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Material Type: Paper; Class: Calculus II; Subject: Mathematics; University: Colgate University; Term: Unknown 1989;
Typology: Papers
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Solutions
dT /(T − 20) =
k dt we get ln(T − 20) = kt + C for some constant C. (We don’t need to write ln |T − 20 | because T will always be at least 20.) Now the initial conditions give
ln(75 − 20) = k · 0 + C =⇒ C = ln(55)
ln(30 − 20) = k · 4 + ln(55) =⇒ k =
ln
(Note that 10/ 55 < 1, so that k really is negative.) So we have
T = 20 + 55e(t/4) ln(10/55)^.
(a) When t = 7, T = 20 + 55e(−^7 /4) ln(55/10)^ ≈ 22 .8. (b) For T = 21, we get
1 = 55e(t/4) ln(10/55)^ =⇒ t = 4
ln(1/55) ln(10/55)
3 s 100 − t
We have gotten practice in the setting up of a differential equation model, but this result seems not to be separable, so we don’t have the means to solve it. (We should note that we don’t need to solve the differential equation to solve the problem, however: Because there are 100 − t gal of water in the tank after t minutes, after 100 min the tank is empty, so there is no salt in it. Obviously the creator of the problem — not me! — was a devious sort.) Let’s rewrite the original problem so that the resulting differential equation is separable:
2a. “A tank contains 100 gal of salt water at a concentration of 6 lb/gal. Salt water at a concentration of 1 lb/gal is pumped into the tank at a rate of 2 gal/min, and the well-stirred mixture runs out at the rate of 2 gal/min. How much salt is in the tank after t min?” Now there are always 100 gal of salt water in the tank, so if s again denotes the number of lb of salt in the tank after t minutes, then the 2 gal of salt water running out of the tank each minute take with it 2s/100 lb of salt. So a differential equation for s is
ds dt
2 s 100
Note that s(0) = 600 as before, and also that, because the water entering the tank has 1 lb/gal of salt, s is never less than 100. Now this differential equation is separable: ds/dt = (200 − 2 s)/100 = −(s − 100)/50, so
dx/(s − 100) =
(− 1 /50)dt gives ln(s − 100) = −t/50 + C, or s = 100 + eC^ · e−t/^50. And the initial condition gives eC^ = 500, so s = 100 + 500e−t/^50.