Problem 16 - Calculus

Question

According to a simple physiological model, an athletic adult male needs 20 calories per day per pound of body weight to maintain his weight. If he consumes more or fewer calories than those required to maintain his weight, his weight changes at a rate proportional to the difference between the number of calories consumed and the number needed to maintain his current weight; the constant of proportionality is \(1 / 3500\) pounds per calorie. Suppose that a particular person has a constant caloric intake of \(I\) calories per day. Let \(W(t)\) be the person's weight in pounds at time \(t\) (measured in days). (a) What differential equation has solution \(W(t) ?\) (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve this differential equation. (d) Graph \(W(t)\) if the person starts out weighing 160 pounds and consumes 3000 calories a day.

Step-by-Step Solution

Answer

W(t) = 150 + 10e^{-t/175} shows how weight converges to equilibrium (150 lbs) over time.

Step 1: Understanding the Problem

The problem involves a differential equation model to describe how a person's weight changes based on caloric intake. We need to derive the differential equation governing this, find the equilibrium solution, solve the differential equation, and graph the function for a specific initial weight and caloric consumption.

Step 1: Derive the Differential Equation

The person requires 20 calories per day per pound of body weight to maintain his weight. Thus, the calories needed to maintain the weight is \(20W\). If \(I\) represents the caloric intake, the rate of weight change is proportional to the difference \(I - 20W\). Given that the proportionality constant is \(\frac{1}{3500}\), the differential equation is:\[\frac{dW}{dt} = \frac{1}{3500}(I - 20W).\]

Step 2: Find the Equilibrium Solution

At equilibrium, the weight does not change with time, which means \(\frac{dW}{dt} = 0\). Setting the differential equation to zero gives:\[0 = \frac{1}{3500}(I - 20W) \implies I = 20W.\]Thus, the equilibrium weight \(W_{eq}\) satisfies \(I = 20W_{eq}\), or \(W_{eq} = \frac{I}{20}\). The equilibrium is expected to be stable as small deviations should return towards equilibrium under normal conditions.

Step 3: Solve the Differential Equation

The differential equation is separable. Rearranging terms gives:\[\frac{dW}{dt} = \frac{1}{3500}(I - 20W) \implies dt = \frac{3500}{I - 20W} dW.\]Integrating both sides, we have:\[t = -\frac{3500}{20} \ln|I - 20W| + C.\]Solving for \(W\), we find:\[W(t) = \frac{I}{20} + (W_0 - \frac{I}{20})e^{-\frac{20}{3500}t},\]where \(W_0\) is the initial weight.

Step 4: Graph the Solution

To graph \(W(t)\), we assume the initial weight \(W_0 = 160\) pounds and the initial caloric intake \(I = 3000\) calories. First, calculate the equilibrium weight: \(W_{eq} = \frac{3000}{20} = 150\) pounds. The weight function is:\[W(t) = 150 + (160 - 150)e^{-\frac{1}{175}t},\]which simplifies to:\[W(t) = 150 + 10e^{-\frac{1}{175}t}.\]Plotting this equation shows how the weight converges to the equilibrium of 150 pounds over time.

Key Concepts

Weight Change Model

The weight change model is a practical application of differential equations used to describe how a person's weight varies over time based on their caloric consumption. The model begins with a simple premise: a person's weight will increase if they consume more calories than required for maintenance and will decrease if they consume fewer. The rate of weight change is proportional to this caloric gap. In the exercise, this is mathematically represented by the equation\[\frac{dW}{dt} = \frac{1}{3500}(I - 20W),\]where \(dW/dt\) is the rate of weight change, \(I\) is the constant daily caloric intake, and \(20W\) denotes the calories required per day based on weight \(W\). The constant \(1/3500\) reflects the fact that 3500 calories are approximately equivalent to one pound of body weight. Understanding this relationship helps us predict weight trends which can be crucial for health and fitness management.
Whenever discussing weight dynamics in physiological models, consider all influencing factors, including metabolism and physical activity, though they are simplified here.

Caloric Intake Model

The caloric intake model plays a crucial role in understanding weight dynamics. It posits that a person's weight stability or change is influenced by their caloric intake relative to the calories required to maintain their body weight. In this context, if a person consumes \(I\) calories daily but requires \(20W\) calories to maintain their weight, any surplus or deficit in caloric intake leads to weight change. The difference \(I - 20W\) directly factors into determining whether weight increases or decreases.
This model simplifies the complexity of human metabolism by focusing solely on caloric intake versus caloric needs. Energy balance is simplified into a clear equation where 20 calories per pound are needed daily. While the model serves as a strong theoretical foundation for weight management, actual energy needs may vary based on additional factors like activity level and metabolic rate.

Equilibrium Solution

In the context of differential equations, an equilibrium solution refers to a state where the variables involved cease to change, signifying a balance. For the weight change model, equilibrium occurs when the weight remains constant over time. Mathematically, this is when the rate of weight change \(\frac{dW}{dt}\) is zero. Substituting into our equation gives us:\[0 = \frac{1}{3500}(I - 20W),\]which simplifies to \(I = 20W\). Thus, the equilibrium weight \(W_{eq}\) is \(\frac{I}{20}\). In our exercise, given a caloric intake \(I\) of 3000 calories, the equilibrium weight would be 150 pounds.
Equilibrium in this context is expected to be stable. This means any small deviations in weight would naturally return to the equilibrium state due to the balancing nature of the equation. Stability arises from the proportional correction exerted by the differential equation, driving the system back to equilibrium when deviated.

Separable Differential Equations

Separable differential equations are a class of equations that can be solved by separating variables on opposite sides of the equation. For weight change, the equation:\[\frac{dW}{dt} = \frac{1}{3500}(I - 20W)\]is separable, meaning it can be rewritten such that all terms involving \(W\) are on one side and those involving \(t\) are on the other, forming:\[dt = \frac{3500}{I - 20W} dW.\]Integrating both sides allows us to find the general solution for \(W(t)\), which describes weight over time. Integration simplifies the task of solving complex differential equations by breaking them down into more manageable parts.
As demonstrated, the solution to the weight change equation, after integration and initial condition application, is:\[W(t) = \frac{I}{20} + (W_0 - \frac{I}{20})e^{-\frac{20}{3500}t},\]where \(W_0\) is the initial weight. This formula not only helps track weight trajectory over time but also demonstrates how initial conditions and caloric intake collectively determine weight patterns.