Transcript T - mlgibbons

Continuous Change  P  t   P0 e
kt
k  0  P  t  grows exponentially & k is called is called the growth constant.
A quantity that decreases exponentially is said to have exponential decay.
P  t   P0 e  kt (with k  0) k is then called the decay constant.
The constant k has units of
“inverse time”; if t is measured in
days, then k has units of (days)−1.
In the laboratory, the number of Escherichia coli bacteria grows
exponentially with growth constant of k = 0.41 (hours)−1. Assume that
1000 bacteria are present at time t = 0.
(a) Find the formula for the number of bacteria P(t) at time t.
P  t   P0e
kt
P  t   1000e
0.41t
(b) How large is the population after 5 hours?
P  5   7768  The # of bacteria is a whole # 
(c) When will the population reach 10,000?
10, 000  1000e
 10  e
ln10  0.41t  t  5.616 h
0.41t
0.41t
The important role played by exponential functions is best
understood in terms of the differential equation y ' = ky. The
function y = P0ekt satisfies this differential equation, as we can
check directly:
kt
y  P0 e
d
kt
kt
y '   P0 e   kP0e  ky
dt
Theorem 1
y (t ) is a differentiable function satisfying the differential equation
y '  ky  y (t )  P0e kt , where P0 is the initial value P0  y (0).
Find all solutions of y ' = 3y. Which solution satisfies y(0) = 9?
y (t )  9e
3t
y '  ky  y (t )  P0e kt
Modeling Penicillin Pharmacologists have shown that penicillin leaves
a person’s bloodstream at a rate proportional to the amount present.
(a) Express this statement as a differential equation.
A '  t   kA  t 
 A  t  is decreasing  k  0 
(b) Find the decay constant if 50 mg of penicillin remains in the
bloodstream 7 hours after an initial injection of 450 mg.
y (t )  P0 e
 kt
(c) At what time was 200 mg
of penicillin present?
200  450e
0.3139 t
 50  450e
 k 7
1
7 k
 e
9
ln 1/ 9 
k 
 0.3139
7
ln  4 / 9 
20
0.3139 t

e
t 
 2.583 h
45
0.3139
Quantities that grow exponentially possess an important property:
There is a doubling time T such that P (t) doubles in size over every
time interval of length T. To prove this, let P (t) = P0ekt and solve for T
in the equation P (t + T) = 2P (t).
P0 e
e
k  t T 
k  t T 
 2e
kt kT
 2e
e e
e
 2 P0 e
kT
kt
kt
QED
kt
 2  kT  ln 2  T   ln 2  / k
Doubling Time
P(t )  P0 e with k  0
kt
 the doubling time of P is
ln 2
T
k
Spread of the Sapphire Worm A computer virus nicknamed the
Sapphire Worm spread throughout the Internet on January 25, 2003.
Studies suggest that during the first few minutes, the population of
infected computer hosts increased exponentially with growth constant
k = 0.0815 s−1.
(a) What was the doubling time of the virus?
ln 2
T
 8.505 s
0.0815
(b) If the virus began in four computers, how many hosts were
infected after 2 minutes? After 3 minutes?
P(t )  P0 e kt  P(t )  4e kt , P 120   4e
P 180   4e
0.0815120 
 70, 700
0.0815180 
 9.4 million
Half-life
P(t )  P0 e
 kt
with k  0
 the half-life of P is
ln 2
T
k
The isotope radon-222 decays exponentially with a half-life of 3.825
days. How long will it take for 80% of the isotope to decay?
ln 2
ln 2
T
k 
 0.1812
k
3.825
0.1812 t
0.1812 t
 R  t   R0 e
 0.2 R0  R0e
ln 0.2
t 
 8.882 days
0.1812
Carbon dating relies on the fact that all living organisms contain carbon
that enters the food chain through the carbon dioxide absorbed by
plants from the atmosphere. Carbon in the atmosphere is made up of
nonradioactive C12 and a minute amount of radioactive C14 that decays
into nitrogen. The ratio of C14 to C12 is approximately Ratm = 10−12.
The carbon in a living organism has the same ratio Ratm because this
carbon originates in the atmosphere, but when the organism dies, its
carbon is no longer replenished. The C14 begins to decay exponentially
while the C12 remains unchanged. Therefore, the ratio of C14 to C12 in
the organism decreases exponentially. By measuring this ratio, we can
determine when the death occurred. The decay constant for C14 is
k = 0.000121 yr−1, so
Ratio of C to C after t years  Ratm e
14
12
0.000121t
Cave Paintings In 1940, a remarkable gallery of prehistoric animal
paintings was discovered in the Lascaux cave in Dordogne, France. A
charcoal sample from the cave walls had a C14-to-C12 ratio equal to
15% of that found in the atmosphere. Approximately how old are the
paintings?
Ratio of C to C after t years  Ratm e
14
12
0.15  e
0.000121t
0.000121t
ln 0.15
t 
 15, 678.678 years
0.000121