Problem 12 - APEX Calculus

Question

A function \(z=f(x, y)\) is given. Find \(\nabla f\). $$ f(x, y)=x^{2} y^{3}-2 x $$

Step-by-Step Solution

Answer

\( \nabla f = \left( 2xy^3 - 2, 3x^2 y^2 \right) \)

Step 1: Understand the Gradient

The gradient of a function, denoted as \( abla f \), is the vector of partial derivatives. That is, \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). To find \( abla f \), we will calculate the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \).

Step 2: Find the Partial Derivative with Respect to x

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

Step 3: Find the Partial Derivative with Respect to y

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

Step 4: Formulate the Gradient Vector

Combine the partial derivatives into a vector to form the gradient: \( abla f = \left( 2xy^3 - 2, 3x^2 y^2 \right) \).

Key Concepts

Partial Derivatives

To truly grasp the gradient of a function, it's essential to understand the concept of partial derivatives. Imagine you're walking on a surface, where the surface is represented by a function of two variables, say, temperature. At different points, the temperature changes based on directions. Partial derivatives help determine how this function changes when you alter only one variable.

The partial derivative of a function with respect to one variable means calculating the derivative while holding other variables constant. This represents the rate of change of the function in that specific direction. In our example, for the function \( f(x, y) = x^2 y^3 - 2x \), to find \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant. Similarly, to find \( \frac{\partial f}{\partial y} \), we consider \( x \) as constant.

• Step 2 showed us finding \( \frac{\partial f}{\partial x} = 2xy^3 - 2 \)
• Step 3 demonstrated finding \( \frac{\partial f}{\partial y} = 3x^2 y^2 \)

Calculating these enables us to dive deeper into how each variable specifically impacts the function's outcome, enhancing our understanding of its behavior.

Function of Two Variables

A function of two variables can be imagined like a surface in three-dimensional space. This kind of function, represented as \( z = f(x, y) \), links two input variables, \( x \) and \( y \), to a single output, \( z \). Think of it as a piece of paper, bending and stretching across a 3D space.

This concept is essential because it shows how one output can be influenced by changes in two different inputs. By studying such functions, we can predict and map out surfaces, optimize processes, and understand complex systems. Such functions can be used in real-world applications like geographical mapping or engineering fields.

In our example function, \( f(x, y) = x^2 y^3 - 2x \), both \( x \) and \( y \) independently influence the value of \( f \), representing a complex surface when plotted graphically. Identifying how each variable influences the outcome individually helps in visualizing the nature of the surface created by the function.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector-valued functions. It plays a critical role when dealing with multi-variable functions, like our previous function, \( z = f(x, y) \). In simple terms, vector calculus allows us to dive into analyzing vector fields and perform operations using vectors.

Central to this is the gradient of a function, which is foundational in vector calculus. The gradient, \( abla f \), is essentially a vector pointing in the direction of the steepest ascent of the function. It's composed of the function's partial derivatives, \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

• In the provided solution, we got \( abla f = (2xy^3 - 2, 3x^2 y^2) \).

Calculating the gradient helps find points of maxima or minima, optimize problems, and analyze vector fields. The gradient vector tells us not only the direction where the output of the function increases most rapidly but also how significant this change is, deepening our comprehension of the landscape of multi-variable functions.