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Quiz #11
Applied Calculus II – Math 132.008 – Spring 2011
solutions
dz
= z + zt2 which satisfies z = 5 when t = 0.
dt
This is a separation of variables problem. The right-hand side is a product of z’s and t’s as
follows: z(1 + t2 ).
Z
Z
dz
= (1 + t2 )dt
z
1. (2 pts) Solve the differential equation
ln z = t + t3 /3 + C
3 /3
z(t) = Aet+t
3 /3
Now, z(0) = 5, so A = 5 and z(t) = 5et+t
.
.
2. (2 pts) The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of Q0 is injected
directly into the blood, 20% is left in the blood after three hours.
(a) Write a differential equation for the quantity Q of the drug in the blood after t hours.
dQ/dt = −kQ
for some positive constant k.
(We solve for k below.)
(b) Suppose the initial dose is 100mg. Find the solution to the differential equation you
wrote in (a). Your answer should not involve any undetermined constants.
The general solution to the above equation is Q(t) = Q0 e−kt , where Q0 is the initial
quantity. We know that 0.2Q0 = Q(3) = Q0 e−k(3) , so solve for k and get k = 0.536 .
Thus Q(t) = 100e−0.536t .
(c) How much of the drug is in a patient’s bloodstream after 7 hours if the patient is given
100mg initially? Round your answer to the nearest milligram.
Simply compute Q(7) ≈ 2.34 mg.
2
The following will be useful in answering the questions below.
Recall that if A and k are constants, the solution to the differential equation
dy
= k(y − A) is y = A + Cekt ,
dt
where C is an arbitrary constant.
3. (2 pts) At 1:00p.m. one winter afternoon, there is a power failure at your house in Calgary,
and your heat does not work without electricity. When the power goes out, it is 68◦ F in
your house. At 10:00p.m. it is 57◦ in the house. Assume that the outside temperature has
been constantly 10◦ during this period. The power is not restored until 10:00a.m. the next
morning. It is known that the rate of change of the temperature in your house is proportional
to the difference between the temperature in your house and the temperature outside.
What is the temperature in your house when the power is restored?
We know that A is the equilibrium temperature, or 10◦ . We need to determine C. We have
68 = y(0) = 10 + Cek·0 ,
so C = 58.
Next, we know that
57 = y(9) = 10 + 58ek·9 ,
so k = −0.02337.
Finally, we need y(21), so
y(21) = 10 + 58e−0.02337(12) ≈ 45◦ .