Problem 274 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Function and Gradient

The function given is \(f(x, y) = x^2 + 2y^2\). The gradient of \(f\), denoted as \(abla f\), is the vector of partial derivatives. To find \(abla f\), compute \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

Step 2: Compute the Gradient

Calculate the partial derivatives: \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = 4y\). Thus, the gradient vector is \(abla f = \langle 2x, 4y \rangle\).

Step 3: Define the Unit Vector

The unit vector is given as \(\mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\). For \(\theta = \frac{\pi}{6}\), calculate \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\) and \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). Thus, \(\mathbf{u} = \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle\).

Step 4: Calculate the Directional Derivative

The directional derivative of \(f\) in the direction of \(\mathbf{u}\) is given by:\[ D_{\mathbf{u}} f = abla f \cdot \mathbf{u} \]Substitute the values: \[ D_{\mathbf{u}} f = \langle 2x, 4y \rangle \cdot \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle = 2x \cdot \frac{\sqrt{3}}{2} + 4y \cdot \frac{1}{2} \]Simplify this to obtain: \[ \sqrt{3}x + 2y \]

Step 5: Conclude with the Expression for the Directional Derivative

The directional derivative of the function \(f(x,y) = x^2 + 2y^2\) in the direction given by \(\theta = \frac{\pi}{6}\) is \(\sqrt{3}x + 2y\).

Key Concepts

Gradient Vector

The gradient vector is a powerful tool in multivariable calculus. It helps you understand how a function changes at a given point. The gradient, denoted as \(abla f\), is made up of partial derivatives. It points in the direction of the greatest rate of increase of the function. In our exercise, the function is \(f(x, y) = x^2 + 2y^2\). The gradient of this function is \(abla f = \langle 2x, 4y \rangle\).
The direction of this vector tells us where the slope is steepest. So, if you wanted to "climb" the fastest along the surface defined by \(f\), follow the gradient vector!

Partial Derivatives

To understand the gradient vector, you need to know about partial derivatives first. A partial derivative represents how a function changes with respect to one variable while keeping others constant.
For the function \(f(x, y) = x^2 + 2y^2\):

• The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2x\).
• The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 4y\).

The computations tell us how sensitive the function is to changes in \(x\) and \(y\).
By assembling these, we construct the gradient vector.

Unit Vector

A unit vector has a magnitude of 1 and indicates direction. In our problem, the unit vector is given by \(\mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\). It characterizes a direction in the plane.
For \(\theta = \frac{\pi}{6}\):

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

Thus, the unit vector \(\mathbf{u}\) is \(\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle\). This vector tells us the specific direction in which we want to compute the change in the function.

Trigonometric Functions

Trigonometric functions such as \(\cos\) and \(\sin\) are fundamental in describing angles and rotations. They are essential in defining our unit vector \(\mathbf{u}\).
These functions relate an angle \(\theta\) to the ratios of sides in a right triangle. For \(\theta = \frac{\pi}{6}\):

• \(\cos \left(\frac{\pi}{6}\right)\) describes the adjacent side's ratio to the hypotenuse, giving us \(\frac{\sqrt{3}}{2}\).
• \(\sin \left(\frac{\pi}{6}\right)\) describes the opposite side's ratio to the hypotenuse, resulting in \(\frac{1}{2}\).

These values become the components of the unit vector, directing us to calculate the directional derivative correctly.