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1. A thermometer reading 75◦F is taken outdoors where the temperature is
20 ◦F. The reading is 30◦F four minutes later. Find
(a) the thermometer reading after seven minutes outdoors, and
(b) the time outdoors until the reading is 21
F.
2. 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 3 gal/min.
How much salt is in the tank after 100 min?
38. (From Stewart, page 608) A certain small country has $10 B in paper
currency in circulation, and each day $50 M comes into the country’s
banks. The government decides to introduce new currency by having the
banks replace the old bills with new ones whenever old currency comes
into the banks. Let x = x(t) denote the amount of new currency in
circulation at time t (in units of $10 M), with x(0) = 0.
(a) Formulate a mathematical model in the form of an initial value prob-
lem that represents the “flow” of the new currency into circulation.
(b) Solve the initial value problem in part (a).
(c) How long with it take for the new bills to account for 90% of the
currency in circulation?
Solutions
- Let T be the thermometer’s temperature reading in degrees Farenheit after t minutes outdoors. Then T (0) = 75 and T (4) = 30, and using Newton’s Law of Cooling, we have dT /dt = k(T − 20) for some negative constant k. From
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)
- Let s denote the number of lb of salt in the tank after t minutes. Then s(0) = 100 · 6 = 600. In order to find the differential equation for s, we note that, after t minutes, there are 100 − t gallons of salt water in the tank, so the number of pounds per gallon of salt water at that time is s/(100 − t). Thus, the 3 gal of salt water that run out of the tank each minute take with it 3s/(100 − t) lb of salt. Because the 2 gal of salt water coming into the tank each minute bring 2 lb of salt with it, here is a differential equation for s: ds dt
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.
- (a) Because 1000 units of paper currency in circulation at any time, when there are x new currency in circulation, it constitutes x/1000 of all currency; so we may assume that (1 − 1000 x )5 of the
