Problem 276 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Components

The directional derivative of a function, denoted as \( D_{\mathbf{u}} f(x, y) \), represents how \( f(x, y) \) changes as you move in the direction of a unit vector \( \mathbf{u} \). We are given the function \( f(x, y)=\cos(3x+y) \) and the direction vector given by the angle \( \theta = \frac{\pi}{4} \). The unit vector \( \mathbf{u} \) is \( \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \).

Step 2: Calculate the Gradient Vector

Calculate the gradient of \( f(x, y) \), denoted as \( abla f(x, y) \). It is obtained by taking the partial derivatives of \( f \) with respect to \( x \) and \( y \). So, \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 3: Partial Derivative with Respect to x

Find \( \frac{\partial f}{\partial x} \). Since \( f(x, y) = \cos(3x+y) \), apply the chain rule: \( \frac{\partial f}{\partial x} = -3\sin(3x+y) \).

Step 4: Partial Derivative with Respect to y

Find \( \frac{\partial f}{\partial y} \). Again, apply the chain rule to get \( \frac{\partial f}{\partial y} = -\sin(3x+y) \).

Step 5: Form the Gradient Vector

The gradient vector \( abla f \) is \( (-3\sin(3x+y), -\sin(3x+y)) \).

Step 6: Evaluate the Unit Vector

The unit vector \( \mathbf{u} \) in terms of the angle \( \theta = \frac{\pi}{4} \) is \( \left( \cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right) \right) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 7: Compute the Dot Product

Calculate the dot product of \( abla f(x, y) \) and \( \mathbf{u} \): \[ D_{\mathbf{u}} f(x, y) = abla f \cdot \mathbf{u} = (-3\sin(3x+y), -\sin(3x+y)) \cdot \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \] \[ = -3\sin(3x+y) \times \frac{\sqrt{2}}{2} + (-\sin(3x+y)) \times \frac{\sqrt{2}}{2} \] \[ = -2\sqrt{2}\sin(3x+y) \].

Step 8: Finalize the Directional Derivative

The directional derivative of the function \( f(x, y) \) in the direction of \( \mathbf{u} \) when \( \theta = \frac{\pi}{4} \) is \( -2\sqrt{2}\sin(3x+y) \).

Key Concepts

Gradient Vector

The gradient vector is a crucial concept in multivariable calculus. It's like a compass pointing in the direction of the steepest ascent of a function. For functions of two variables, the gradient vector is expressed as \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).This vector gives us a multi-dimensional rate of change, showing us how much the function increases with small changes in both the \(x\) and \(y\) directions.

• The first component is the partial derivative with respect to \(x\), showing how sensitive the function is to \(x\).
• The second component is the partial derivative with respect to \(y\), indicating how sensitive it is to changes in \(y\).

Think of the gradient vector as the most efficient push to increase the output of our function, guided by the terrain shaped by inputs \(x\) and \(y\). Understanding the gradient helps solve optimization problems and analyze level curves of functions.

Partial Derivatives

Partial derivatives offer insight into how a multivariable function changes when altering one independent variable, keeping the others constant. In essence, they isolate the sensitivity of the function to a single variable.
To find the partial derivative of a function, say \(f(x, y)\) with respect to \(x\), the notation \(\frac{\partial f}{\partial x}\) is used.

• For our given function \(f(x, y) = \cos(3x + y)\), the partial derivative with respect to \(x\) was found using the chain rule, yielding \(-3\sin(3x+y)\).
• Similarly, the partial derivative with respect to \(y\) is \(-\sin(3x+y)\).

Partial derivatives are foundational in building gradient vectors, as they contribute each component of the vector. Understanding them separately allows us to determine how each variable independently influences a function's value.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It enables us to tackle complex functions by breaking them down into compositions of simpler functions. This rule states that if a variable \(z\) depends on a variable \(u\), which in turn depends on another variable \(x\), then the rate of change of \(z\) with respect to \(x\) can be found by multiplying the derivatives:\[\frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} \].
In the context of multivariable functions like \(f(x, y) = \cos(3x+y)\), the chain rule is essential:

• We used the chain rule to find partial derivatives for both \(x\) and \(y\).
• For \(\frac{\partial f}{\partial x}\), this involved differentiating the inner function \((3x+y)\) and multiplying by the derivative of the outer function, \(-\sin\).

By mastering the chain rule, one gains the ability to unravel more intricate functions, providing a powerful tool in both pure and applied mathematics for solving real-world problems.