Expert: Paul Klarreich - 5/6/2007

Question
An economist studying the price changes of a particular commodity uses the model for which  dP/dt =kPt  where P(pounds) is the cost at the time  t  months and  k  is a positive constant.
a) Given that P =100 when t=0, show that P=100e^(0.5kt^2)
b)Find the price of the commodity, predicted by the model at time 9months, given that the price after 2 months is 102(pounds)  

Answer
Questioner:   Maureen
Category:  Calculus
 
Question:  An economist studying the price changes of a particular commodity uses the model for which  dP/dt =kPt  where P(pounds) is the cost at the time  t  months and  k  is a positive constant.
a) Given that P =100 when t=0, show that P=100e^(0.5kt^2)
b)Find the price of the commodity predicted by the model at time 9 months, given that the price after 2 months is 102(pounds)
.....................................
Hi, Maureen,

This is your standard 'exponential growth (and decay)' situation.  To solve your differential equation:

dP
-- = kPt
dt  

you do a Separation of Variables:

dP = kPt dt
dP
-- = k t dt
P

Integrate each side:

ln P = kt^2/2 + c    
ln P = 0.5 kt^2 + c

Solve for P:

P =  e^(0.5 kt^2 + c)

P =  e^c e^(0.5 kt^2)

P =  C e^(0.5 kt^2)   << big C, now.

Enter initial condition:  P = 100  when  t = 0:

100 = C e^(0.5 k(0)^2)

100 = C e^0 = C

So C = 100 and your equation is complete: (sort of, because you don't have a k yet)

P =  100 e^(0.5 kt^2)

To find k, you need another piece of data, such as:

"given that the price after 2 months is 102(pounds)"

P(2) = 100 e^(0.5 k (2)^2)

102 = 100 e^(2k)

e^2k = 102/100

2k = ln(102/100)

k = ln(102/100)/2

Now use your calculator on that.  Your Windows calculator should do fine.  Mine gives:

k = 0.0099013136480898565130145334425502

So your equation is now REALLY complete:

P =  100 e^(0.0049506568240449282565072667212751 t^2)

and you can use it to find P for any time t, such as 9 months.  I'll leave that to you.