Problem 1 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=x^{2} y+x y^{2} $$

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = 2xy + y^2; \frac{\partial f}{\partial y} = x^2 + 2xy\)

Step 1: Identify Terms of the Function

Examine the function \[f(x, y) = x^2y + xy^2\]. Notice that this is a function of two variables \(x\) and \(y\), with two terms: \(x^2y\) and \(xy^2\).

Step 2: Calculate \( \partial f / \partial x \)

To find the partial derivative of \(f\) with respect to \(x\), treat \(y\) as a constant. Hence, differentiate each term:- For \(x^2y\), differentiate with respect to \(x\): \(2xy\).- For \(xy^2\), differentiate with respect to \(x\): \(y^2\).Combine the results to get:\[\frac{\partial f}{\partial x} = 2xy + y^2\]

Step 3: Calculate \( \partial f / \partial y \)

To find the partial derivative of \(f\) with respect to \(y\), treat \(x\) as a constant. Hence, differentiate each term:- For \(x^2y\), differentiate with respect to \(y\): \(x^2\).- For \(xy^2\), differentiate with respect to \(y\): \(2xy\).Combine the results to get:\[\frac{\partial f}{\partial y} = x^2 + 2xy\]

Key Concepts

Function of Two Variables

A function of two variables, like the one in this problem, is an expression where the output is determined by two different input variables, typically denoted as \(x\) and \(y\). This is an extension of a single-variable function, where the output depends only on one independent variable. In our example, the function is \(f(x, y) = x^2y + xy^2\). Here, both \(x\) and \(y\) influence the value of \(f\).

Such functions are crucial in various fields like economics, physics, and engineering, where systems naturally depend on more than one changing variable. When dealing with these functions, the goal is often to analyze how one variable affects the function while keeping the other constant. This analysis is done using partial derivatives, which are a fundamental concept in multivariable calculus. If you imagine a surface plotted in three dimensions, partial derivatives help you understand how the surface slopes as you move along the \(x\) or \(y\) axis alone. This is essential for optimization and understanding the behavior of complex systems.

Differentiation Rules

Differentiation rules are the tools that help us compute derivatives. In calculus, these rules allow us to find out how fast a function changes with respect to one of its variables. For functions of two variables, partial differentiation is used. When computing the partial derivative of a function \(f(x, y)\) with respect to \(x\), we treat \(y\) as a constant. Conversely, when differentiating with respect to \(y\), \(x\) is treated as a constant.

This selective differentiation makes it easier to determine the rate of change of the function concerning just one variable at a time. In the given problem, for the partial derivative \(\frac{\partial f}{\partial x}\), you'll differentiate each term where \(x\) appears, treating \(y\) as constant:

• For the term \(x^2y\), treat \(y\) as constant, and derive \(x^2\) to get \(2xy\).
• For the term \(xy^2\), treat \(y^2\) as constant, deriving it with respect to \(x\) gives \(y^2\).

Then you combine these results for your final partial derivative with respect to \(x\). The same logic applies for \(\frac{\partial f}{\partial y}\), but here, treat \(x\) as constant, and differentiate accordingly. Differentiation rules help streamline this process significantly.

Calculus Problem Solving

Calculus problem solving involves breaking down a problem into manageable parts and employing methods like differentiation to solve them systematically. Tackling a calculus problem, especially one involving multiple variables, can seem daunting at first. That's why it's essential to approach it step by step.

For the function \(f(x, y) = x^2y + xy^2\), problem solving starts with identifying and separating terms. Then, use partial derivatives, a key concept in multivariable calculus, to explore how the function changes as each variable changes. Calculus problem-solving often uses derivative calculations to:

• Determine rates of change\(\rightarrow\) An effective use in physics to describe velocity and acceleration, for example.
• Find critical points\(\rightarrow\) Crucial for understanding minima, maxima, and saddle points in graphs.

After computing derivatives, interpreting them in the context of the problem is the final but crucial step. Visualizing the impact of each variable separately provides insight into the function's behavior, guiding effective problem-solving in diverse scientific and engineering applications.