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Trinket's question at Yahoo! Answers regarding the modeling of weight loss

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MarkFL

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Feb 24, 2012
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Here is the question:

Differential equation about weight loss?

A person's weight depends on both amount of calories consumed and the energy used. Moreover, the amount of energy used depends on the person's weight - the average amount of energy used by a person is 17.5 calories per pound a day. Thus, the more weight the person loses, the less energy the person uses (assuming that the person maintains a constant level of activity). An equation that can be used to model weight loss is

dw/dt = (C/3500) - (17.5w/3500)

Where w is the person's weight in pounds, t is the time in days and C is the constant daily calorie consumption.

(A) find the general solution of the differential equation
(B) consider a person who weighs 180lb and begins a diet of 2500 calories per day. How long will it take the person to lose 10lb? 35?
(C) what is the limiting weight of the person?
(D) repeat (B) for a person who weighs 200lb when the diet started
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Trinket,

We are told to model a person's weight with the IVP:

\(\displaystyle \frac{dw}{dt}=\frac{C}{3500}-\frac{17.5}{3500}w\) where \(\displaystyle w(0)=w_0\)

I would choose the write the ODE in standard linear form:

\(\displaystyle \frac{dw}{dt}+\frac{1}{200}w=\frac{C}{3500}\)

Next, we may compute the integrating factor:

\(\displaystyle \mu(t)=e^{\frac{1}{200}\int\,dt}=e^{\frac{1}{200}t}\)

Multiplying the ODE by the integrating factor, we obtain:

\(\displaystyle e^{\frac{1}{200}t}\frac{dw}{dt}+\frac{1}{200}e^{ \frac{1}{200}t}w=\frac{C}{3500}e^{ \frac{1}{200}t}\)

Now, this allows us the express the left side of the ODE as the differentiation of a product:

\(\displaystyle \frac{d}{dt}\left(e^{\frac{1}{200}t}w \right)=\frac{C}{3500}e^{\frac{1}{200}t}\)

Integrate with respect to $t$:

\(\displaystyle \int \frac{d}{dt}\left(e^{\frac{1}{200}t}w \right)\,dt=\frac{C}{3500}\int e^{\frac{1}{200}t}\,dt\)

\(\displaystyle e^{\frac{1}{200}t}w=\frac{2C}{35}e^{\frac{1}{200}t}+c_1\)

Solve for $w(t)$:

\(\displaystyle w(t)=\frac{2C}{35}+c_1e^{-\frac{1}{200}t}\)

We may determine the parameter $c_1$ by using the initial value:

\(\displaystyle w(0)=\frac{2C}{35}+c_1=w_0\,\therefore\,c_1=\frac{35w_0-2C}{35}\)

And thus, the solution to the IVP is:

(1) \(\displaystyle w(t)=\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}\)

The amount of weight lost $L$ is the initial weight minus the current weight, and so we may write:

\(\displaystyle L(t)=w_0-\frac{2C}{35}-\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}\)

\(\displaystyle L(t)=\frac{35w_0-2C}{35}\left(1-e^{-\frac{1}{200}t} \right)\)

Solving this for $t$, we find:

\(\displaystyle \frac{35L(t)}{35w_0-2C}=1-e^{-\frac{1}{200}t}\)

\(\displaystyle e^{-\frac{1}{200}t}=\frac{35\left(w_0-L(t) \right)-2C}{35w_0-2C}\)

(2) \(\displaystyle t=200\ln\left(\frac{35w_0-2C}{35\left(w_0-L(t) \right)-2C} \right)\)

The limiting weight $w_L$ of the person is:

(3) \(\displaystyle w_L=\lim_{t\to\infty}\left(\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t} \right)=\frac{2C}{35}\)

Now we have formulas to answer the questions.

(A) Find the general solution of the differential equation.

\(\displaystyle w(t)=\frac{2C}{35}+\frac{35w_0-2C}{35}e^{-\frac{1}{200}t}\)

(B) Consider a person who weighs 180lb and begins a diet of 2500 calories per day. How long will it take the person to lose 10lb? 35?

Using the given data:

\(\displaystyle w_0=180,\,C=2500\)

The formula in (2) becomes:

\(\displaystyle t=200\ln\left(\frac{35\cdot180-2\cdot2500}{35\left(180-L(t) \right)-2\cdot2500} \right)=200\ln\left(\frac{1300}{1300-35L(t)} \right)\)

Now to compute the time to lose 10 lbs, we use $L(t)=10$ to get:

\(\displaystyle t=200\ln\left(\frac{1300}{1300-35\cdot10} \right)=200\ln\left(\frac{1300}{950} \right)=200\ln\left(\frac{26}{19} \right)\approx62.7315117710083\)

And to compute the time to lose 35 lbs, we use $L(t)=35$ to get:

\(\displaystyle t=200\ln\left(\frac{1300}{1300-35\cdot35} \right)=200\ln\left(\frac{1300}{75} \right)=200\ln\left(\frac{52}{3} \right)\approx570.526285982664\)

(C) What is the limiting weight of the person?

\(\displaystyle w_L=\frac{2\cdot2500}{35}=\frac{1000}{7}=142. \overline{857142}\)

(D) Repeat (B) for a person who weighs 200lb when the diet started.

Using the given data:

\(\displaystyle w_0=200,\,C=2500\)

The formula in (2) becomes:

\(\displaystyle t=200\ln\left(\frac{35\cdot200-2\cdot2500}{35\left(200-L(t) \right)-2\cdot2500} \right)=200\ln\left(\frac{2000}{2000-35L(t)} \right)\)

Now to compute the time to lose 10 lbs, we use $L(t)=10$ to get:

\(\displaystyle t=200\ln\left(\frac{2000}{2000-35\cdot10} \right)=200\ln\left(\frac{2000}{1650} \right)=200\ln\left(\frac{40}{33} \right)\approx38.4743785294912\)

And to compute the time to lose 35 lbs, we use $L(t)=35$ to get:

\(\displaystyle t=200\ln\left(\frac{2000}{2000-35\cdot35} \right)=200\ln\left(\frac{2000}{775} \right)=200\ln\left(\frac{80}{31} \right)\approx189.607886037747\)