when we calculate the directional derivative, it is purely in the direction desired, giving the exact rate of change of the function in that direction.