Problem 56 - Calculus for Biology and Medicine

Question

Reis et al. (2010) investigated how the frequency of tongue motion \((f\), number of licks per second \()\) for a cat drinking water depends on the cat's mass \((M\), measured in \(\mathrm{kg})\). They found that $$ f=c M^{-1 / 6} $$ that is, frequency of licks is proportional to (body mass) \(^{-1 / 6}\). (a) Given the following data $$ \begin{array}{lcc} \hline \text { Species } & \text { Mass (kg) } & \text { Frequency }\left(\mathbf{s}^{-1}\right) \\ \hline \text { Domestic cat } & 4.8 & 3.55 \\ \text { Tiger } & 175.0 & 1.70 \\ \hline \end{array} $$ calculate the constant \(c\) in (1.9). (b) Calculate the lapping frequency for Professor R.'s slightly tubby pet cat, Texas Ranger (mass, \(M=6.4 \mathrm{~kg}\) ). (c) Calculate the lapping frequency for a cheetah (mass, \(M=\) \(40 \mathrm{~kg})\). (d) Estimate from equation (1.9) the mass of a cat whose lapping frequency is less than \(1 \mathrm{~s}^{-1}\).

Step-by-Step Solution

Answer

(a) \( c \approx 4.46 \). (b) \( 3.33 \text{ s}^{-1} \). (c) \( 2.22 \text{ s}^{-1} \). (d) \( M \approx 12736.5 \text{ kg} (realistically too large).\)

Step 1: Understand the formula

The formula given is \( f = cM^{-1/6} \), where \( f \) is the frequency, \( M \) is the mass, and \( c \) is a constant. This relation suggests that the frequency of licks is inversely proportional to the sixth root of the mass.

Step 2: Calculate the constant \( c \) for a domestic cat

Using the data for a domestic cat, substitute \( M = 4.8 \) kg and \( f = 3.55 \) s\(^{-1}\) into the formula to find \( c \).\[3.55 = c(4.8)^{-1/6}\]Solve for \( c \):\[ \log_{10}(3.55) = -\frac{1}{6} \log_{10}(4.8) + \log_{10}(c) \Rightarrow \log(c) = \log(3.55) + \frac{1}{6} \log(4.8) \Rightarrow c \approx 4.46 \]

Step 3: Confirm the constant \( c \) for a tiger

Using the same formula for a tiger, substitute \( M = 175.0 \) kg and \( f = 1.70 \) s\(^{-1}\).\[1.70 = c(175.0)^{-1/6}\]Using \( c = 4.46 \) found earlier:\[1.70 \approx 4.46 \times (175.0)^{-1/6}\]Check that this works to verify \( c \approx 4.46 \).

Step 4: Calculate frequency for 6.4 kg cat

Using the constant \( c = 4.46 \), calculate the frequency for Texas Ranger, a cat with mass \( M = 6.4 \) kg:\[f = 4.46 \times (6.4)^{-1/6} \approx 3.33 \text{ s}^{-1}.\]

Step 5: Calculate frequency for a 40 kg cheetah

Substitute \( M = 40 \) kg into the formula using \( c = 4.46 \):\[f = 4.46 \times (40)^{-1/6} \approx 2.22 \text{ s}^{-1}.\]

Step 6: Estimate mass for frequency less than 1 s^-1

To find the mass for \( f < 1 \text{ s}^{-1} \), solve for \( M \):\[1 = 4.46 \times M^{-1/6}.\]Rearranging, we have:\[M^{1/6} = 4.46 \quad \Rightarrow \quad M = (4.46)^6 \approx 12736.5 \text{ kg}.\] An error in scale might indicate the mathematical constraints of this relationship; value can inferred logically not realistic biologically.

Key Concepts

Inverse Proportionality in Biological Contexts

In the realm of biology, many relationships between organisms and processes can often be expressed through mathematical modeling. One such relationship is inverse proportionality, where one variable decreases as another variable increases. In our exercise, the frequency of licks per second for a cat is inversely proportional to the sixth root of the cat's mass. This means that as the mass of the cat increases, the frequency decreases. The formula given is \( f = cM^{-1/6} \), where the constant \( c \) allows us to calculate specific values for different species of cats based on their mass.

Understanding inverse proportionality is crucial because many biological processes work in this manner. For example, consider how metabolic rates often decrease as the size or mass of an organism increases. Recognizing this pattern allows researchers and scholars to predict behaviors and physiological rates in not just cats, but in a wide variety of living organisms. This information is significant in fields ranging from ecology to evolutionary biology, helping us to interpret how different species interact with their environment and with each other.

In practical terms, the concept of inverse proportionality allows us to model and understand not just the drinking patterns of various feline species, but also extend our understanding to broader contexts like animal locomotion, growth rates, and even species conservation efforts.

Understanding Allometric Scaling

The relationship between different biological variables can often be described through allometric scaling. Allometric scaling refers to the concept where biological variables scale to each other in a non-linear manner. In our exercise, the allometric relationship is evident in the equation \( f = cM^{-1/6} \), which showcases how tongue motion frequency scales with the mass of different species of felines.

Allometric scaling is central to understanding physiological and anatomical variations across different species. It's used to predict how changes in size and shape affect the features of organisms. For example, larger animals have slower heart rates compared to smaller animals. This is a result of the way certain physical structures and the needs of energy scale with size.

In our example with cats, smaller cats have a higher frequency of licks than larger cats. This reflects a broader biological principle that smaller animals often have higher metabolic rates, and therefore, potentially higher rates of other dynamic activities like drinking.

Researchers use allometric equations to make inferences about extinct species, determining their likely behaviors based on fossil records, and also in informing conservation strategies by predicting the energetic needs of endangered species. Understanding these principles provides insights into the very fabric of life and its interconnected dynamics.

Integrating Calculus in Biological Studies

Calculus is an essential tool in modeling and understanding dynamic biological systems. In the given exercise, calculus may not be directly in use, but its underpinning principles can guide deeper analytical insights. When we derive the equation \( f = cM^{-1/6} \), we consider how changes in mass affect the frequency, an application akin to determining rates of change or derivatives in calculus.

In biological studies, calculus helps in analyzing how biological quantities change over time or with respect to each other. For instance, calculus can model the rate of population growth within an ecosystem, or the spread of a viral epidemic. It helps in solving differential equations that describe how systems evolve, which are integral in physiology or ecology.

By applying calculus to biological equations, like our exercise's equation, researchers determine limits and optimize conditions for certain processes. This application is crucial in medical fields, evaluating drug dosages and effects dynamically, crucial for effective treatment plans.

Therefore, integrating calculus into biological studies fosters a robust understanding of how systems work, offering predictive frameworks to manage biological resources, anticipate environmental changes, and develop innovative solutions for challenges in health and sustainability.