Problem 1 - Advanced Engineering Mathematics

Question

Compute the gradient for the given function. $$ f(x, y)=x^{2}-x^{3} y^{2}+y^{4} $$

Step-by-Step Solution

Answer

Gradient is \((2x - 3x^2y^2, -2x^3y + 4y^3)\).

Step 1: Understand the function

We have the function \( f(x, y) = x^2 - x^3y^2 + y^4 \). We need to compute its gradient, which involves calculating partial derivatives with respect to each variable \( x \) and \( y \).

Step 2: Compute the partial derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate \( f(x, y) \) with respect to \( x \) treating \( y \) as a constant:\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2 - x^3y^2 + y^4) = 2x - 3x^2y^2. \]

Step 3: Compute the partial derivative with respect to y

Now, find \( \frac{\partial f}{\partial y} \) by differentiating \( f(x, y) \) with respect to \( y \), treating \( x \) as a constant:\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^2 - x^3y^2 + y^4) = -2x^3y + 4y^3. \]

Step 4: Write the gradient vector

Combine the partial derivatives to write the gradient vector \( abla f(x, y) \):\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2x - 3x^2y^2, -2x^3y + 4y^3). \]

Key Concepts

Partial Derivatives

When dealing with functions of several variables, such as \( f(x, y) = x^2 - x^3y^2 + y^4 \), partial derivatives are used to analyze how the function changes with respect to one variable at a time. This is crucial for understanding the behavior of the function in a multi-dimensional context. To compute the partial derivative with respect to one variable, like \( x \), you treat all the other variables, here \( y \), as constants. For example, to find the partial derivative of \( f \) with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), you differentiate each term of the function involving \( x \) similarly as you would in single-variable calculus, while holding \( y \) constant.

• The term \( x^2 \) becomes \( 2x \).
• The term \( -x^3y^2 \) becomes \( -3x^2y^2 \) because \( y^2 \) is treated as a constant multiplier.
• The term \( y^4 \) vanishes because its derivative with respect to \( x \) is zero.

So, \( \frac{\partial f}{\partial x} = 2x - 3x^2y^2 \). Similarly, you find \( \frac{\partial f}{\partial y} \) by treating \( x \) as a constant.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable, allowing us to explore complex relationships among several variables. In this field, we handle functions like \( f(x, y) \) that depend on two or more variables.This branch of calculus involves:

• Understanding curves and surfaces in three-dimensional space.
• Analyzing rates of change in multiple directions simultaneously.
• Calculating partial derivatives to understand the gradient and use it to optimize the function or find tangent planes and normal lines to surfaces.

For example, when working with the function \( f(x, y) = x^2 - x^3y^2 + y^4 \), multivariable calculus allows us to compute the gradient vector. This vector gives us a powerful tool to study the function's behavior at different points.

Gradient Vector

The gradient vector is a fundamental concept in multivariable calculus, representing the vector of partial derivatives of a multivariable function. For a function \( f(x, y) \), the gradient vector, denoted by \( abla f(x, y) \), combines the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).In our example, we computed these partial derivatives as:- \( \frac{\partial f}{\partial x} = 2x - 3x^2y^2 \)- \( \frac{\partial f}{\partial y} = -2x^3y + 4y^3 \)These are components of the gradient vector:\[abla f(x, y) = \left( 2x - 3x^2y^2, -2x^3y + 4y^3 \right)\]
The gradient vector points in the direction of the greatest rate of increase of the function, making it useful in optimization problems. Its magnitude indicates how fast the function increases in that direction. Thus, knowing the gradient helps us understand and navigate the topography of the function's surface.