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: Gradient of the function

First, find the gradient of the function \( f(x, y) = e^{-x} \cos y \). The gradient is computed as the vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate these partial derivatives:- \( \frac{\partial f}{\partial x} = -e^{-x} \cos y \)- \( \frac{\partial f}{\partial y} = -e^{-x} \sin y \)Thus, \( abla f(x, y) = \left( -e^{-x} \cos y, -e^{-x} \sin y \right) \).

Step 2: Evaluate the Gradient at the Point

Next, evaluate the gradient at the point \((0, \pi/3)\). Substitute \( x = 0 \) and \( y = \pi/3 \) into \( abla f(x, y) \):\[ abla f(0, \pi/3) = \left( -e^{0} \cos(\pi/3), -e^{0} \sin(\pi/3) \right) \]Simplify using the known values: \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):\[ abla f(0, \pi/3) = \left( -1 \cdot \frac{1}{2}, -1 \cdot \frac{\sqrt{3}}{2} \right) = \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \]

Step 3: Determine the Unit Vector in the Direction of Interest

To find the unit vector in the direction from \((0, \pi/3)\) to the origin \((0, 0)\), calculate the vector pointing toward the origin:\[ \mathbf{v} = (0 - 0, 0 - \pi/3) = (0, -\pi/3) \]Now, convert \( \mathbf{v} \) into a unit vector by dividing by its magnitude:Magnitude of \( \mathbf{v} \) is \( \left\|(0, -\pi/3)\right\| = \frac{\pi}{3} \).The unit vector \( \mathbf{u} \) is:\[ \mathbf{u} = \left( 0, -1 \right) \]

Step 4: Compute the Directional Derivative

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

Step 5: Conclusion

The directional derivative of the function \( f(x, y) = e^{-x} \cos y \) at the point \((0, \pi/3)\) in the direction toward the origin is \( \frac{\sqrt{3}}{2} \).

Key Concepts

Gradient

The gradient of a function provides critical information about the rate and direction of the greatest increase. For a function of two variables, like the one given as \( f(x, y) = e^{-x} \cos y \), the gradient is a vector composed of the partial derivatives with respect to each variable. This vector is often denoted as \( abla f(x, y) \) and represented as \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

The gradient points in the direction of the steepest ascent of the function. This makes it extremely useful when determining the directional derivative, which indicates how much the function increases as you move in a particular direction.
In the given example, the calculated gradient vector \( abla f(x, y) \) at the point \((0, \pi/3)\) is \( \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \). This specific vector provides the direction and magnitude of the steepest climb from that point.

Partial Derivatives

Partial derivatives are fundamental when working with functions of multiple variables. They represent the rate of change of the function as one variable changes while keeping others constant. In our function, \( f(x, y) \), we compute two partial derivatives: one with respect to \( x \) and one with respect to \( y \).

The calculation:

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

These expressions help form the gradient vector \( abla f(x, y) \). Each component of the gradient gives us insight into how sensitive \( f \) is to changes in \( x \) and \( y \).
Understanding partial derivatives is crucial as they help us comprehend how a function behaves in a multidimensional space.

Unit Vector

A unit vector defines direction while having a magnitude of one. When calculating a directional derivative, the unit vector tells you which direction you're moving in and ensures that the directional derivative's magnitude reflects only the rate of change in that direction.

To find the unit vector from \((0, \pi/3)\) towards the origin, we first identify the direction vector, \( \mathbf{v} = (0, -\pi/3) \). Then, we compute its magnitude: \( \left\|(0, -\pi/3)\right\| = \frac{\pi}{3} \).
Afterwards, we scale the direction vector by its magnitude to get the unit vector: \( \mathbf{u} = \left( 0, -1 \right) \).
It's important to use a unit vector so the directional derivative indicates the pure rate of change, un-influenced by varying lengths of the directional vectors.

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a scalar. This operation reveals how much of one vector goes in the direction of another. For directional derivatives, the dot product is used to project the gradient vector onto the unit direction vector.

In our example, the dot product \( abla f(0, \pi/3) \cdot \mathbf{u} \) is calculated as follows:

• Multiply corresponding components: \(-\frac{1}{2} \cdot 0\) and \(-\frac{\sqrt{3}}{2} \cdot (-1)\)
• Sum the products: \(0 + \frac{\sqrt{3}}{2}\)

Thus, the result is \( \frac{\sqrt{3}}{2} \), which is the directional derivative.
This scalar encodes how quickly the function \( f(x, y) \) increases or decreases in the direction defined by the unit vector, when moving from the given point on the function.