Quiz II
1. The differential form ω = (y cos(xy) − x)dx + (x cos(xy) + y)dy is exact. Find
all functions F with dF = ω.
Since ∂F/∂x = y cos(xy) − x, we have
F (x, y) =
and that makes
Z
(y cos(xy) − x)dx + g(y) = sin(xy) − x2 /2 + g(y),
∂F
= x cos(xy) + g ′ (y).
∂y
Since ∂F/∂y = x cos(xy) + y, we have g ′ (y) = y and g(y) = y 2 /2 + C. So the
answer is F = sin(xy) − x2 /2 + y 2 /2 + C.
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2. A tank contains 50 gal of pure water. Starting at time zero a salt-water solution
containing 0.4 lb/gal of salt enters the tank at a rate of 2 gal/min. At the same
time a drain is opened at the bottom of the tank allowing the salt solution to leave
at a rate of 2 gal/min. Assume that the solution is kept perfectly mixed at all
times. How long will it take for the amount of salt (in lbs) in the tank to reach 10
lbs? Leave your answer in exact form (in terms of logarithms).
The volume of liquid in the tank is constant at 50 gal. Letting S(t) be the
amount of salt in the tank at time t (in lbs), we have
S(t) lbs
dS
= [rate in] − [ rate out] = (0.4 lbs/gal)(2 gal/min) − (
)(2 gal/min).
dt
50 gal
Rearranging and simplifying that gives you
S
dS
+
= 0.8.
dt
25
The integrating factor for that equation is just et/25 , but this is a good place to
solve the equation by guessing a particular solution, and adding on the general
solution to the homogeneous equation. Note that S ≡ 20 is a particular solution,
so the general solution to the inhomogeneous equation is S(t) = 20 + C exp(−t/25).
Since S(0) = 0, C = −20. So to answer the problem you have to solve 10 =
20(1 − exp(−T /25) for T . That means
e−T /25 =
1
which leads right to T = 25 ln 2.
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