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Question

According to a simple physiological model, an athletic adult 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.

(a) Suppose a person has a constant caloric intake of 3000 calories per day. Write a differential equation for W(t).

dW/dt = ?

(b) What is the equilibrium solution?

= ? pounds

(c) If a person initially weighs 160 pounds, what will happen to their weight if their constant caloric intake is 3000 calories per day?
Increase?
Remain constant?
Decrease?

Answer 1

Cal cons per day = C*dt

Cal excess = (C - 20*W )*dt

Weight change = [(C - 20*W ) / 3500]*dt = dW

dW/dt = (C - 20*W ) / 3500 this is the differential equation.

To solve, separate variables

dW/[(C - 20*W )] = dt/3500

Integrate

-1/20 * ln(C - 20*W) = t/3500 + K

ln(C - 20*W) = -(20/3500)*t - K

C - 20*W = K'*e^-(20/3500)*t

20*W = C - K'*e^-(20/3500)*t

W = C/20 - K''*e^-(20/3500)*t

If the initial weight is 160, then W = 160 at t = 0:

160 = C/20 - K'';

K'' = C/20 - 160

W = C/20 -(C/20 - 160) *e^-(20/3500)*t

If C/20 > 160, the coefficient of the exponential is negative and weight decreases. 3000/20 = 150. so weight will increase.

For a steady-state condition, C/20 - 160 must be 0.

Answer 2

qa
let c be calory intake, W be weight and k be the contant of proportionality.

dW/dt = k(c-20W)

now k = 1/3500 and c = 3000
dW/dt = (3000-20W) / 3500
= (150-W)/175

qb
equilibrium at dW/dt = 0
0 = (150-W)/175
W = 150 pounds

qc
dW/dt = (150-160)/175
= -10/175
= -2/35
negative means decrease.

Answer 3

(a) dW/dt = (3000 - 20W) / 3500.

(b) Put dW/dt = 0, solve for W.

(c) Calculate dW/dt = (3000 - 20x160) / 3500. If it's negative, he is losing, if positive, he's gaining.