Problem 23 - Advanced Engineering Mathematics

Question

Find a vector that gives the direction in which the given function increases most rapidly at the indicated point. Find the maximum rate. $$ f(x, y)=e^{2 x} \sin y ;(0, \pi / 4) $$

Step-by-Step Solution

Answer

The gradient at \((0, \pi/4)\) is \( (\sqrt{2}, \frac{\sqrt{2}}{2}) \), with a maximum rate of \( \sqrt{\frac{5}{2}} \)."

Step 1: Compute the Gradient Vector

The gradient of a function \( f(x, y) \) is given by the vector of its partial derivatives: \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right). \] First, we find the partial derivatives for \( f(x, y) = e^{2x} \sin y \). - \( \frac{\partial f}{\partial x} = 2e^{2x} \sin y \). - \( \frac{\partial f}{\partial y} = e^{2x} \cos y \). Thus, the gradient vector is \[ abla f(x, y) = \left( 2e^{2x} \sin y, e^{2x} \cos y \right). \]

Step 2: Evaluate the Gradient at the Given Point

We need to evaluate the gradient vector at the point \((0, \pi/4)\). Substitute \(x = 0\) and \(y = \pi/4\) into the gradient: \[ \frac{\partial f}{\partial x} = 2e^{0} \sin(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}, \]\[ \frac{\partial f}{\partial y} = e^{0} \cos(\pi/4) = \frac{\sqrt{2}}{2}. \] Thus, \( abla f(0, \pi/4) = \left( \sqrt{2}, \frac{\sqrt{2}}{2} \right) \).

Step 3: Find the Maximum Rate of Increase

The maximum rate of increase is the magnitude of the gradient vector at the point \((0, \pi/4)\). Calculate the magnitude: \[ \left| abla f(0, \pi/4) \right| = \sqrt{ (\sqrt{2})^2 + \left( \frac{\sqrt{2}}{2} \right)^2 } = \sqrt{2 + \frac{1}{2}} = \sqrt{\frac{5}{2}}. \] This value represents the maximum rate of increase of the function at the given point.

Key Concepts

Partial Derivatives

When we're dealing with functions of multiple variables, like our function \( f(x, y) = e^{2x} \sin y \), it's important to understand how the function changes with respect to each variable. This is where partial derivatives come into play.
A partial derivative measures the rate of change of the function as you change one variable, keeping the others constant. In our case, we have two variables: \( x \) and \( y \).
Here's how we find the partial derivatives for our function:

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), involves differentiating \( f \) as if \( y \) is a constant. For \( f(x, y) = e^{2x} \sin y \), it results in \( 2e^{2x} \sin y \).
• The partial derivative with respect to \( y \), denoted as \( \frac{\partial f}{\partial y} \), is found by differentiating \( f \) as if \( x \) is constant. Here, it comes out to be \( e^{2x} \cos y \).

Understanding partial derivatives is crucial in vector calculus, as they help us form the gradient vector, which indicates how our function changes at any given point.

Maximum Rate of Increase

Once we have the gradient vector, we can easily determine the direction in which our function increases the fastest at a specific point. The gradient vector not only helps us find the direction, but its magnitude also tells us the maximum rate of change.
In simpler terms, the magnitude of the gradient vector at a particular point gives the steepest slope of the function's surface at that point.
Here's how we find the maximum rate of increase:

• First, calculate the gradient vector using the partial derivatives and substitute in our specific point, which is \((0, \pi/4)\) for the exercise.
• Next, compute the magnitude of this vector. For our function, the gradient vector at \((0, \pi/4)\) is \( (\sqrt{2}, \frac{\sqrt{2}}{2}) \).
• Finally, use the formula for the magnitude of a vector \((a, b)\): \( \sqrt{a^2 + b^2} \). This will give us the maximum rate of increase, which in this case is \( \sqrt{\frac{5}{2}} \).

Understanding this concept is important for tackling problems in fields like physics and engineering, where understanding rapid changes is crucial.

Vector Calculus

Vector calculus is a field of mathematics that deals with vector fields and the different operations that can be performed on vectors. In our problem, vector calculus plays a crucial role, especially in finding the direction and rate of change for our function.
One key operation in vector calculus is finding the gradient vector. For a function of two variables \( f(x, y) \), the gradient is a vector of its partial derivatives \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
The gradient has some interesting properties:

• It points in the direction of maximum rate of increase of the function.
• The magnitude of the gradient vector provides the maximum rate of increase at a given point.
• The gradient is normal to level curves (surfaces of constant value) in a multi-variable function.

By learning vector calculus, we develop tools to navigate and manipulate fields that are essential in modeling and understanding physical phenomena. The gradient, combined with the concepts of divergence and curl, enables us to solve more complex problems in three-dimensional space.