Problem 8 - Active Calculus

Question

During the first few years of life, the rate at which a baby gains weight is proportional to the reciprocal of its weight. a. Express this fact as a differential equation. b. Suppose that a baby weighs 8 pounds at birth and 9 pounds one month later. How much will he weigh at one year? c. Do you think this is a realistic model for a long time?

Step-by-Step Solution

Answer

W(t) = sqrt(34t + 64). The weight after one year is approximately 16.37 lbs. The model is not realistic long-term.

Step 1: Express the proportionality relationship

Given that the rate at which a baby gains weight is proportional to the reciprocal of its weight, we can express this as \[ \frac{dW}{dt} = k \frac{1}{W} \] where \(W\) is the weight of the baby at time \(t\), and \(k\) is the proportionality constant.

Step 2: Set up the initial conditions

We know that the baby weighs 8 pounds at birth (\(t = 0\)), so \(W(0) = 8\). Additionally, the baby weighs 9 pounds one month later (\(t = 1\)), so \(W(1) = 9\).

Step 3: Separate variables and integrate

Separate variables in the differential equation: \[ W \frac{dW}{dt} = k \] Now integrate both sides with respect to their respective variables: \[ \frac{1}{2} W^2 = kt + C \]

Step 4: Apply initial conditions to find constants

Using \(W(0) = 8\), we find the constant \(C\): \[ \frac{1}{2} (8)^2 = C \] \[ C = 32 \] Using \(W(1) = 9\), we find the constant \(k\): \[ \frac{1}{2} (9)^2 = k \times 1 + 32 \] \[ \frac{81}{2} = k + 32 \] \[ k = \frac{81}{2} - 32 = \frac{17}{2} \]

Step 5: Find the baby’s weight after one year

Substitute back to find the weight \(W\) at \(t = 12\) months: \[ \frac{1}{2} W^2 = \frac{17}{2} \times 12 + 32 \] Simplify and solve for \(W\): \[ \frac{1}{2} W^2 = 102 + 32 \] \[ \frac{1}{2} W^2 = 134 \] \[ W^2 = 268 \] \[ W = \frac{\text{sqrt}(268)}{\text{lbs}} \] \[ W \text{ (approx)} = 16.37 \text{ lbs} \]

Step 6: Evaluate the long-term realism of the model

Considering the model long-term, the weight would continue to increase rapidly according to \(W^2 = 2kt + 2C\). Realistically, biological growth rates slow down and stabilize, thus this mathematical model is not suitable for a long term.

Key Concepts

Rate of Change

The rate of change is a fundamental concept in calculus. It describes how one quantity changes in relation to another.
In this exercise, the rate at which a baby gains weight is given by the differential equation: \(\frac{dW}{dt} = k \frac{1}{W}\).
This means the rate of weight gain (\frac{dW}{dt}) is proportional to the reciprocal of the baby's weight (1/W).
Understanding this concept helps in setting up and solving differential equations that describe real-world phenomena.

Initial Conditions

Initial conditions are values provided at the start of a problem to help solve differential equations.
They are crucial in determining the specific solution among a family of solutions. By substituting these values into the integrated equation, we find the constants needed for a particular solution.
This allows us to make accurate predictions about the baby's weight over time.

Biological Growth Modeling

Biological growth modeling uses mathematical formulas to describe how living organisms grow over time.
These models can be simple or complex, depending on the factors influencing growth.
In this exercise, the model assumes the rate of weight gain is proportional to the inverse of weight.
Such a model works well for short-term predictions but may not be realistic for long periods.
Over the long term, growth rates typically slow and stabilize. Thus, more sophisticated models might be required for long-term scenarios.