Question

About how much (in percent) does an error of \(1 \%\) in \(x\) and \(y\) affect \(x^{3} y^{2} ?\)

Short answer

The error of 1% in x and y causes a 5% error in x^3 y^2.

Step-by-step solution

Define the function

Let the function be defined as \((f(x, y) = x^{3} y^{2})\). Given that there is an error in both x and y, we need to determine how these errors propagate.

Calculate the differential

Find the partial derivatives of the function with respect to both x and y. The partial derivative with respect to x is: \( \frac{\partial f}{\partial x} = 3x^2 y^2\) The partial derivative with respect to y is: \( \frac{\partial f}{\partial y} = 2x^3 y\)

Estimate the total differential

The total differential (df) is given by: \( df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \) Substitute the partial derivatives into this equation: \( df = 3x^2 y^2\,dx + 2x^3 y\,dy\)

Calculate relative errors

Relative errors in x and y are given as 1%, which means: \( \frac{dx}{x} = 0.01\) and \( \frac{dy}{y} = 0.01\) To find the relative error in f, express df as a function of f: \( df = 3x^2 y^2\,(0.01x) + 2x^3 y\,(0.01y)\) = \(0.03x^3 y^2 + 0.02x^3 y^2\) = \(0.05x^3 y^2 \).

Calculate the relative percentage error

The relative error as a percentage is given by: \( \frac{df}{f} \times 100\% = \frac{0.05x^3 y^2}{x^3 y^2} \times 100\% = 0.05 \times 100\% = 5\% \).

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when working with functions of multiple variables. In simple terms, a partial derivative measures how a function changes as one of its input variables changes, while keeping the other input variables constant.

For example, consider the function in our problem: \( f(x, y) = x^{3} y^{2} \). Here, we are interested in how changing \( x \) and \( y \) will affect the output of the function. To do this, we find the partial derivatives:

• \( \frac{\textstyle \text {∂} f}{\textstyle \text {∂} x} = 3x^2 y^2 \)
• \( \frac{\textstyle \text {∂} f}{\textstyle \text {∂} y} = 2x^3 y \)

Each partial derivative tells us how \( f \) changes when either \( x \) or \( y \) changes, respectively.

The partial derivative with respect to \( x \) tells us that for small changes in \( x \), the function value changes by approximately \( 3x^2 y^2 \times \text{change in} \ x \) (represented mathematically as \( dx \)). Similarly, the partial derivative with respect to \( y \) tells us how \( f \) changes when \( y \) changes.

Differential Calculus

Differential calculus is the branch of mathematics that deals with the study of rates at which quantities change. It provides tools, like derivatives, to understand and quantify change.

In our exercise, we use differential calculus to understand the total differential \( (df) \). The total differential helps us estimate how small changes in multiple variables affect the function's output. If we have a function \( f(x, y) \), the total differential is calculated as:
\[ df = \frac{\textstyle \text {∂} f}{\textstyle \text {∂} x} dx + \frac{\textstyle \text {∂} f}{\textstyle \text {∂} y} dy \]
This formula combines the partial derivatives of the function with the small changes in \( x \) and \( y \) (represented as \( dx \) and \( dy \)).

For our function, the total differential is:
\[ df = 3x^2 y^2\,dx + 2x^3 y\,dy \]
This allows us to estimate how the function value will change as both \( x \) and \( y \) change. This is a powerful tool for analyzing the impact of small errors or variations in input variables on the overall function value.

Relative Error

Relative error is a measure that quantifies the accuracy of a measurement or calculation by comparing the size of the error to the size of the actual measurement. It is often expressed as a percentage. In our problem, we are given errors in \( x \) and \( y \) as percentages and we want to find how these errors propagate to the function \( f(x, y) = x^{3} y^{2} \).

The relative error in \( x \) and \( y \) given as 1% can be written as:

• \( \frac{\textstyle dx}{\textstyle x} = 0.01 \)
• \( \frac{\textstyle dy}{\textstyle y} = 0.01 \)

To find the relative error in \( f \), we first express the total differential \( (df) \) in terms of \( x \) and \( y \):
\[ df = 3x^2 y^2 \times (0.01x) + 2x^3 y \times (0.01y) = 0.03x^3 y^2 + 0.02x^3 y^2 = 0.05x^3 y^2 \]
We then compare \( df \) with \( f \):
\