Problem 269 - Calculus 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: Understand the Directional Derivative Formula

The directional derivative of a function \( f(x, y) \) at a point \( P \) in the direction of a vector \( \mathbf{v} \) is given by the dot product of the gradient of \( f \) at \( P \) and the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \). The formula is: \[ D_{\mathbf{u}} f = abla f(x, y) \cdot \mathbf{u} \] where \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Calculate the Gradient of the Function

Calculate the partial derivatives of \( f(x, y) = e^x \cos y \). \[ \frac{\partial f}{\partial x} = e^x \cos y \]\[ \frac{\partial f}{\partial y} = -e^x \sin y \] Thus, the gradient is \( abla f(x, y) = \langle e^x \cos y, -e^x \sin y \rangle \).

Step 3: Evaluate the Gradient at Point \( P \)

Substitute point \( P = \left(0, \frac{\pi}{2}\right) \) into the gradient.\[ \frac{\partial f}{\partial x}\mid_{(0, \pi/2)} = e^0 \cos \left(\frac{\pi}{2}\right) = 1 \times 0 = 0 \]\[ \frac{\partial f}{\partial y}\mid_{(0, \pi/2)} = -e^0 \sin \left(\frac{\pi}{2}\right) = -1 \]Thus, \( abla f\mid_{(0, \pi/2)} = \langle 0, -1 \rangle \).

Step 4: Ensure \( \mathbf{u} \) is a Unit Vector

Verify that \( \mathbf{u} = \langle 0, 1 \rangle \) is a unit vector. The magnitude of \( \mathbf{u} \) is \[ \|\mathbf{u}\| = \sqrt{0^2 + 1^2} = 1 \]. Since its magnitude is 1, \( \mathbf{u} \) is indeed a unit vector.

Step 5: Compute the Dot Product for the Directional Derivative

Calculate the dot product of the gradient at \( P \) and the unit vector \( \mathbf{u} \).\[ D_{\mathbf{u}} f = abla f\mid_{(0, \pi/2)} \cdot \mathbf{u} = \langle 0, -1 \rangle \cdot \langle 0, 1 \rangle = 0 \times 0 + (-1) \times 1 = -1 \]

Step 6: Conclusion

The directional derivative of the function \( f(x, y) = e^x \cos y \) at the point \( \left(0, \frac{\pi}{2}\right) \) in the direction of \( \mathbf{u} = \langle 0, 1 \rangle \) is \(-1\).

Key Concepts

Gradient

The gradient is a fundamental concept in multivariable calculus. It provides the direction and rate of the fastest increase of a function. Imagine a hill: the steepest path upward is akin to the gradient in terrain. The gradient of a function, denoted as \( abla f(x, y) \), is a vector comprising all its partial derivatives. For a function \( f(x, y) \), the gradient can be expressed as:

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

This vector points towards the direction of steepest ascent. In our exercise, the function \( f(x, y) = e^x \cos y \) has a gradient \( abla f(x, y) = \langle e^x \cos y, -e^x \sin y \rangle \). Evaluating this at the point \( P = \left(0, \frac{\pi}{2}\right) \) gives \( \langle 0, -1 \rangle \), representing the rate of change at that specific point.

Partial Derivatives

Partial derivatives break down functions of several variables. They measure how a function changes as each variable changes, keeping others constant. Consider \( f(x, y) = e^x \cos y \). To find its partial derivatives, follow these steps:

• With respect to \( x \): Treat \( y \) as a constant. Compute: \( \frac{\partial f}{\partial x} = e^x \cos y \).
• With respect to \( y \): Here, treat \( x \) as constant. Find \( \frac{\partial f}{\partial y} = -e^x \sin y \).

These results form the gradient vector: \( \langle e^x \cos y, -e^x \sin y \rangle \). Partial derivatives are crucial for understanding how a function behaves in multi-dimensional spaces.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. This number provides insights into how much one vector goes in the direction of another.

• Mathematically, for vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is computed as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).

In our context, to find the directional derivative, we calculate the dot product of the gradient at point \( P \) and the unit direction vector \( \mathbf{u} \). With \( abla f \mid_{(0, \pi/2)} = \langle 0, -1 \rangle \) and \( \mathbf{u} = \langle 0, 1 \rangle \), the dot product becomes \( 0 \times 0 + (-1) \times 1 = -1 \). This shows the rate of change of the function in the specified direction.

Unit Vector

A unit vector serves as a fundamental element in vector mathematics and physics. It has a magnitude of exactly one and indicates direction only.

• To ascertain if a vector is a unit vector, calculate its magnitude using the formula: \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \).

For the vector \( \mathbf{u} = \langle 0, 1 \rangle \), its magnitude is \( \sqrt{0^2 + 1^2} = 1 \), confirming it is indeed a unit vector. When calculating the directional derivative, a unit vector such as \( \mathbf{u} \) ensures the directional measurement is unbiased by the vector's length, focusing purely on the direction.