Problem 279 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\). $$f(x, y)=\ln (x+2 y), \quad \theta=\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{1 + 2\sqrt{3}}{2(x + 2y)} \).

Step 1: Calculate the Gradient of the Function

The first step is to calculate the gradient of the function \( f(x, y) = \ln(x + 2y) \). The gradient, \( abla f(x, y) \), is a vector of the partial derivatives of the function with respect to \( x \) and \( y \).\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \].Calculate \( \frac{\partial f}{\partial x} \):\[ \frac{\partial f}{\partial x} = \frac{1}{x + 2y} \cdot 1 = \frac{1}{x + 2y} \].Calculate \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial y} = \frac{1}{x + 2y} \cdot 2 = \frac{2}{x + 2y} \].Therefore, the gradient is:\[ abla f(x, y) = \left( \frac{1}{x + 2y}, \frac{2}{x + 2y} \right) \].

Step 2: Find the Unit Vector

The unit vector \( \mathbf{u} \) is given in terms of \( \theta \). For \( \theta = \frac{\pi}{3} \), substitute \( \theta \) to find \( \mathbf{u} \):Calculate \( \cos(\theta) \):\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \].Calculate \( \sin(\theta) \):\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \].Thus, the unit vector is:\[ \mathbf{u} = \left( \frac{1}{2} \right) \mathbf{i} + \left( \frac{\sqrt{3}}{2} \right) \mathbf{j} \].

Step 3: Calculate the Directional Derivative

The directional derivative \( D_{\mathbf{u}}f(x, y) \) of the function \( f(x, y) \) in the direction of the unit vector \( \mathbf{u} \) is given by the dot product of \( abla f(x, y) \) and \( \mathbf{u} \):\[ D_{\mathbf{u}}f(x, y) = abla f(x, y) \cdot \mathbf{u} \].Compute the dot product:\[ D_{\mathbf{u}}f(x, y) = \left( \frac{1}{x+2y}, \frac{2}{x+2y} \right) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \].This simplifies to:\[ D_{\mathbf{u}}f(x, y) = \frac{1 \cdot 1 + 2 \cdot \sqrt{3}}{2(x+2y)} \].\[ D_{\mathbf{u}}f(x, y) = \frac{1 + 2\sqrt{3}}{2(x + 2y)} \].

Key Concepts

Gradient

The gradient is a fundamental concept when dealing with functions of several variables. Think of the gradient as a generalization of the derivative for multi-variable functions. For a function like \( f(x, y) = \ln(x + 2y) \), the gradient can be visualized as a vector, pointing in the direction of the steepest increase of the function. It is not just a direction indicator, but also gives the function's rate of change in that direction. The gradient vector, denoted as \( abla f(x, y) \), is made up of partial derivatives of the function concerning each of its variables, in this case, \( x \) and \( y \).

• To find the gradient, calculate the partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = \frac{1}{x + 2y} \).
• Similarly, the partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = \frac{2}{x + 2y} \).

Combined, these results give the gradient as a vector: \( abla f(x, y) = \left( \frac{1}{x + 2y}, \frac{2}{x + 2y} \right) \). This vector guides us on how the value of the function would change most rapidly from a given point.

Partial Derivative

The partial derivative is a crucial part of understanding change in multi-variable calculus. When a function involves several variables, a regular derivative doesn't suffice; instead, partial derivatives come into play. Imagine a function \( f(x, y) \) and consider its behavior concerning only one of its variables at a time.

• For \( f(x, y) = \ln(x + 2y) \), isolate the change with respect to \( x \), keeping \( y \) constant. This is the partial derivative \( \frac{\partial f}{\partial x} = \frac{1}{x + 2y} \).
• Next, consider the change concerning \( y \) while treating \( x \) as constant. The result is \( \frac{\partial f}{\partial y} = \frac{2}{x + 2y} \).

Each derivative reveals how the function reacts to incremental changes in one variable, holding the others fixed. When combined into the gradient, they show how the function changes collectively with respect to all its variables.

Unit Vector

A unit vector is a vector with a magnitude of one. It is used primarily to indicate a direction, rather than magnitude. In the context of directional derivatives, the unit vector specifies the direction in which the derivative is being computed. For our exercise, the unit vector \( \mathbf{u} \) is given by the formula \( \mathbf{u} = \cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j} \). Here, \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the directions of the \( x \)-axis and \( y \)-axis, respectively.

• The angle \( \theta = \frac{\pi}{3} \) translates to \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).

Therefore, our unit vector becomes \( \mathbf{u} = \left( \frac{1}{2} \right) \mathbf{i} + \left( \frac{\sqrt{3}}{2} \right) \mathbf{j} \). This vector plays a key role in calculating the directional derivative, linking the direction of change quantified by the gradient with the specific direction in question.