course mth174
previous problem: Oil is pumped continuously from a well at a rate proportional to the amount of the oil left in the well. Initially, there were 1 million barrels of oil in the well' six years later 500,000 barrels remain. At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining. When will will there be 50,000 barrels remaining?
dQ/dt=-kQ
to solve get all variables on side
dQ/Q= -kdt
take the integral
and get Q=Be^-tk
integration gives you ln | Q | = e^(-k t) + c. Be sure you see how this gets you to the form Q = B e^(-kt).
the initial conditions are that at t=0 there is 1,000,000 barrels of oil currently, so we can solve for B
and get B=1,000,000
then Q=1,000,000e^-tk
the problem tells us that in 6 years there are 500,000 barrells left, so then we can find k
500,000=1000000e^-6k
then we can solve for k
and get k= .1155
it asks us to find at what rate is production when there are 600,000 barrels left
600,000=1000000e^-.1155t
then solving for t we find that its equal to 4.42, which seems about correct, so it wants to know the rate, so should we take d'q/ d't and get the rate oil production is changing wrt to time (My answer was 92000 something; the book was 62000 something)
e^(-.1155 t) would be .6
dQ/dt would be -.1155 * 1000000 e^(-.1155t); since e^(-.1155 t) = .6 we should get
dQ/dt = -.1155 * 1000000 * .6,
which is about 67,000. So I appear to disagree with both you and the book, but I'm closer to the book.
But i know I have the right equation for modeling because in the second part it wants to know in how many years will there be 50,000 barrels left
50,000=1000000e^-.1155 t
and solve for t and get 25 years.
Maybe a little off. Production halves every 6 years, so in four halvings (24 years) you'll be down to 62,500 barrels. Another year won't be enough to bring it down to 50,000; that will take just about 2 more years, for a total of 26 years.
I don't have time to check the actual numbers, but everything in your solution appears plausible. So, however, does the answer in the book. See my notes.