Problem 5 - Advanced Engineering Mathematics

Question

Find the gradient of the given function at the indicated point. $$ f(x, y)=x^{2}-4 y^{2} ;(2,4) $$

Step-by-Step Solution

Answer

The gradient at (2, 4) is \([4, -32]\).

Step 1: Understand the Gradient

The gradient of a function \( f(x, y) \) is a vector of its partial derivatives. It is denoted as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This vector points in the direction of the greatest rate of increase of the function.

Step 2: Differentiate with Respect to x

Calculate the partial derivative of the function \( f(x, y) = x^2 - 4y^2 \) with respect to \( x \). \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 - 4y^2) = 2x \]

Step 3: Differentiate with Respect to y

Calculate the partial derivative of the function \( f(x, y) = x^2 - 4y^2 \) with respect to \( y \). \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 - 4y^2) = -8y \]

Step 4: Evaluate the Gradient at the Point (2, 4)

Substitute \( x = 2 \) and \( y = 4 \) into the partial derivatives to find the gradient at the given point. \[ abla f(2, 4) = (2 \times 2, -8 \times 4) = (4, -32) \]

Step 5: Express the Gradient as a Vector

Write the gradient as a vector. Thus, the gradient of the function at the point (2, 4) is \[ abla f(2, 4) = \begin{bmatrix} 4 \ -32 \end{bmatrix} \]

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding functions of multiple variables. When you have a function like \( f(x, y) = x^2 - 4y^2 \), you can think of \( f \) as a surface in a 3D space. A partial derivative measures how the function changes as you change just one of the variables, keeping the others constant.

• To find the partial derivative with respect to \( x \), you differentiate the function as if \( y \) were a constant. Hence, you find that \( \frac{\partial f}{\partial x} = 2x \).
• Similarly, for \( y \), treat \( x \) as constant, obtaining \( \frac{\partial f}{\partial y} = -8y \).

Partial derivatives help locate the gradient, which provides important information about the behavior of the function, such as its slope and direction of greatest increase.

Vector Calculus

Vector calculus is a branch that uses vector-valued functions and introduces the concept of vectors like the gradient. The gradient itself, \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \), is a vector derived from partial derivatives that points in the direction of the steepest ascent of the function.

• In our example, the gradient is \( abla f(2, 4) = (4, -32) \). This is a vector in the plane, showing how \( f(x, y) \) changes most rapidly at the point \( (2, 4) \).
• The components \( (4, -32) \) reveal the rate and direction of change along the \( x \) and \( y \) axes, respectively.

Understanding vectors in calculus allows us to describe changes in multidimensional surfaces efficiently and aids in optimization problems in engineering and physics.

Multivariable Calculus

Multivariable calculus extends single-variable calculus to functions with more than one variable, like \( f(x, y) \). In this context, the concept of the gradient is crucial because it provides a way to understand how functions change in multi-dimensional spaces.

• Using multivariable calculus, you're able to find maxima and minima for functions of several variables, helping in tasks ranging from data fitting to economic modeling.
• The gradient vector you computed, \( abla f(2, 4) = (4, -32) \), is an essential tool for these processes, indicating how to move in space to increase the function value most rapidly.

This approach is very powerful when dealing with real-world phenomena, where multiple factors affect outcomes simultaneously, making multivariable calculus an indispensable tool in various fields.