Problem 276 - Calculus 3

Question

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) $$ f(x, y)=\cos (3 x+y), \theta=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

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

Step 1: Determine the Gradient of the Function

The gradient of a function \( f(x, y) \), denoted by \( abla f \), is a vector composed of its partial derivatives. First, we find the partial derivative of \( f \) with respect to \( x \):\[ f_x(x, y) = \frac{\partial}{\partial x} \cos(3x + y) = -3\sin(3x + y) \]. Next, find the partial derivative with respect to \( y \): \[ f_y(x, y) = \frac{\partial}{\partial y} \cos(3x + y) = -\sin(3x + y) \]. Thus, the gradient is: \[ abla f = (-3\sin(3x + y), -\sin(3x + y)) \].

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

We are given the direction in which to take the derivative as a unit vector \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \). With \( \theta = \frac{\pi}{4} \), we compute \( \cos \theta = \frac{\sqrt{2}}{2} \) and \( \sin \theta = \frac{\sqrt{2}}{2} \). Thus, \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \).

Step 3: Compute the Directional Derivative

The directional derivative of \( f \) in the direction of \( \mathbf{u} \) is given by the dot product \( abla f \cdot \mathbf{u} \). Thus, we have: \[ abla f \cdot \mathbf{u} = (-3\sin(3x + y), -\sin(3x + y)) \cdot \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \]. Carrying out the dot product: \[ abla f \cdot \mathbf{u} = -3\sin(3x + y) \cdot \frac{\sqrt{2}}{2} - \sin(3x + y) \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} (3\sin(3x + y) + \sin(3x + y)) \]. Simplifying: \[ = -\frac{\sqrt{2}}{2} \cdot 4\sin(3x + y) = -2\sqrt{2}\sin(3x + y) \].

Step 4: Summary of Results

The directional derivative of \( f(x, y) = \cos(3x + y) \) in the direction given by \( \mathbf{u} = \frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j} \) is \( -2\sqrt{2}\sin(3x + y) \).

Key Concepts

Gradient Vector

In mathematics, a gradient vector is a crucial tool for understanding how a function changes at any given point. Consider a function, say, \( f(x, y) \). This function maps a point from a two-dimensional space (i.e., \( x \) and \( y \)) to a scalar value. The gradient vector, denoted as \( abla f \), is essentially a vector comprised of the partial derivatives of the function. It points in the direction of the steepest increase of the function.The components of the gradient vector \( abla f \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \). Mathematically, it's expressed as:

• \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

In our exercise, the function is \( f(x, y) = \cos(3x+y) \). The gradient vector becomes \( abla f = (-3\sin(3x+y), -\sin(3x+y)) \). Each component tells us how the function changes as you adjust \( x \) or \( y \), respectively.Understanding the gradient vector helps us in calculating the directional derivative as it tells us the function's rate of change along a particular direction.

Partial Derivative

Partial derivatives are foundational in multivariable calculus. They measure how a function changes as you vary one of its variables while keeping others constant. This concept is crucial when dealing with functions of several variables, such as \( f(x, y) \).Let's look at the function \( f(x, y) = \cos(3x + y) \). To compute the partial derivative with respect to \( x \), denoted \( f_x(x, y) \), we differentiate \( f \) while keeping \( y \) constant. The result is \( -3\sin(3x+y) \).Similarly, the partial derivative with respect to \( y \), denoted \( f_y(x, y) \), is found by differentiating \( f \) while treating \( x \) as a constant. This results in \( -\sin(3x+y) \).Using partial derivatives, we build gradient vectors and explore how functions behave locally. They are foundational to computing the gradient vector and essential for finding the directional derivative.

Unit Vector

A unit vector is a vector with a magnitude of 1. It points in a specific direction but has no other dimension like size. Unit vectors are especially important when calculating directional derivatives because they determine the exact direction in which we examine the rate of change of a function.In the given exercise, we're asked to calculate the directional derivative using a unit vector \( \mathbf{u} = \cos\theta \mathbf{i} + \sin\theta \mathbf{j} \). Since \( \theta = \frac{\pi}{4} \), we find:

• \( \cos\theta = \frac{\sqrt{2}}{2} \)
• \( \sin\theta = \frac{\sqrt{2}}{2} \)

Thus, the unit vector becomes \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \). As a unit vector, it confirms that the magnitude is exactly 1, ensuring that our directional derivative is a pure rate without directional bias.

Dot Product

The dot product is a fundamental operation involving two vectors, resulting in a scalar. It is especially significant in calculating directional derivatives because it measures the component of one vector in the direction of another.For two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is given by:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In our problem, to find the directional derivative, we compute the dot product of the gradient \( abla f = (-3\sin(3x+y), -\sin(3x+y)) \) and the unit vector \( \mathbf{u} = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \).Carrying out the calculations:

• The dot product results in \( -\frac{\sqrt{2}}{2} (3\sin(3x+y) + \sin(3x+y)) \)
• Simplified to \( -2\sqrt{2}\sin(3x+y) \)

This dot product gives the directional derivative, indicating how rapidly the function's value changes in the direction specified by the vector \( \mathbf{u} \).