Problem 2 - Advanced Engineering Mathematics

Question

In Problems, compute the gradient for the given function. $$ f(x, y)=y-e^{-2 x^{2} y} $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f = \left( -4xye^{-2x^2 y}, 1 + 2x^2 e^{-2x^2 y} \right) \).

Step 1: Understand the Gradient

The gradient of a function is a vector of its partial derivatives. For the function \( f(x, y) = y - e^{-2x^2 y} \), we will find the partial derivative of \( f \) with respect to \( x \) and \( y \). The gradient is denoted by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Calculate \( \frac{\partial f}{\partial x} \)

To find the partial derivative of \( f \) with respect to \( x \), we differentiate the function treating \( y \) as a constant:\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( y - e^{-2x^2 y} \right) = 0 + \frac{\partial}{\partial x} \left( -e^{-2x^2 y} \right)\]Using the chain rule, we find:\[\frac{\partial}{\partial x} \left( -e^{-2x^2 y} \right) = e^{-2x^2 y} \cdot \frac{\partial}{\partial x} (-2x^2 y)\]\[= e^{-2x^2 y} \cdot (-4xy)\]Thus, \( \frac{\partial f}{\partial x} = -4xye^{-2x^2 y} \).

Step 3: Calculate \( \frac{\partial f}{\partial y} \)

To compute the partial derivative of \( f \) with respect to \( y \), differentiate the function treating \( x \) as a constant:\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( y - e^{-2x^2 y} \right) = 1 + \frac{\partial}{\partial y} \left( -e^{-2x^2 y} \right)\]Using the chain rule again:\[\frac{\partial}{\partial y} \left( -e^{-2x^2 y} \right) = e^{-2x^2 y} \cdot \frac{\partial}{\partial y} (-2x^2 y)\]\[= e^{-2x^2 y} \cdot (-2x^2)\]Thus, \( \frac{\partial f}{\partial y} = 1 + 2x^2 e^{-2x^2 y} \).

Step 4: Write the Gradient \( \nabla f \)

Now that we have both partial derivatives, the gradient is:\[abla f = \left( -4xye^{-2x^2 y}, 1 + 2x^2 e^{-2x^2 y} \right)\]This vector represents the direction and rate of the steepest ascent of the function at each point \((x, y)\).

Key Concepts

Partial Derivatives

Partial derivatives are fundamental concepts in calculus, especially when dealing with functions of several variables. Consider a function with more than one variable, like \( f(x, y) \). Instead of observing how the function changes overall, we may want to know how it changes with respect to just one of its variables at a time. This is where partial derivatives come in handy.

When calculating a partial derivative with respect to one variable, we treat all other variables as constants. For instance, in our function \( f(x, y) = y - e^{-2x^2 y} \), the partial derivative with respect to \( x \) \( \left( \frac{\partial f}{\partial x} \right) \) is calculated by treating \( y \) as a constant. Similarly, the partial derivative with respect to \( y \) \( \left( \frac{\partial f}{\partial y} \right) \) treats \( x \) as a fixed value.

This careful separation allows us to understand how each variable individually influences the function's behavior. In many applications, partial derivatives help model multi-dimensional phenomena, like weather predictions or economic forecasts.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function based on the derivatives of its component functions. In the context of partial derivatives, the chain rule helps us tackle expressions where one variable is naturally dependent on another.

Let's break it down further using our example function \( f(x, y) = y - e^{-2x^2 y} \). In steps to find \( \frac{\partial f}{\partial x} \), we encounter \( -e^{-2x^2 y} \), which is a composite function. To differentiate it with respect to \( x \), we need to acknowledge that \( -2x^2 y \) itself depends on \( x \).

• The direct differentiation of the external function, \( -e^{-2x^2 y} \), gives us \( e^{-2x^2 y} \).
• The chain rule requires that we also differentiate the exponent \( -2x^2 y \) with respect to \( x \), which results in \( -4xy \).

Multiply these results together for the final derivative: \( e^{-2x^2 y} (-4xy) \).

This approach simplifies dealing with complex functions, ensuring we capture all relational changes among variables. Understanding the chain rule is crucial for solving intricate problems in mathematics and applied sciences.

Vector Calculus

Vector calculus extends the principles of calculus to vector fields and is pivotal in fields such as physics and engineering. It provides the tools to analyze quantities that have both magnitude and direction.

The gradient is a central concept in vector calculus. It represents how a scalar function changes at each point in space, effectively pointing in the direction of the greatest rate of increase. For a function \( f(x, y) \), the gradient is a vector consisting of its partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

In our exercise, the calculated gradient \( abla f = \left( -4xye^{-2x^2 y}, 1 + 2x^2 e^{-2x^2 y} \right) \) demonstrates how the function \( f(x, y) \) increases most steeply at different locations \( (x, y) \).

• The first component \( -4xye^{-2x^2 y} \) indicates how steeply \( f \) changes concerning \( x \).
• The second component \( 1 + 2x^2 e^{-2x^2 y} \) highlights the rate of change with respect to \( y \).

Using gradients, scientists and engineers can model and optimize complex systems, such as analyzing electromagnetic fields or predicting fluid flows in pipelines. This makes vector calculus indispensable for addressing real-world challenges.