Problem 48 - Applied Calculus

Question

Kleiber's Law states that the daily calorie requirement, \(C(w)\), of a mammal is proportional to the mammal's body weight \(w\) raised to the \(0.75\) power. \({ }^{1}\) If body weight is measured in pounds, the constant of proportionality is approximately 42 . (a) Give formulas for \(C(w)\) and \(C^{\prime}(w)\). (b) Find and interpret (i) \(C(10)\) and \(C^{\prime}(10)\) (ii) \(C(100)\) and \(C^{\prime}(100)\) \(\begin{array}{ll}\text { (iii) } C(1000) \text { and } C^{\prime}(1000) & 58 .\end{array}\)

Step-by-Step Solution

Answer

Formulas: \(C(w) = 42w^{0.75}\) and \(C'(w) = 31.5w^{-0.25}\). Values: \(C(10)\approx236.57\), \(C'(10)\approx17.80\), \(C(100)\approx1328.88\), \(C'(100)\approx9.95\), \(C(1000)\approx7476.50\), \(C'(1000)\approx5.58\).

Step 1: Understanding the problem

Kleiber's Law suggests that a mammal's daily calorie requirement \(C(w)\) is proportional to its body weight \(w\) raised to the power of 0.75. Given the constant of proportionality is 42, we need to determine the formula for \(C(w)\) and its derivative. Then, we will find the calories required and rate of change for different weights.

Step 2: Finding the formula for C(w)

Since the calorie requirement is proportional to body weight raised to the power of 0.75, we can express this as \(C(w) = 42w^{0.75}\), where 42 is the constant of proportionality.

Step 3: Calculating the derivative C'(w)

To find the derivative \(C'(w)\), apply the power rule to \(C(w) = 42w^{0.75}\). The derivative is \(C'(w) = 42 \times 0.75 \times w^{-0.25} = 31.5w^{-0.25}\). This represents the rate of change of calories required with respect to weight.

Step 4: Calculating C(10) and C'(10)

Substitute \(w = 10\) into the formulas.\[ C(10) = 42 \times 10^{0.75} \approx 236.57 \]\[ C'(10) = 31.5 \times 10^{-0.25} \approx 17.80 \]This interpretation suggests a 10-pound animal requires about 236.57 calories, and small increases in weight will increase the calorie requirement by about 17.80 calories per pound.

Step 5: Calculating C(100) and C'(100)

Substitute \(w = 100\) into the formulas.\[ C(100) = 42 \times 100^{0.75} \approx 1328.88 \]\[ C'(100) = 31.5 \times 100^{-0.25} \approx 9.95 \]A 100-pound animal needs approximately 1328.88 calories, and each additional pound increases the calorie requirement by about 9.95 calories.

Step 6: Calculating C(1000) and C'(1000)

Substitute \(w = 1000\) into the formulas.\[ C(1000) = 42 \times 1000^{0.75} \approx 7476.50 \]\[ C'(1000) = 31.5 \times 1000^{-0.25} \approx 5.58 \]This suggests a 1000-pound animal needs around 7476.50 calories, and each additional pound increases the calorie requirement by about 5.58 calories.

Key Concepts

Calorie Requirement

Understanding how much energy an animal needs is crucial for its health and growth. Daily calorie requirements for mammals are governed by Kleiber's Law. This law highlights that the calorie need, frequently noted as \( C(w) \), is directly linked to the animal's body weight \( w \). However, this relationship isn't straightforward; it's not a simple linear relationship where calorie needs increase equally with weight.
Instead, the connection is governed by a specific power of body weight. In the case of mammals, the calorie requirement grows according to the weight raised to the power of 0.75.

• This means as animals grow larger, they require more calories, but not in direct proportion to their increase in size.
• For example, a mammal weighing twice as much as another doesn't need twice the calories, but something less than that.

This nuanced understanding helps in maintaining the right balance of energy intake for animals, ensuring they grow but do not face issues related to overconsumption or energy deficit.

Body Weight

Body weight plays a critical role in determining the energy requirements as outlined by Kleiber's Law. Rather than having a straightforward, linear relationship with calorie needs, as one might assume, the connection is non-linear.
Unlike a mere linear scale, where doubling the body size would double the calorie needs, the 0.75 power adds a complex dimension. This exponent reflects high weight but diminishing returns in energy necessity.

• The practical outcome of this is that larger animals, although having greater total energy needs, become more "caloric" efficient as their size increases.
• A 1000-pound mammal will need more total calories than a 100-pound one, but not ten times more.

This means that body weight is truly essential to understanding how energy is metabolically managed in animals, according to Kleiber's Law.

Power Rule

The power rule is a fundamental tool in calculus, especially useful in differentiating functions like the calorie-weight relationship described by Kleiber's Law. The power rule provides a straightforward way to find the derivative of a function.
For a function of the form \( f(x) = ax^n \), the power rule asserts that its derivative is \( f'(x) = nax^{n-1} \). This is particularly helpful when dealing with functions where a variable is raised to a power, just as in \( C(w) = 42w^{0.75} \).

• By applying the power rule here, the derivative \( C'(w) \) can be swiftly calculated as \( 31.5w^{-0.25} \).
• This derivative helps explain how the calorie need changes as the body weight of the animal changes.

Understanding the power rule allows for efficient computation and insight into how different factors affect one another dynamically across a breadth of study problems.

Derivative

Derivatives represent the rate at which a quantity changes. In the context of Kleiber's Law, we use derivatives to see how the calorie requirement changes with respect to body weight.
In our example, \( C'(w) = 31.5w^{-0.25} \) is the derivative, indicating the rate of change in calorie needs per change in weight. This tells us how much the calorie requirement increases as the mammal's weight goes up slightly.

• A higher derivative value corresponds to a larger increase in calories needed for a small increase in body weight.
• Significant changes in body weight lead to changes in calorie needs, but as we see, the rate slows down as the weight increases due to the negative exponent.

Derivatives, therefore, offer vital insights into the dynamic changes occurring in biological systems, making them an invaluable tool for modeling and understanding complex relationships.