Problem 269 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=e^{x} \cos y, \quad \mathbf{u}=\langle 0,1\rangle, \quad P=\left(0, \frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The directional derivative is -1.

Step 1: Calculate the Gradient

The directional derivative involves taking the dot product between the gradient of the function, \( abla f(x, y) \), and the unit vector \( \mathbf{u} \). First, calculate \( abla f(x, y) \), which consists of the partial derivatives with respect to \( x \) and \( y \). For \( f(x, y) = e^x \cos y \), we have:\[ f_x(x, y) = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y \]\[ f_y(x, y) = \frac{\partial}{\partial y}(e^x \cos y) = -e^x \sin y \]\.

Step 2: Evaluate the Gradient at Point P

Substitute the coordinates of the point \( P = \left(0, \frac{\pi}{2}\right) \) into the gradient. We have:\[ abla f(0, \frac{\pi}{2}) = \left( e^0 \cos \frac{\pi}{2}, -e^0 \sin \frac{\pi}{2} \right) = (\cos \frac{\pi}{2}, -\sin \frac{\pi}{2}) \].Since \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), it simplifies to:\[ abla f(0, \frac{\pi}{2}) = (0, -1) \].

Step 3: Verify the Direction Vector is a Unit Vector

Check if the given direction vector \( \mathbf{u} = \langle 0, 1 \rangle \) is a unit vector. Calculate its magnitude:\[ \| \mathbf{u} \| = \sqrt{0^2 + 1^2} = 1 \].Since its magnitude is 1, \( \mathbf{u} \) is indeed a unit vector.

Step 4: Compute the Directional Derivative

The directional derivative \( D_{\mathbf{u}}f \) in the direction of \( \mathbf{u} \) is given by the dot product of \( abla f(0, \frac{\pi}{2}) \) and \( \mathbf{u} \):\[ D_{\mathbf{u}}f = abla f(0, \frac{\pi}{2}) \cdot \mathbf{u} = (0, -1) \cdot (0, 1) = 0 \times 0 + (-1) \times 1 = -1 \].The directional derivative at point \( P \) in the direction of \( \mathbf{u} \) is \( -1 \).

Key Concepts

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. It's like a compass showing you the steepest uphill direction from any given point in a multivariable function. The gradient is denoted by \(abla f\) and comprises the partial derivatives of the function with respect to each of its variables. In the function \(f(x, y) = e^x \cos y\), the gradient can be calculated by finding the partial derivatives:

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

Thus, the gradient is \(abla f(x, y) = (e^x \cos y, -e^x \sin y)\). At any point, substituting the coordinates into this gives the gradient vector at that point.

Partial Derivatives

Partial derivatives are the derivatives of functions with multiple variables taken with respect to one variable, holding the others constant. They tell us how the function changes as we tweak one variable at a time. They are crucial for understanding how multi-variable functions behave, and are the building blocks of the gradient.For the function \(f(x, y) = e^x \cos y\), the partial derivative with respect to \(x\), denoted \(f_x\), calculates how \(f\) changes as \(x\) changes while \(y\) is held constant. Similarly, \(f_y\) shows the change in \(f\) with respect to \(y\), keeping \(x\) constant. Understanding these concepts is essential for calculating the gradient and thus the directional derivative.

Unit Vector

A unit vector is a vector with a magnitude of 1, pointing in a specific direction. It is essential in directional derivatives because it provides the direction for which we want the rate of change of a function to be calculated. To check if a vector is a unit vector, calculate its magnitude. If the magnitude is 1, it indeed is a unit vector.In this context, the vector \(\mathbf{u} = \langle 0, 1 \rangle\) was verified as a unit vector. Its magnitude is \(\sqrt{0^2 + 1^2} = 1\), confirming that \(\mathbf{u}\) maintains a constant distance from the origin, making it suitable for use in finding the directional derivative.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This number is the product of the two lengths of vectors and the cosine of the angle between them. For directional derivatives, the dot product helps combine the direction of the gradient and the unit vector to give the rate of change in the desired direction.In our exercise, the dot product of the gradient \(abla f(0, \frac{\pi}{2}) = (0, -1)\) and the unit vector \(\mathbf{u} = \langle 0, 1 \rangle\) is calculated as \(0 \times 0 + (-1) \times 1 = -1\). This result, \( -1\), is the directional derivative, which tells us that at point \(P\) in the direction of \(\mathbf{u}\), the function \(f\) decreases.