Problem 41 - Calculus Early Transcendentals

Question

A model for the surface area of a human body is given by \(S=0.1091 w^{0.425} h^{0.725}\) , where \(w\) is the weight (in pounds), \(h\) is the height (in inches), and \(S\) is measured in square feet. If the errors in measurement of \(w\) and \(h\) are at most \(2 \%,\) use differ- entials to estimate the maximum percentage error in the calculated surface area

Step-by-Step Solution

Answer

The maximum percentage error in surface area is approximately 0.25%.

Step 1: Understand the Formula

We are given the formula for the surface area, \( S = 0.1091 w^{0.425} h^{0.725} \). Here \( w \) represents weight, \( h \) represents height, and \( S \) is the surface area.

Step 2: Calculate Differentials

We will use differentials to find the approximate error in \( S \). The differential \( dS \) is given by \( dS = \frac{\partial S}{\partial w} dw + \frac{\partial S}{\partial h} dh \). First, calculate the partial derivatives: \( \frac{\partial S}{\partial w} = 0.1091 \times 0.425 w^{-0.575} h^{0.725} \) and \( \frac{\partial S}{\partial h} = 0.1091 \times 0.725 w^{0.425} h^{-0.275} \).

Step 3: Substitute the Maximum Errors

Consider the maximum errors: \( dw = 0.02w \) and \( dh = 0.02h \). Substituting these into the expression for \( dS \), we get:\[dS = 0.1091 \times 0.425 \times w^{-0.575} h^{0.725} \times 0.02w + 0.1091 \times 0.725 \times w^{0.425} h^{-0.275} \times 0.02h\]

Step 4: Simplify the Expression

Factor out the common \( 0.1091 \times 0.02 \) to simplify the expression:\[dS = 0.1091 \times 0.02 \left(0.425 \times h^{0.725} \times w^{0.425} + 0.725 \times h^{0.425} \times w^{0.425}\right)\]We then have:\[dS = 0.002182 \left(0.425 + 0.725\right)\]Since \( h^{0.725} \times w^{-0.575} \) and \( h^{-0.275} \times w^{0.425} \) factor out and sum equals approximately 1.

Step 5: Calculate Percentage Error

The percentage error in the surface area is:\[\frac{dS}{S} = 0.002182 \times (0.425 + 0.725) = 0.002182 \times 1.15 = 0.0025093\]Thus, the maximum percentage error in the calculated surface area is approximately \( 0.25 \% \).

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in calculus that allows us to understand how a function changes as one of its variables changes, while all other variables are held constant. In the surface area model given by the formula \( S = 0.1091 w^{0.425} h^{0.725} \), we have a function of two variables: weight \( w \) and height \( h \). To compute the partial derivative of \( S \) with respect to \( w \), denoted \( \frac{\partial S}{\partial w} \), we differentiate \( S \) considering \( h \) as a constant. This gives us the expression \( 0.1091 \times 0.425 \times w^{-0.575} \times h^{0.725} \). Similarly, the partial derivative with respect to \( h \), \( \frac{\partial S}{\partial h} \), is calculated by differentiating \( S \) as though \( w \) is constant, resulting in \( 0.1091 \times 0.725 \times w^{0.425} \times h^{-0.275} \). These partial derivatives are critical as they help us evaluate how small changes in weight and height affect the overall surface area, which is key for estimation using differentials.

Percentage Error

The concept of percentage error is crucial for understanding the potential inaccuracies that may arise from measuring physical quantities. It expresses the relative error as a percentage of the actual value. In the context of the surface area model, if the weight \( w \) and height \( h \) are each measured with a possible error of up to 2%, the task is to assess how this error propagates to the calculated surface area. When we use differentials, we first find the differential change in \( S \), denoted by \( dS \). Then we calculate the percentage error by considering the ratio \( \frac{dS}{S} \), which gives the fraction of error over the original quantity. We then multiply this fraction by 100 to convert it into a percentage. Understanding percentage error is essential as it offers a tangible way to gauge measurement uncertainty, especially significant in scientific and engineering contexts.

Surface Area Model

In the exercise, the surface area model described by \( S = 0.1091 w^{0.425} h^{0.725} \) serves as a practical tool for estimating the human body's surface area based on weight and height. This model is motivated by empirical fitting to real-life data, balancing simplicity and accuracy. The powers of weight and height in the formula were determined through studies to provide the best approximation for a large variety of individuals, making this model widely accepted. The constants and exponents were chosen to accurately reflect the complex relationship between weight, height, and surface area in a mathematically manageable way. Using this model, scientists and health professionals can quickly estimate surface area, which is relevant for various applications, such as calculating dosages for certain medications and evaluating metabolic rates.

Measurement Errors

Measurement errors are unavoidable discrepancies that occur when measuring physical quantities. In the exercise, the weight \( w \) and height \( h \) of an individual are subject to measurement errors, potentially affecting our calculations. The exercise assumes these errors to be at most 2%. Differentials help approximate the effect of these errors on related calculations by determining how much variation in \( S \) will occur due to a small change in \( w \) or \( h \). By substituting the maximum errors \( dw = 0.02w \) and \( dh = 0.02h \) into the differential equation, we obtain an estimate of how much the errors in measurement will impact the surface area calculation. Understanding measurement errors and their potential impact is crucial for anyone who needs to rely on precise and accurate data, such as engineers, scientists, and healthcare providers. This awareness helps in designing experiments, interpreting data, and making informed decisions.