Problem 86 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Empirical data suggest that the surface area of a \(180-\mathrm{cm}\) -tall human body changes at the rate of $$ S^{\prime}(W)=0.131773 W^{-0.575} $$ square meters/kilogram, where \(W\) is the weight of the body in kilograms. If the surface area of a 180 -cm-tall human body weighing \(70 \mathrm{~kg}\) is \(1.886277 \mathrm{~m}^{2}\), what is the surface area of a human body of the same height weighing \(75 \mathrm{~kg}\) ?

Step-by-Step Solution

Answer

The surface area of a 180cm-tall human body weighing 75 kg is approximately \(1.932295 \mathrm{~m}^2\).

Step 1: Integrate the rate of change function

Integrate \(S^{\prime}(W)\) with respect to \(W\) to get the surface area function \(S(W)\). $$ S(W) = \int S^{\prime}(W) \, dW = \int 0.131773 W^{-0.575} \, dW. $$

Step 2: Solve the integral

Applying the power rule and adding the integration constant \(C\), we get: $$ S(W) = \frac{0.131773}{1-(-0.575)} W^{1 - 0.575} + C = \frac{0.131773}{1.575} W^{0.425} + C. $$

Step 3: Determine the constant C using the given information

We know that when \(W = 70 \mathrm{~kg}\), the surface area is \(1.886277 \mathrm{~m}^2\). So, we have: $$ 1.886277 = \frac{0.131773}{1.575} \times 70^{0.425} + C. $$ Solve for the integration constant \(C\): $$ C = 1.886277 - \frac{0.131773}{1.575} \times 70^{0.425}. $$

Step 4: Calculate the constant C

Compute the value of \(C\): $$ C = 1.886277 - \frac{0.131773}{1.575} \times 70^{0.425} \approx -1.827115. $$

Step 5: Obtain the surface area function

Write the final \(S(W)\) function using the value of \(C\): $$ S(W) = \frac{0.131773}{1.575} W^{0.425} - 1.827115. $$

Step 6: Calculate the surface area for W = 75 kg

Evaluate the function \(S(W)\) for \(W = 75 \mathrm{~kg}\): $$ S(75) = \frac{0.131773}{1.575} (75)^{0.425} - 1.827115 \approx 1.932295. $$ Therefore, the surface area of a 180cm-tall human body weighing 75 kg is approximately 1.932295 m².

Key Concepts

Understanding Integration

Integration in mathematics helps us find the accumulation of quantities. In the context of our exercise, we use integration to calculate the total surface area of a human body over a range of weights. Consider it the opposite of differentiation, where we are essentially 'undoing' the derivative to find the original function.

• We start with the derivative of a function, known as the rate of change, denoted by \( S^{\prime}(W) \).
• By integrating, we determine the surface area function \( S(W) \) from its rate of change form.

Integration may involve constants, which are accounted for through the constant of integration \( C \). In real-world applications, like calculating body surface area, integration helps us summarize data over a range rather than just at individual points.

Applying the Power Rule

The power rule is a straightforward method used in both differentiation and integration. When integrating a function like \( W^n \), you increase the exponent by one and divide by the new exponent. This rule makes it simpler to solve integrals that involve powers of variables.

• For example, when integrating \( 0.131773 W^{-0.575} \), according to the power rule, we add one to the exponent \(-0.575\), making it \(0.425\).
• We then divide the coefficient \(0.131773\) by the new exponent \(0.425\).

Eventually, this gives the integrated function, which includes adding the constant of integration. It’s a fundamental technique that often appears in applied mathematics.

Surface Area Calculation

Calculating the surface area of a human body for a specific weight involves deducing the value of the integration constant \( C \), which aligns mathematically determined values with empirical data. We use a known data point, such as a body weight of 70 kg with a surface area of 1.886277 m², to plug into our integrated equation and solve for \( C \).

• This step translates the general surface area function into a specific and practical tool for estimating other weights.
• Having \( C \) allows recalibrations of the formula for any similar empirical data range.

Thus, surface area calculation through integration becomes a precise tool, useful for various practical applications including medicine and physiology.

Exploring Rate of Change

The concept of rate of change encapsulates how one quantity varies with respect to another. In this exercise, \( S^{\prime}(W) \) represents how the surface area changes in relation to weight. It is expressed as a derivative function, detailing the infinitesimal changes in body surface area as weight increases or decreases.

• Understanding the rate of change helps in predicting how small weight fluctuations influence surface area.
• It allows for the calculation of overall changes over larger weight ranges via integration.

Rate of change is crucial in diverse fields, providing insights into dynamics of physical, economic, and even social systems, by modeling them mathematically.