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Dy/dt=ky gallons of oil

  • Thread starter lreichardt
  • Start date
1. Oil is being pumped continuously from a certain oil well at a rate proportional to the amount of oil left in the well; that is, dy/dt=ky, where y is the amount left in the well at any time t. Initially, there were 1,000,000 gallons of oil in the well, and 6 years later there were 500,000 gallons remaining. It will no longer be profitable to pump oil when there are fewer than 50,000 gallons remaining.

1.a Write an equation for y, the amount of oil remaining in the well at any time t.

Attempt at solution:

y=y0e^(kt)
I know I need to solve for k, I just don't know how.
At t=0, y0=1,000,000
t=6 (Half life=??)

1. b At what rate is the amount of oil in the well decreasing when there are 600,000 gallons of oil remaining?

solution: haven't quite gotten here yet...

1. c In order not to lose money, at what time t should oil no longer be pumped from the well?

solution: see above. :-(

Thank you to anyone who can point me in the right direction, here.
 

Answers and Replies

cepheid
Staff Emeritus
Science Advisor
Gold Member
5,183
35
Right, so at t = 6,

y(t) = y0e6k

so obviously e6k = 1/2

Solve for k
 
HallsofIvy
Science Advisor
Homework Helper
41,734
893
By the way, you don't really need to solve for k. Since
[tex]e^{kt}= (e^k)^t[/tex]
You really only need [itex]e^k[/tex].
 

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