Problem 279 - Calculus 3

Question

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

Step-by-Step Solution

Answer

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

Step 1: Calculate the Gradient of the Function

The gradient of the function \(f(x, y) = \ln(x + 2y)\) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). 1. Calculate \( \frac{\partial f}{\partial x} \): \( \frac{\partial}{\partial x} \ln(x + 2y) = \frac{1}{x+2y} \).2. Calculate \( \frac{\partial f}{\partial y} \): \( \frac{\partial}{\partial y} \ln(x + 2y) = \frac{2}{x+2y} \).So, the gradient is \( abla f = \left( \frac{1}{x+2y}, \frac{2}{x+2y} \right) \).

Step 2: Find the Unit Vector Components

The directional vector is given by \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \).Substitute \( \theta = \frac{\pi}{3} \):1. \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).2. \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \).Thus, \( \mathbf{u} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \).

Step 3: Compute the Directional Derivative

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

Step 4: Final Result

Hence, the directional derivative of \( f(x, y) = \ln(x + 2y) \) in the direction of the unit vector \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \) with \( \theta = \frac{\pi}{3} \) is:\[ D_{\mathbf{u}} f = \frac{1 + \sqrt{3}}{2(x+2y)} \].This expression gives the rate of change of the function in the direction specified by the given angle.

Key Concepts

Understanding Gradient Vectors

The gradient vector is an essential concept when dealing with functions of several variables. It provides the direction and rate of the steepest increase of a function. For a given function of two variables, such as \(f(x, y) = \ln(x + 2y)\), the gradient is denoted by \( abla f\). This vector is comprised of the partial derivatives of the function with respect to each variable.

• The gradient vector \(abla f\) is found by calculating the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
• Each component of the gradient corresponds to the influence of that variable on the function's output.

For our example function, the gradient is calculated as \( abla f = \left( \frac{1}{x+2y}, \frac{2}{x+2y} \right) \). This shows how changes in \(x\) and \(y\) affect the function \(f(x, y)\).

Exploring Partial Derivatives

Partial derivatives look at how a function changes as one variable is varied while all other variables are held constant. They are crucial for understanding the behavior of multivariable functions.

• The partial derivative \(\frac{\partial f}{\partial x}\) shows how the function changes with variations in \(x\).
• Similarly, \(\frac{\partial f}{\partial y}\) indicates changes in the function as \(y\) varies.

For the function \(f(x, y) = \ln(x + 2y)\), we compute the partial derivatives as follows:

• \(\frac{\partial}{\partial x} \ln(x + 2y) = \frac{1}{x+2y}\)
• \(\frac{\partial}{\partial y} \ln(x + 2y) = \frac{2}{x+2y}\)

These derivatives build the gradient and help determine how sensitive the function is to changes in \(x\) and \(y\).

Defining Unit Vectors

A unit vector is a vector with a magnitude of one. It specifies a direction without scaling. In vector analysis, unit vectors are often used to represent direction.

• For two-dimensional space \(\mathbb{R}^2\), a unit vector \(\mathbf{u}\) can be expressed as \(\cos \theta \mathbf{i} + \sin \theta \mathbf{j}\), where \(\theta\) is the angle relative to the positive x-axis.

In our exercise, we substitute \(\theta = \frac{\pi}{3}\) to get components of the unit vector:

• \(\cos(\frac{\pi}{3}) = \frac{1}{2}\)
• \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)

Thus, the unit vector \(\mathbf{u}\) becomes \(\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j}\). This vector indicates the direction in which we evaluate the directional derivative.

Applying the Dot Product

The dot product is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It's a way to multiply vectors like multiplying scalar numbers, but there's more depth to it.

• The dot product of two vectors \((a, b)\) and \((c, d)\) is calculated as \(a \cdot c + b \cdot d\).
• It gives us a measure of how much one vector extends in the direction of another.

To find the directional derivative of a function in the direction of a unit vector, we compute the dot product of the gradient vector and the unit vector.
For our problem, we find:

• \(D_{\mathbf{u}} f = abla f \cdot \mathbf{u}\)
• This results in \(D_{\mathbf{u}} f = \frac{1}{2(x+2y)} + \frac{2\sqrt{3}}{2(x+2y)}\)
• Ultimately simplifying to \(\frac{1 + \sqrt{3}}{2(x+2y)}\)

Thus, the dot product helps in calculating how the function changes as you move in a specified direction. It's essential for understanding the orientation and influence of changes in multivariable contexts.