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# Assignment 7 Solutions.pdf

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Western University

Calculus

Calculus 1000A/B

Chris Brandl

Fall

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MATH 127 Fall 2012
Assignment 7 Solutions
1. (a) The half-life of morphine in the bloodstream is 3 hours. Suppose that there’s
initially 0.4 mg of morphine in the system, how long does it take until there’s
only 0.01 mg of morphine remaining in the bloodstream?
Let f(t) be the amount of morphine in milligrams in the bloodstream at time t
where t is in hours. Since the morphine undergoes exponential decay and there
is initially 0.4 mg in the system we have that f(t) = 0:4e . Since the half-life is
▯1▯
3k ln 2
3 hours we have that 0:2 = f(3) = 0:4e and from this we get that k = 3 .
ln 1
32)t
Thus, we have that f(t) = 0:4e . To ▯nd the amount of time it takes for
there to be only 0.01 mg left, we solve
ln(2)
0:01 = 0:4e 3 t
ln 1
0:01 32)t
0:4 = e
▯ ▯ ▯ 1▯
1 ln 2
ln = t
40 3
▯1 ▯
ln ▯ ▯
t = 3 1
ln 2
ln(40)
= 3
ln(2)
It would take 3 ln(40) hours for there to be only 0.01 mg of morphine remaining.
ln(2)
(b) Suppose that there is initially x gr0ms of Kool-Aid powder in a glass of water
and that it is dissolving at an exponential rate. After 1 minute there are 3 grams
remaining and after 3 minutes there is only 1 gram remaining. Find x and the 0
amount of Kool-Aid powder remaining after 5 minutes.
Let f(t) be the amount in grams of the Kool-Aid powder at time t where t is
in minutes. Since it is dissolving at an exponential rate and x is the0original
amount, we have that f(t) = x e . W0 also have that 3 = f(1) = x e and 0 k
1 1 = f(3) = x 0 . Dividing the two equations we get
f(1)
3 = = e▯2k
f(3)
ln(3) ln (3)
Solving for k we get k = ▯ and so, f(t) = x e ▯ 2 t. To ▯nd the initial
2 0
amount of Kool-Aid powder we solve
▯ ln (3)
3 = f(1) = x0e 2
ln2(3)
x0= 3e
p
= 3 3
p
There are initially 3 3 grams of Kool-Aid powder. To ▯nd the amount after 5
minutes we evaluate at t = 5.
p ▯5 ln (3)
f(5) = 3 3e 2
p
3 3
=
3 2
1
= 3
1
Thus, there are 3 grams of Kool-Aid powder left after 5 minutes.
(c) Suppose a cell culture undergoing exponential growth originally has 100 cells.
Further suppose that after 2 hours there are 300 cells. What will the population
be after 7 hours?
Let f(t) be the number of cells after t hours. Since the cells are growing expo-
nentially we have that f(t) = 100e . Now as there are 300 cells after 2 hours we
2k
get 300 = f(2) = 100e , solving for k we get
ln(3) p
k = = ln( 3)
2
ln (3)t
and so, f(t) = 100e . To ▯nd the population after 7 hours we evaluate at
t = 7.
p
f(7) = 100e 7 ln 3)
7
= 100(3 )
2 There are 100(3 ) cells after 7 hours.
2. At 2:00 pm boat A is 25 km north of boat B. If boat A is sailing 6 km/h west and boat
B is sailing north at 20 km/h, what is the rate at which the distance between the two
boats is changing at 2:30?
Drawing a picture we get
Boat A
x
z 25-y
Boat B
where x is the distance travelled by boat A and y is the distance travelled by boat
dx dy
B and so, = 6 and = 20. Let z be the distance between the two boats. We
dt dt
3 2 2 2
have that z = x + (25 ▯ y) . To ▯nd the rate at which the distance is changing
dz p p
we need to ▯nd and evaluate at x = 3, y = 10, and z = 3 + 15 = 234 (the
dt
distance travelled by each boat and the distance between the boats after 30 minutes).
Di▯erentiating we get
dz dx dy
2zdt = 2x dt ▯ 2(25 ▯ y)dt
dx dy
dz 2x dt▯ 2(25 ▯ y) dt
=
dt 2z
2(3)(6) ▯ 2(15)(20)
= p
2 234
282
= ▯ p
234
282
Thus, the distance between the boats is changing at a rate of ▯ p km/h at 2:30
234
pm (they’re getting closer).
3. A cone shaped funnel is 8 cm across the top and 12 cm deep. Water is
owing in at 60
cm =s and
owing out at 40 cm =s. How fast is the water level rising when the water
is 6 cm deep?
Drawing a picture we get
3
60 cm /s
8 cm
4 cm
r
12 cm
h
6 cm
40 cm /s
The diagram on the right shows that the cross section of the c

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