Problem 194 - Calculus Volume 3

Question

Complete each task. Find the total differential of the function \(z=\frac{x y}{y+x}\) where \(x\) changes from 10 to 10.5 and \(y\) changes from 15 to \(13 .\)

Step-by-Step Solution

Answer

The total differential is \( dz = -0.14 \).

Step 1: Identify the function

The function given is \( z = \frac{xy}{y + x} \). We need to find the total differential of this function as changes are made to \(x\) and \(y\).

Step 2: Compute the partial derivatives

First, find the partial derivative with respect to \(x\): \( \frac{\partial z}{\partial x} = \frac{y(y + x) - xy}{(y + x)^2} = \frac{y^2}{(y + x)^2} \).Next, find the partial derivative with respect to \(y\): \( \frac{\partial z}{\partial y} = \frac{x(x + y) - xy}{(y + x)^2} = \frac{x^2}{(y + x)^2} \).

Step 3: Evaluate the partial derivatives at the initial point

The initial values are \( x = 10 \) and \( y = 15 \). Evaluate:- \( \frac{\partial z}{\partial x} \bigg|_{(10,15)} = \frac{15^2}{(15 + 10)^2} = \frac{225}{625} = 0.36 \).- \( \frac{\partial z}{\partial y} \bigg|_{(10,15)} = \frac{10^2}{(15 + 10)^2} = \frac{100}{625} = 0.16 \).

Step 4: Calculate the changes in variables

The changes in the variables are: \( dx = 10.5 - 10 = 0.5 \) and \( dy = 13 - 15 = -2 \).

Step 5: Apply the total differential formula

The total differential \( dz \) is given by: \( dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy \).Substitute the evaluated derivatives and changes:\( dz = 0.36 \times 0.5 + 0.16 \times (-2) \).Calculate: \( dz = 0.18 - 0.32 = -0.14 \).

Step 6: Calculate the approximate change in \( z \) and describe the result

The total differential \( dz = -0.14 \) gives an approximation of the change in \( z \). This calculation indicates that as \( x \) increases by 0.5 and \( y \) decreases by 2, the function \( z \) decreases by approximately 0.14.

Key Concepts

Partial Derivative

Partial derivatives are a fundamental tool in calculus that help us understand how a function changes with respect to one variable while keeping other variables constant. They're especially useful for analyzing functions of multiple variables.
When computing partial derivatives, we follow these steps:

• Identify the function and the variables involved. In our exercise, this is the function \( z = \frac{xy}{y+x} \) with variables \( x \) and \( y \).
• To find the partial derivative with respect to \( x \), treat \( y \) as a constant and differentiate the function as you would with a single-variable function. The result gives us \( \frac{\partial z}{\partial x} \).
• Then, to find the partial derivative with respect to \( y \), treat \( x \) as constant and differentiate, yielding \( \frac{\partial z}{\partial y} \).

The values of these derivatives provide us with the rates of change of \( z \) with respect to \( x \) and \( y \). To find specific rate values, we substitute the initial values of \( x \) and \( y \) into these partial derivatives, as seen in the exercise where \( \frac{\partial z}{\partial x} \bigg|_{(10,15)} = 0.36 \) and \( \frac{\partial z}{\partial y} \bigg|_{(10,15)} = 0.16 \).

Function Change

Understanding how a function changes is crucial to many applications in mathematics and real-world scenarios. Calculus gives us various tools to explore these changes, and in the context of this exercise, we focus on the total differential of a function.
The total differential helps us approximate the change in a function based on small changes in the input variables. Consider a function \( z(x, y) \). If \( x \) changes by \( dx \) and \( y \) changes by \( dy \), the total change in \( z \), or \( dz \), can be estimated as:

• \( dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \)

This powerful formula captures how small variations in \( x \) and \( y \) collectively alter the value of \( z \). In the exercise, by using the computed derivatives and changes \( dx = 0.5 \) and \( dy = -2 \), we predict that the function \( z \) decreases by approximately 0.14.

Calculus

Calculus is the branch of mathematics that gives us the tools to analyze and understand changes, slopes, areas, and the behavior of functions. It provides the framework through which we study rates of change using derivatives and accumulation through integrals.
In this context, derivatives let us investigate how a function changes based on its variables. When it comes to partial derivatives, calculus enlightens us on the individual impact each variable has on a function. It separates the changes to give us a clearer picture of how each component contributes to the overall result.
This exercise uses calculus to analyze the total differential of the function \( z = \frac{xy}{y+x} \). By calculating partial derivatives and applying the differential formula, we can approximate how small changes in \( x \) and \( y \) affect \( z \). This approach illustrates the predictive power of calculus in estimating functional changes in real-time, providing insights that are crucial in fields ranging from physics to economics.