Problem 5 - Calculus Early Transcendentals

Question

\(4-6\) Find the directional derivative of \(f\) at the given point in the direction indicated by the angle \(\theta\) . $$f(x, y)=y e^{-x}, \quad(0,4), \quad \theta=2 \pi / 3$$

Step-by-Step Solution

Answer

The directional derivative is approximately 2.866.

Step 1: Find the Gradient of the Function

The first step in finding the directional derivative is to compute the gradient of the function, \( f(x, y) = y e^{-x} \). The gradient is a vector of the partial derivatives of the function. Thus, we find:\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right). \]

Step 2: Compute the Partial Derivatives

For the function \( f(x, y) = y e^{-x} \), let's find the partial derivatives:1. \( \frac{\partial f}{\partial x} = -y e^{-x} \).2. \( \frac{\partial f}{\partial y} = e^{-x} \).So the gradient is:\[ abla f(x, y) = (-y e^{-x}, e^{-x}). \]

Step 3: Evaluate the Gradient at the Given Point

Now, evaluate the gradient at the point \((0, 4)\).\[ abla f(0, 4) = (-4e^{0}, e^{0}) = (-4, 1). \]

Step 4: Determine the Unit Vector in Direction of \(\theta\)

The direction given is by angle \(\theta = \frac{2\pi}{3}\). Find the unit vector \( \mathbf{u} \) pointing in this direction:\[ \mathbf{u} = (\cos \theta, \sin \theta) = \left(\cos \frac{2\pi}{3}, \sin \frac{2\pi}{3}\right). \]Calculating, \( \cos \frac{2\pi}{3} = -\frac{1}{2} \) and \( \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2} \), so:\[ \mathbf{u} = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right). \]

Step 5: Calculate the Directional Derivative

Finally, compute the directional derivative \( D_{\mathbf{u}}f(0,4) \), which is the dot product of the gradient and the unit vector \( \mathbf{u} \):\[ D_{\mathbf{u}}f(0, 4) = abla f(0, 4) \cdot \mathbf{u} = (-4, 1) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right). \]The dot product is:\[ = (-4)(-\frac{1}{2}) + (1)(\frac{\sqrt{3}}{2}) = 2 + \frac{\sqrt{3}}{2} = 2 + 0.866. \]

Step 6: Simplify the Result

Compute the final result:\[ D_{\mathbf{u}}f(0, 4) = 2 + \frac{\sqrt{3}}{2} \approx 2.866. \]

Key Concepts

Gradient Vector

In the realm of multivariable calculus, the gradient vector is an essential tool that helps us understand the behavior of functions. When we talk about the gradient of a scalar function, we're essentially referring to a vector composed of the function's partial derivatives with respect to each variable.

For a function of two variables, such as \( f(x, y) = y e^{-x} \), the gradient vector is expressed as:

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

The gradient vector points in the direction of the greatest rate of increase of the function. Its components highlight how changes in each variable impact the function's value.

In practice, calculating the gradient involves finding the partial derivatives. Once computed, the vector can be evaluated at a specific point to give us vital directional information. Thus, the gradient is indispensable for understanding the intricate tapestry of change a function undergoes at any given point.

Partial Derivatives

Partial derivatives lie at the heart of multivariable calculus, offering a lens through which to view the rate of change of a function with respect to one of its variables while keeping other variables constant. For the function \( f(x, y) = y e^{-x} \), we need to compute partial derivatives with respect to \( x \) and \( y \).

• \( \frac{\partial f}{\partial x} = -y e^{-x} \)
• \( \frac{\partial f}{\partial y} = e^{-x} \)

By isolating each variable during differentiation, partial derivatives provide insights into how changes in a single aspect of our inputs affect the whole function. These derivatives are the building blocks of the gradient vector and are crucial for calculating directional derivatives and other advanced calculus concepts.

When evaluated, they offer concrete numbers that can be used to assess how the function shifts, molds, and reshapes as each variable in the equation changes.

Unit Vector

Unit vectors are integral in vector mathematics, especially when dealing with directionality. They provide a way to focus solely on direction without the influence of magnitude. A unit vector has a length of 1, making it ideal for expressing directions in space.

To determine a unit vector in any direction given by an angle \( \theta \), we use trigonometric functions:

• \( \mathbf{u} = (\cos \theta, \sin \theta) \)

In our problem, the angle \( \theta = \frac{2\pi}{3} \), translates to the unit vector:

• \( \mathbf{u} = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)

This ability to describe direction using unit vectors allows us to calculate the directional derivative by focusing on where we want to measure the rate of change, without magnifying the result arbitrarily.

In any vector calculation involving unit vectors, the outcome's integrity remains purely about direction, simplifying and focusing subsequent computations.

Dot Product

The dot product is a fundamental operation on two vectors, giving us a scalar that represents the product of their magnitudes and the cosine of the angle between them. This operation finds frequent use in various applications, particularly in calculating the directional derivative.

• The formula for the dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the context of our exercise, the dot product is used to compute the directional derivative:

• \( ( abla f(0,4) \cdot \mathbf{u} = (-4, 1) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
• This calculation yields a value that quantifies the rate of change of the function in that specific direction.

The dot product simplifies how we measure and interpret changes in functions, encapsulating the relation between changes in variables directly influenced by the directional vector. It is this product that ultimately offers insights into how the function evolves along a given path.