Homework SourseOil is pumped continuously from a well at a rate proportiona
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 5 million barrels of oil in the well; six years later 2,500,000 barrels remain.
(A):Let Q(t) be the number of barrels left in the well after t years, measured in millions. Write a differential equation for Q that captures the information in the problem. Use k>0 for any proportionality constant you need. Be careful about signs.
Q\'=
(B):Solve the above differential equation, without yet determining k.
Q(t)=
(C):Determine k.
k=
(D):At what rate was the amount of oil in the well decreasing when there were 3,000,000 barrels remaining? [Hint: Use the equation in (a). Be careful about units.]
rate= barrels/year
(E):When will there be 250,000 barrels remaining?
years=
(A):Let Q(t) be the number of barrels left in the well after t years, measured in millions. Write a differential equation for Q that captures the information in the problem. Use k>0 for any proportionality constant you need. Be careful about signs.
Q\'=
(B):Solve the above differential equation, without yet determining k.
Q(t)=
(C):Determine k.
k=
(D):At what rate was the amount of oil in the well decreasing when there were 3,000,000 barrels remaining? [Hint: Use the equation in (a). Be careful about units.]
rate= barrels/year
(E):When will there be 250,000 barrels remaining?
years=
Solution
A) The rate that Q(t) is changing is proportional to Q, and that Q is decreasing, so dQ/dt = -kQ Q\' = -kQ where k is a constant. B) dQ/Q = - kdt ln(Q) - ln(Qo) = -kt where Qo is the amount of oil in the subsurface at time t = 0. ln(Q/Qo) = -kt Q(t) = Qo exp(-kt) c) ln(2.5/5) = -k(6 yr) ln(1/2) = -k(6 yr) ln(2) = k(6 yr) k = ln(2)/(6 yr) = 0.116/yr d) N(t) = 5 exp(-0.116t/yr) The rate at which the oil in the well is decreasing when there were 3 million barrels remaining is - dN/dt = -(0.116/yr)(3 million barrels) = -0.347 million barrels/year e) N(t) = 5 exp(-0.116t/yr) 250,000 = 5,000,000 exp(-0.116t/yr) t = 25.93 yrs
