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Differential Calculus Problem?
Mary weighs 60 kg and is on a diet of 1600 calories per day . Each day her body uses up 850 calories for basic metobolism . Exercise uses up calories at a rate propotional to her weight : 15 calories per kilogram per day . Assume 1 kg of body mass is equivalent to 10000 calories
a) Write a differential equation for Mary's weight in kilograms
b) Solve the equation to find Mary's weight as a funtion of time
c) What will happen to her weight after a long time?
( Hint : Convert data on work in kg , not in calories )
1 Answer
- AfOreUgNyLv 68 years agoFavorite Answer
Converting calories to kg :
Diet is the rate at which mass is added.
r in = 1600 cal/day / 10000 cal/kg = 0.16 kg/day
Metabolism is the (constant) rate at which mass is consumed.
r out1 = 850 / 10000 = 0.085 kg/day
Exercise is the (variable) rate at which mass is consumed.
Let m be the mass, then
r out2 = ( 15 cal/kg*day / 10000 cal/kg ) * m kg = 0.0015 m kg/day
a)
dm/dt = 0.16 - 0.085 - 0.0015m
dm/dt = 0.075 - 0.0015m
b)
dm/dt + 0.0015m = 0.075
The integrating factor
µ = e^∫ 0.0015 dt = e^0.0015t
Multiplying the equation by µ
e^0.0015t dm/dt + 0.0015 m e^0.0015t = 0.075 e^0.0015t
The LHS is now the derivative d(m e^0.0015t) / dt
d(m e^0.0015t) / dt = 0.075 e^0.0015t
Separating the variables
d(m e^0.0015t) = 0.075 e^0.0015t dt
Integrating
m e^0.0015t = C + 50 e^0.0015t
m(t) = 50 + Ce^(- 0.0015t)
The initial condition is m(0) = 60
60 = 50 + C ==> C = 10
m(t) = 50 + 10 e^(- 0.0015t)
c)
lim(t ⟶ ∞) 50 + 10 e(- 0.0015t) = 50
She'll reach 50 kg after a very long time.
