Problem 9 - Advanced Engineering Mathematics

Question

In Problems, use Definition \(9.5 .1\) to find \(D_{\mathrm{a}} f(x, y)\) given that \(\mathbf{u}\) makes the indicated angle with the positive \(x\) -axis. $$ f(x, y)=x^{2}+y^{2} ; \theta=30^{\circ} $$

Step-by-Step Solution

Answer

The directional derivative \( D_{\mathbf{a}} f(x, y) = x\sqrt{3} + y \).

Step 1: Vector \\ \( \mathbf{u} \) Definition

The vector \( \mathbf{u} \) represents the direction in which the derivative is taken. Given an angle \( \theta = 30^\circ \) with the positive x-axis, the vector in terms of its components is \( \mathbf{u} = (\cos \theta, \sin \theta) \). Substituting the value of \( \theta \), we get \( \mathbf{u} = \left( \cos 30^\circ, \sin 30^\circ \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

Step 2: Compute Partial Derivatives

Find the partial derivatives of the function \( f(x, y) = x^2 + y^2 \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x \), and with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y \).

Step 3: Apply the Directional Derivative Formula

The directional derivative \( D_{\mathbf{a}}f(x, y) \) is given by the dot product of the gradient of \( f \) and the unit vector \( \mathbf{u} \). The gradient \( abla f = (2x, 2y) \). Thus, \[D_{\mathbf{a}} f(x, y) = abla f \cdot \mathbf{u} = (2x, 2y) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)\].

Step 4: Compute the Dot Product

Calculate \( D_{\mathbf{a}} f(x, y) = 2x \times \frac{\sqrt{3}}{2} + 2y \times \frac{1}{2} \). Simplifying gives you: \( x\sqrt{3} + y \).

Step 5: Write the Final Result

The directional derivative \( D_{\mathbf{a}} f(x, y) \) in the direction of the unit vector given by the angle \( 30^\circ \) with the positive x-axis for the function \( f(x, y) = x^2 + y^2 \) is \( x\sqrt{3} + y \).

Key Concepts

Partial Derivatives

Partial derivatives are used to find the rate at which a function changes with respect to one variable while keeping other variables constant. When dealing with functions of multiple variables, such as \( f(x, y) = x^2 + y^2 \), partial derivatives allow us to explore how changes in one input, like \( x \), influence the output while \( y \) remains unchanged.

For the function given, the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \). It involves differentiating only the \( x \)-related part of the function, treating other variables as constants. Similarly, \( \frac{\partial f}{\partial y} \) is calculated by changing only \( y \), treating \( x \) as constant.

So, in this problem:

• \( \frac{\partial f}{\partial x} = 2x \)
• \( \frac{\partial f}{\partial y} = 2y \)

By finding these derivatives, you unravel how each individual component of the function affects the overall behavior of \( f \). This foundational concept is critical for applying directional derivatives.

Gradient Vector

The gradient vector is a powerful tool in calculus, representing the multi-variable derivative of a function. For a function like \( f(x, y) \), the gradient is a vector composed of all its partial derivatives.

The gradient vector \( abla f \) of the function \( f(x, y) = x^2 + y^2 \) includes:

• The partial derivative with respect to \( x \), \( 2x \)
• The partial derivative with respect to \( y \), \( 2y \)

The gradient is expressed as \( abla f = (2x, 2y) \).
This vector points in the direction of the steepest increase of the function. The magnitude or length of this vector indicates the rate of that increase.

By understanding the gradient, you see how changes in \( x \) and \( y \) interact to affect the outcome of the function \( f \), guiding you in navigating the directional derivative.

Dot Product

The dot product is a central concept in vector mathematics, often used to compute directional derivatives. It is essentially a way to multiply two vectors together to produce a single number.

For two vectors, say \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is defined as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the context of this exercise, the dot product is employed to calculate the directional derivative \( D_{\mathbf{a}} f(x, y) \) by combining the gradient vector \( abla f = (2x, 2y) \) with the direction vector \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

This involves:

• Multiplying corresponding components: \( 2x \times \frac{\sqrt{3}}{2} \) and \( 2y \times \frac{1}{2} \)
• Adding them together to yield: \( x\sqrt{3} + y \)

The result of the dot product gives you a scalar that tells you how fast the function \( f \) changes in the direction of \( \mathbf{u} \).

Trigonometric Functions

Trigonometric functions play a crucial role in describing angles and oscillations. These functions, like sine and cosine, are essential in converting angular directions into vector components, as they relate angles with the unit circle.

In this problem, you are given an angle \( \theta = 30^\circ \) to define a direction. Using trigonometric functions, you find the corresponding unit vector \( \mathbf{u} \):

• \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
• \( \sin 30^\circ = \frac{1}{2} \)

Combining these values, the direction vector becomes \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

By leveraging trigonometric functions, which are deeply rooted in circular relationships, you effectively translate the given angle into precise vector components. This conversion allows you to apply the directional derivative in the specified direction.