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Chapter 7: Problem 8

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?

Short Answer

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

Step by step solution

01

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.
02

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 \(W(1) = 9\).
03

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: \[ \frac{1}{2} W^2 = kt + C \]
04

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 \(k\): \[ \frac{1}{2} (9)^2 = k \times 1 + 32 \] \[ \frac{81}{2} = k + 32 \] \[ k = \frac{81}{2} - 32 = \frac{17}{2} \]
05

Find the baby鈥檚 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 \(W\): \[ \frac{1}{2} W^2 = 102 + 32 \] \[ \frac{1}{2} W^2 = 134 \] \[ W^2 = 268 \] \[ W = \frac{\text{sqrt}(268)}{\text{lbs}} \] \[ W \text{ (approx)} = 16.37 \text{ lbs} \]
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Newton's Law of Cooling says that the rate at which an object, such as a cup of coffee, cools is proportional to the difference in the object's temperature and room temperature. If \(T(t)\) is the object's temperature and \(T_{r}\) is room temperature, this law is expressed at $$ \frac{d T}{d t}=-k\left(T-T_{r}\right) $$ where \(k\) is a constant of proportionality. In this problem, temperature is measured in degrees Fahrenheit and time in minutes. a. Two calculus students, Alice and Bob, enter a \(70^{\circ}\) classroom at the same time. Each has a cup of coffee that is \(100^{\circ} .\) The differential equation for Alice has a constant of proportionality \(k=0.5,\) while the constant of proportionality for Bob is \(k=0.1 .\) What is the initial rate of change for Alice's coffee? What is the initial rate of change for Bob's coffee? b. What feature of Alice's and Bob's cups of coffee could explain this difference? c. As the heating unit turns on and off in the room, the temperature in the room is $$ T_{r}=70+10 \sin t $$ Implement Euler's method with a step size of \(\Delta t=0.1\) to approximate the temperature of Alice's coffee over the time interval \(0 \leq t \leq 50 .\) This will most easily be performed using a spreadsheet such as Excel. Graph the temperature of her coffee and room temperature over this interval. d. In the same way, implement Euler's method to approximate the temperature of Bob's coffee over the same time interval. Graph the temperature of his coffee and room temperature over the interval. e. Explain the similarities and differences that you see in the behavior of Alice's and Bob's cups of coffee.

The Gompertz equation is a model that is used to describe the growth of certain populations. Suppose that \(P(t)\) is the population of some organism and that $$ \frac{d P}{d t}=-P \ln \left(\frac{P}{3}\right)=-P(\ln P-\ln 3) $$ a. Sketch a slope field for \(P(t)\) over the range \(0 \leq P \leq 6\). b. Identify any equilibrium solutions and determine whether they are stable or unstable. c. Find the population \(P(t)\) assuming that \(P(0)=1\) and sketch its graph. What happens to \(P(t)\) after a very long time? d. Find the population \(P(t)\) assuming that \(P(0)=6\) and sketch its graph. What happens to \(P(t)\) after a very long time? e. Verify that the long-term behavior of your solutions agrees with what you predicted by looking at the slope field.

We have seen that the error in approximating the solution to an initial value problem is proportional to \(\Delta t .\) That is, if \(E_{\Delta t}\) is the Euler's method approximation to the solution to an initial value problem at \(\bar{t},\) then $$ y(\bar{t})-E_{\Delta t} \approx K \Delta t $$ for some constant of proportionality \(K .\) In this problem, we will see how to use this fact to improve our estimates, using an idea called accelerated convergence. a. We will create a new approximation by assuming the error is exactly proportional to \(\Delta t,\) according to the formula $$ y(\bar{t})-E_{\Delta t}=K \Delta t . $$ Using our earlier results from the initial value problem \(d y / d t=y\) and \(y(0)=1\) with \(\Delta t=0.2\) and \(\Delta t=0.1,\) we have $$ \begin{array}{l} y(1)-2.4883=0.2 \mathrm{~K} \\ y(1)-2.5937=0.1 \mathrm{~K} \end{array} $$ This is a system of two linear equations in the unknowns \(y(1)\) and \(K .\) Solve this system to find a new approximation for \(y(1)\). (You may remember that the exact value is \(y(1)=\) \(e=2.71828 \ldots . .)\) b. Use the other data, \(E_{0.05}=2.6533\) and \(E_{0.025}=2.6851\) to do similar work as in (a) to obtain another approximation. Which gives the better approximation? Why do you think this is? c. Let's now study the initial value problem $$ \frac{d y}{d t}=t-y, y(0)=0 $$ Approximate \(y(0.3)\) by applying Euler's method to find approximations \(E_{0.1}\) and \(E_{0.05}\). Now use the idea of accelerated convergence to obtain a better approximation. (For the sake of comparison, you want to note that the actual value is \(y(0.3)=0.0408 .\) )

Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn \(5 \%\) each year. a. Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to grow-first, you are depositing one millon dollars per year and second, you are earning \(5 \%\) interest. b. If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years? c. The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years? d. Do you prefer the first or second option? Explain your thinking. e. At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?

The mass of a radioactive sample decays at a rate that is proportional to its mass. a. Express this fact as a differential equation for the mass \(M(t)\) using \(k\) for the constant of proportionality. b. If the initial mass is \(M_{0}\), find an expression for the mass \(M(t)\). c. The half-life of the sample is the amount of time required for half of the mass to decay. Knowing that the half-life of Carbon-14 is 5730 years, find the value of \(k\) for a sample of Carbon-14. d. How long does it take for a sample of Carbon-14 to be reduced to one- quarter its original mass? e. Carbon-14 naturally occurs in our environment; any living organism takes in Carbon14 when it eats and breathes. Upon dying, however, the organism no longer takes in Carbon-14. Suppose that you find remnants of a pre-historic firepit. By analyzing the charred wood in the pit, you determine that the amount of Carbon-14 is only \(30 \%\) of the amount in living trees. Estimate the age of the firepit.
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