Problem 275 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\). $$f(x, y)=\frac{y}{x+2 y}, \theta=-\frac{\pi}{4}$$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{\sqrt{2}x}{2(x + 2y)^2} \).

Step 1: Understand the Function and Notation

The given function is \( f(x, y) = \frac{y}{x + 2y} \). We are asked to find the directional derivative in the direction of the unit vector \( \mathbf{u} = \cos(\theta)\mathbf{i} + \sin(\theta)\mathbf{j} \) where \( \theta = -\frac{\pi}{4} \).

Step 2: Compute the Gradient of the Function

The gradient of the function \( f(x, y) \), denoted as \( abla f \), is computed as follows:- Calculate the partial derivative with respect to \( x \): \( f_x = -\frac{y}{(x + 2y)^2} \).- Calculate the partial derivative with respect to \( y \): \( f_y = \frac{x + y}{(x + 2y)^2} \).Thus, \( abla f = \left< -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right> \).

Step 3: Determine the Unit Vector \( \mathbf{u} \)

With \( \theta = -\frac{\pi}{4} \), compute \( \mathbf{u} = \cos(-\frac{\pi}{4})\mathbf{i} + \sin(-\frac{\pi}{4})\mathbf{j} = \frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j} \).

Step 4: Calculate Dot Product for Directional Derivative

The directional derivative \( D_{\mathbf{u}}f \) is the dot product of \( abla f \) and \( \mathbf{u} \):\[D_{\mathbf{u}}f = \left< -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right> \cdot \left< \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right>.\]

Step 5: Perform and Simplify the Dot Product

Calculate the dot product:\[ D_{\mathbf{u}}f = \left( -\frac{y}{(x + 2y)^2} \cdot \frac{\sqrt{2}}{2} \right) + \left( \frac{x + y}{(x + 2y)^2} \cdot \left(-\frac{\sqrt{2}}{2}\right) \right) \]\[= \frac{-\sqrt{2}y}{2(x + 2y)^2} - \frac{-\sqrt{2}(x+y)}{2(x + 2y)^2} \]\[= \frac{-\sqrt{2}y + \sqrt{2}(x+y)}{2(x + 2y)^2} = \frac{\sqrt{2}x}{2(x + 2y)^2} \].

Key Concepts

Gradient of a Function

The gradient of a function is like a roadmap indicating the direction of the steepest ascent for multi-variable functions. When you're dealing with the function \( f(x, y) = \frac{y}{x + 2y} \), the gradient is expressed as a vector of its partial derivatives.
The gradient is often symbolized by \( abla f \) and calculated as follows:

• Find the partial derivative with respect to \( x \): \( f_x = -\frac{y}{(x + 2y)^2} \).
• Find the partial derivative with respect to \( y \): \( f_y = \frac{x + y}{(x + 2y)^2} \).

Thus, the gradient vector \( abla f \) becomes \( \left< -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right> \).
This vector points in the direction where the function increases the fastest, providing essential information for calculating the directional derivative.

Unit Vector

Unit vectors serve an essential purpose as they signify direction, being vectors of length one. A unit vector in a specific direction can be derived from any vector by dividing the vector by its magnitude. In this exercise, the unit vector \( \mathbf{u} \) guides us to compute the directional derivative of the function.
Given \( \theta = -\frac{\pi}{4} \), the components of the unit vector are:

• \( \cos(-\frac{\pi}{4}) \) aligning with the \( x \)-axis.
• \( \sin(-\frac{\pi}{4}) \) aligning with the \( y \)-axis.

This leads to \[\mathbf{u} = \frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j}\]because both components equal \( \frac{\sqrt{2}}{2} \) or its negative due to the angle given.
These computations ensure that \( \mathbf{u} \) remains a unit vector for proper use in further calculations.

Dot Product

The dot product is a crucial operation that combines two vectors into a single number, encapsulating both their magnitudes and the cosine of the angle between them. It's the key to obtaining a directional derivative.
For this exercise, the directional derivative \( D_{\mathbf{u}}f \) is calculated as a dot product between the gradient \( abla f \) and the unit vector \( \mathbf{u} \):\[ D_{\mathbf{u}}f = \left< -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right> \cdot \left< \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right> \]This calculation involves multiplying corresponding components and summing the results:

• Multiply \( -\frac{y}{(x + 2y)^2} \) with \( \frac{\sqrt{2}}{2} \).
• Multiply \( \frac{x + y}{(x + 2y)^2} \) with \( -\frac{\sqrt{2}}{2} \).
• Add these products together.

The resulting expression gives the rate of change of the function along the direction dictated by \( \mathbf{u} \).

Partial Derivative

Partial derivatives are derivatives of multi-variable functions, keeping all variables constant except one. They reveal how a function behaves with respect to individual variables.
To find the gradient, we start by calculating partial derivatives:

• \( f_x = -\frac{y}{(x + 2y)^2} \), showing how \( f(x, y) \) changes as \( x \) changes while \( y \) stays constant.
• \( f_y = \frac{x + y}{(x + 2y)^2} \), indicating the change in \( f(x, y) \) with respect to \( y \).

These partials are the building blocks for the gradient vector \( abla f \) and are vital in computing the directional derivative. When carefully applied, partial derivatives unravel the variation of a function along any path defined by the variables.