Problem 31 - Calculus for Biology and Medicine

Question

In Problems \(31-38\), find \(\partial f / \partial x, \partial f / \partial y\), and \(\partial f / \partial z\) for the given functions. $$ f(x, y, z)=x^{2} z+y z^{2}-x y $$

Step-by-Step Solution

Answer

The partial derivatives are: \(\frac{\partial f}{\partial x} = 2xz - y\), \(\frac{\partial f}{\partial y} = z^2 - x\), \(\frac{\partial f}{\partial z} = x^2 + 2yz\).

Step 1: Understanding Partial Derivatives

Partial derivatives are used to find the rate of change of a function with respect to one variable while keeping others constant. For the function given, we need to find how it changes with respect to each of its variables: \(x\), \(y\), and \(z\).

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

To find \(\partial f / \partial x\), treat \(y\) and \(z\) as constants and differentiate the function with respect to \(x\). This gives us: \[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2z + yz^2 - xy) = 2xz - y\].

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

For \(\partial f / \partial y\), treat \(x\) and \(z\) as constants and differentiate the function with respect to \(y\). This calculation results in: \[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^2z + yz^2 - xy) = z^2 - x\].

Step 4: Finding \(\partial f / \partial z\)

To find \(\partial f / \partial z\), treat \(x\) and \(y\) as constants and differentiate the function with respect to \(z\). The derivative is: \[\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (x^2z + yz^2 - xy) = x^2 + 2yz\].

Key Concepts

Understanding Calculus in Partial Derivatives

Calculus plays an essential role in analyzing changes in functions. In particular, partial derivatives aid in understanding how a multivariable function varies with respect to one of its variables. A partial derivative is similar to an ordinary derivative, but
targets functions with more than one variable. It shows the rate of change of the function concerning one variable while keeping others fixed. This approach allows us to analyze one specific dimension of change in a complex system.
In more practical terms:

• If you have a function that depends on multiple variables, like weather models depending on temperature, pressure, and humidity, calculus through partial derivatives can help predict changes in one specific variable by fixing others.
• This concept can be visualized by imagining the slope on a hill by looking at one axis direction only, ignoring the rest of the landscape. Thus, calculus plays a fundamental role in both scientific and engineering fields, where such precise change predictions are essential.

Explaining Multivariable Functions

Multivariable functions involve two or more variables and their interdependent relationships. Unlike single-variable functions that depend solely on one variable, such as time or distance, multivariable functions can involve multiple parameters, making them much more complex.
For example:

• Consider a scenario where we are measuring the temperature in a room. Here, the function could depend on more than one variable, like time and the position within the room (latitudinal and longitudinal coordinates).
• In our exercise, the function \[f(x, y, z) = x^{2}z + yz^{2} - xy\] depends on three variables. Understanding how each variable contributes to the function's output requires considering each of them independently through partial derivatives.

Multivariable calculus extends our ability to create models and solutions applicable to real-world phenomena where more than one factor must be considered.

The Process of Differentiation in Context

Differentiation in the context of partial derivatives refers to finding how a function changes as one of its input variables changes, while keeping other variables constant. It differs from regular differentiation in that it delves into changes in sections of a multidimensional topology created by the function.
Steps in differentiation involve :

• Identifying the function's variables and targeting one for differentiation, while treating others as constants.
• This often simplifies the mathematical complexity, turning the differentiation process around one variable only. For example, differentiating with respect to \(x\) in our function, treating \(y\) and \(z\) as constants, results in: \(\partial f / \partial x = 2xz - y\).
• Repetition of this process with other variables allows us to construct a complete picture of how changes in each variable affect the overall function.

Differentiation is a tool for dissecting the intricacies of multivariable functions and is crucial for optimization, modeling, and simulation in different scientific fields.