a quantity that decreases exponentially is said to have exponential decay
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A quantity that decreases exponentially is said to have exponential decay. The constant k has units of “inverse time”; if t is measured in days, then k has units of (days) −1. - PowerPoint PPT PresentationTRANSCRIPT
0Continuous Change ktP t P e 0 grows exponentially & is called is called the growth constant.k P t k
A quantity that decreases exponentially is said to have exponential decay.
0 (with 0) is then called the decay constant.ktP kt P ke
The constant k has units of “inverse time”; if t is measured in days, then k has units of (days)−1.
0.41 0.41
5.6110,000 1000 10ln1 1 6 h0 0.4
t tet
et
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.
0ktP t P e 0.411000 tP t e
(b) How large is the population after 5 hours?
The # of bacteria is a 77 wh68 5 ole #P (c) When will the population reach 10,000?
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:
'
0
0 0'
kt
kt kt
y P edy P e kP e kydt
0 0 0
( ) is a differentiable function satisfying the differential equation
' ( ) ,
The
where is the i
o
n
rem
itial value (0).
1
kt
y t
y ky y t P e P P y
Find all solutions of y = 3y. Which solution satisfies y(0) = 9?'
0' ( ) kty ky y t P e
3( ) 9 ty t e
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 is decreasing 0A t k (b) Find the decay constant if 50 mg of penicillin remains in the bloodstream 7 hours after an initial injection of 450 mg.
0( ) kty t P e
7 7
0.
150 4
3139
509
ln 1/ 97
k ke e
k
(c) At what time was 200 mg of penicillin present?
0.3139
0.3139
200 450
2.5ln 4 / 920
4583
0.3h
13
9
t
t
e
e t
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).
0 0
l
2
2
nl 2
2
n 2 /2
k t T kt
k t T kt
kt kT kt
kT
P e P e
e e
e e e
e kT T k
QED
0( ) with 0the doubling time of is
Doubling Tim
e
ktP t P e kP
ln 2Tk
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.
ln 20.081
8.5 5 s5
0T
(a) What was the doubling time of the virus?
(b) If the virus began in four computers, how many hosts were infected after 2 minutes? After 3 minutes?
0.0815 1200 70,7( ) ( ) 4 , 4 0120 0kt ktP t P e P t e P e
0.0815 180 9.4 18 mi0 4 llionP e
0( ) with 0the half-life of is
Half-lif
e
ktP t P e kP
ln 2Tk
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?
0.1812 0.18120 0 0
ln 2 0.18123.825
0.
l
8.882 days
2ln
n
0.20.1812
2
t t
k
R t R e R R e
t
Tk
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
14 12 0.000121atmRatio of to after years tC C t R e
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?
14 12 0.000121atmRatio of to after years tC C t R e
0.000121
ln 0.150.0001
15,678.678 years
0
2
1
1
. 5 t
t
e