According to a simple physiological model, an athletic adult male needs 20 calor

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 H calories per day. Let W(t) be the person's weight in pounds at time t (measured in days).
What differential equation has solution W(t)? Solve this differential equation, if the person starts out weighing 175 pounds and consumes 4000 calories a day.

Explanation / Answer

The number of calories per day required to maintain a given body mass (W(t)) is a*W, where a = 20 cal/(lb*day). (Here, I'm using pounds as a unit of mass, not force). The rate of mass change is proportional to the difference between the calories consumed, and the calories required to maintain a given body mass: dW(t)/dt = -k*(a*W(t) - H) where H is the calories consumed per day, which we will assume is constant (consistent with what we are told in the question). k is a positive constant of proportionality. If k is positive, then there must be a minus sign in front of the right hand side as I have written it, because dW/dt must be negative if a*W(t) > C. In terms of the values given in the question, the differential equation to be solved is: dW(t)/dt = (-1 lb/(3500 cal))*((20 cal/(lb*day) * W(t) - H) This is the answer to part (a). ---------------- It's easier to solve the solution "generally" in terms of k and a, rather than using the numerical values given in the question. Once we have a solution, we can plug in the values relevant here. We have (where W is understood to be a function of time, i.e. W(t)): dW/dt = -k*(a*W - H) = -a*k*(W - H/a) This is a separable, first-order, linear equation. Separate the variables: dW/(W - H/a) = -a*k dt Integrate: ln(W - H/a) = -a*k*t + c where c is the constant of integration. Exponentiate both sides: W(t) = H/a + exp(c - a*k*t) Remember that exp(y + z) = exp(y)*exp(z), so we can write the solution as: W(t) = H/a + C*exp(-a*k*t) where C = exp(c) is just another way of writing the integration constant. If the person's mass at time t = 0 is Wo, then: W(0) = Wo = H/a + C C = Wo - H/a So the general solution is given by: W(t) = H/a + (Wo - H/a)*exp(-k*a*t) We are told that a = 20 cal/(day*lb), k = (1 lb)/(3500 cal), Wo = 170 lb, and H = 3800 cal/day, so: H/a = 190 lb, and k*a = 1/(175 day) W(t) = 190 lb - (20 lb)*exp(-t/(175 day)) This is the answer to part (b)

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