Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the directional derivative of \(f(x, y)=e^{-x} \cos y\) at \((0, \pi / 3)\) in the direction toward the origin.

Step-by-Step Solution

Answer

The directional derivative is \( \frac{\sqrt{3}}{2} \).

Step 1: Find the Gradient of f

The gradient of a function \( f(x, y) \) is given by \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = e^{-x} \cos y \), compute the partial derivatives:1. \( \frac{\partial f}{\partial x} = -e^{-x} \cos y \)2. \( \frac{\partial f}{\partial y} = -e^{-x} \sin y \)Therefore, the gradient is \( abla f(x, y) = (-e^{-x} \cos y, -e^{-x} \sin y) \).

Step 2: Evaluate the Gradient at the Given Point

Substitute \((x, y) = (0, \pi/3)\) into the gradient:- \( \frac{\partial f}{\partial x} \) evaluated at \((0, \pi/3)\): \[ -e^{-0} \cos \left(\frac{\pi}{3}\right) = -1 \cdot \frac{1}{2} = -\frac{1}{2}. \]- \( \frac{\partial f}{\partial y} \) evaluated at \((0, \pi/3)\): \[ -e^{-0} \sin \left(\frac{\pi}{3}\right) = -1 \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}. \]So, \( abla f(0, \pi/3) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \).

Step 3: Identify the Direction Vector

The direction is from \((0, \pi/3)\) to the origin \((0, 0)\). The direction vector is \((0 - 0, 0 - \pi/3) = (0, -\pi/3)\).

Step 4: Normalize the Direction Vector

To find the unit vector in the direction of \((0, -\pi/3)\), calculate its magnitude:\[ \|\mathbf{v}\| = \sqrt{0^2 + \left(-\frac{\pi}{3}\right)^2} = \frac{\pi}{3}. \]The unit vector \( \mathbf{u} \) is: \[ \mathbf{u} = \left(0, -\frac{\pi}{3} \right) \times \frac{1}{\frac{\pi}{3}} = (0, -1). \]

Step 5: Calculate the Directional Derivative

The directional derivative of \( f \) at \((0, \pi/3)\) in the direction of \( \mathbf{u} \) is given by\[ D_{\mathbf{u}} f = abla f(0, \pi/3) \cdot \mathbf{u} = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \cdot (0, -1). \]Calculate the dot product:\[ D_{\mathbf{u}} f = \left(-\frac{1}{2} \cdot 0\right) + \left(-\frac{\sqrt{3}}{2} \cdot (-1)\right) = 0 + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}. \]

Key Concepts

Gradient

The gradient of a function is a crucial concept when considering the directional derivative. It essentially represents the direction and rate of the fastest increase of the function value.
Think of the gradient as a vector that points in the direction of the steepest ascent of the function's surface.

• For a function of two variables, say \( f(x, y) \), the gradient is a two-dimensional vector.
• The components of this vector are the partial derivatives of the function: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

The gradient tells us how much the function \( f \) changes for a small change in directions along the axes of the input variables.
For the exercise at hand, the function given is \( f(x, y) = e^{-x} \cos y \), and after computing the partial derivatives, we find the gradient to be \( abla f(x, y) = (-e^{-x} \cos y, -e^{-x} \sin y) \).
This plays an important role as it gets evaluated at the specific point \((0, \pi/3)\), providing the basis for calculating the directional derivative.

Partial Derivatives

Partial derivatives are a fundamental concept when studying multivariable calculus, specifically in the context of functions with several variables.
They measure how the function changes as only one of the input variables is varied, keeping the others constant.

• For a function \( f(x, y) \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \).
• Similarly, the partial derivative with respect to \( y \) is noted as \( \frac{\partial f}{\partial y} \).

These derivatives give us slope information along the respective axes of the input space.
In the given problem, we found:

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

These components form the backbone of the gradient, which is essential to finding the directional derivative.

Unit Vector

A unit vector is an essential tool when determining directional derivatives. It gives the direction of a vector and has a magnitude of one.
To find the unit vector from a given vector, you divide each component of the vector by the magnitude of the vector.

• For a vector \( \mathbf{v} = (a, b) \), the magnitude is \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \).
• The unit vector \( \mathbf{u} \) is then \( \left( \frac{a}{\|\mathbf{v}\|}, \frac{b}{\|\mathbf{v}\|} \right) \).

In this exercise, the direction vector from the point \((0, \pi/3)\) to the origin is \((0, -\pi/3)\). By dividing this vector by its magnitude, \( \frac{\pi}{3} \), we get the unit vector \((0, -1)\).
This ensures that we use a direction with consistent scaling for calculating the directional derivative.

Dot Product

The dot product is a key operation when calculating the directional derivative. It takes two vectors and returns a single scalar value representing the product of their magnitudes and the cosine of the angle between them.

• For vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \).

The result of the dot product gives us a measure of how much one vector aligns with another.
In the context of the directional derivative, the dot product of the gradient and the unit vector yields the derivative along the specified direction.
In the problem solution, we perform the dot product of \( abla f(0, \pi/3) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \) and the unit vector \( \mathbf{u} = (0, -1) \).
The calculation \( 0 \cdot -\frac{1}{2} + -1 \cdot -\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \) provides the directional derivative, indicating the rate of function change toward the origin.