Problem 275 - Calculus 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 \(-\frac{\sqrt{2}}{2(x + 2y)}\).

Step 1: Understand the Function

The function provided is \( f(x, y) = \frac{y}{x + 2y} \). This is a function of two variables \( x \) and \( y \). We will calculate the directional derivative based on the formula and the direction provided.

Step 2: Directional Vector from Angle \( \theta \)

The problem provides the unit vector \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \) with \( \theta = -\frac{\pi}{4} \). Calculate the components of the unit vector: \( \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \) and \( \sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \), making \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \).

Step 3: Calculate the Gradient of \( f \)

The gradient vector \( abla f(x, y) \) is composed of partial derivatives: \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Calculate these partial derivatives: \[ \frac{\partial f}{\partial x} = -\frac{y}{(x + 2y)^2} \] and \[ \frac{\partial f}{\partial y} = \frac{x + y}{(x + 2y)^2} \]. Therefore, \( abla f(x, y) = \left( -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right) \).

Step 4: Compute the Directional Derivative

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

Key Concepts

Gradient Vector

The gradient vector is a crucial concept in understanding directional derivatives. It gives us the direction of the steepest increase of a function at any given point. For a function like \( f(x, y) \), which depends on both \( x \) and \( y \), the gradient vector \( abla f \) is composed of the partial derivatives with respect to each variable.

In our example, the function is \( f(x, y) = \frac{y}{x+2y} \). To find the gradient vector \( abla f \), we calculate the partial derivatives, \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). These derivatives tell how the function changes as you vary \( x \) and \( y \), respectively.

The calculated gradient is:

• \( \frac{\partial f}{\partial x} = -\frac{y}{(x + 2y)^2} \)
• \( \frac{\partial f}{\partial y} = \frac{x + y}{(x + 2y)^2} \)

This gives the gradient vector: \( abla f(x, y) = \left( -\frac{y}{(x + 2y)^2}, \frac{x + y}{(x + 2y)^2} \right) \).

The gradient vector not only informs us about the rate of change but also helps calculate the directional derivative, highlighting its importance in the analysis of functions of several variables.

Partial Derivatives

Partial derivatives play a fundamental role in multi-variable calculus, specifically for functions like \( f(x, y) \) that depend on more than one variable. They measure how a function changes as we change one variable while keeping others constant.

To delve deeper, consider the function \( f(x, y) = \frac{y}{x + 2y} \). The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), assesses how \( f \) changes as \( x \) increases, keeping \( y \) fixed. Similarly, \( \frac{\partial f}{\partial y} \) examines the change in \( f \) as \( y \) varies with \( x \) constant.

In our example:

• \( \frac{\partial f}{\partial x} = -\frac{y}{(x + 2y)^2} \), meaning as \( x \) increases, the function generally decreases.
• \( \frac{\partial f}{\partial y} = \frac{x + y}{(x + 2y)^2} \), indicating an increase with \( y \).

Partial derivatives are foundational because they contribute to forming the gradient vector, enabling us to explore the function's behavior through directional derivatives. Understanding these small changes in each direction lets us grasp the broader changes in multivariable functions.

Unit Vector

Unit vectors are essential in directional calculus because they define the direction in which we measure the rate of change of a function. A unit vector has a magnitude of one, making it a perfect "direction marker" without altering the function's magnitude changes.

In the context of directional derivatives, you find the derivative of a function in the direction of a given unit vector \( \mathbf{u} \).

For the exercise given, the unit vector is defined as \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \), where \( \theta = -\frac{\pi}{4} \). Calculation of its components yields:

• \( \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
• \( \sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \)

Thus, \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \).

Having a standardized scale, these unit vectors guide us in measuring how a function's magnitude changes in any specified direction, crucial for figuring out directional derivatives. Here, they ensure that the rate of change is signified strictly by the direction, not by any additional scaling.