Problem 274 - Calculus 3

Question

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) $$ f(x, y)=x^{2}+2 y^{2}, \theta=\frac{\pi}{6} $$

Step-by-Step Solution

Answer

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

Step 1: Find the Gradient of the Function

First, we need to find the gradient of the function \( f(x, y) = x^2 + 2y^2 \). The gradient is a vector composed of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate each partial derivative: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = 4y \). Therefore, \( abla f = (2x, 4y) \).

Step 2: Evaluate the Unit Vector \( \mathbf{u} \)

The unit vector \( \mathbf{u} \) is given by \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \) with \( \theta = \frac{\pi}{6} \). Calculate \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \). Thus, \( \mathbf{u} = \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \).

Step 3: Calculate the Directional Derivative

The directional derivative of a function \( f \) in the direction of a unit vector \( \mathbf{u} \) is given by \( D_{\mathbf{u}} f = abla f \cdot \mathbf{u} \). Substitute the expressions of \( abla f = (2x, 4y) \) and \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \) into the formula: \[ D_{\mathbf{u}} f = (2x, 4y) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) = 2x \cdot \frac{\sqrt{3}}{2} + 4y \cdot \frac{1}{2} \].

Step 4: Simplify the Expression

Simplify the calculated expression from Step 3: \[ 2x \cdot \frac{\sqrt{3}}{2} + 4y \cdot \frac{1}{2} = x \sqrt{3} + 2y \]. Therefore, the directional derivative \( D_{\mathbf{u}} f \) is \( x \sqrt{3} + 2y \).

Key Concepts

Gradient of a Function

The gradient of a function is an essential concept in multivariable calculus. It represents the direction and rate of the steepest increase of a function. For a function of two variables, such as \( f(x, y) \), the gradient is a vector that comprises the function's partial derivatives with respect to each variable.

In our problem, we are working with the function \( f(x, y) = x^2 + 2y^2 \). To find its gradient, noted as \( abla f \), we need to compute the partial derivatives:

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x \). This indicates how much the function changes as \( x \) changes, while \( y \) is kept constant.
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 4y \). This shows the rate of change of the function in the \( y \)-direction.

Hence, the gradient vector \( abla f = (2x, 4y) \) provides both the direction and magnitude of the maximum increase of the function as we move away from a given point \((x, y)\). It plays a critical role in understanding how the function behaves in space.

Partial Derivatives

Partial derivatives are derivatives of functions with more than one variable, taken with respect to just one of these variables at a time, while all other variables are held constant. This is a crucial technique, especially in fields like physics and engineering, where systems often depend on several variables.

In our example with the function \( f(x, y) = x^2 + 2y^2 \), identifying the partial derivatives helps us understand how the function changes along each axis:

• The partial derivative \( \frac{\partial f}{\partial x} = 2x \) indicates the rate of change of \( f \) with respect to \( x \), reflecting how sensitive our function is to changes in the \( x \)-direction.
• Similarly, \( \frac{\partial f}{\partial y} = 4y \) shows how the function shifts as \( y \) varies, giving insight into the behavior of the function in the \( y \)-direction.

By calculating these partial derivatives, we gain detailed information about the local behavior of the function, making it a vital step in analyzing multivariable functions and solving problems involving gradients and directional derivatives.

Unit Vector

A unit vector is a vector with a length of 1, and it is primarily used to indicate direction. When calculating directional derivatives, the unit vector specifies the direction in which you want to measure the function's rate of change.

In our given task, the unit vector \( \mathbf{u} \) is expressed as \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \), where \( \theta \) represents the angle that vector \( \mathbf{u} \) makes with the positive \( x \)-axis.

For our specific case, \( \theta = \frac{\pi}{6} \). We calculate:

• \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
• \( \sin \frac{\pi}{6} = \frac{1}{2} \)

Thus, the unit vector becomes \( \mathbf{u} = \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \). This unit vector guides us in evaluating the directional derivative as it tells us the precise direction to measure the rate of change of the function.