3.8 EXPONENTIAL GROWTH AND DECAY 1. This population grows according to the function
0.02 t
Exponential Growth Model f (t)=200 e , where t is measured in minutes. How many
bacteria are present in the population after 5 hours (300
Many systems exhibit exponential growth. These minutes)? When does the population reach 100,000 bacteria?
systems follow a model of the form
y(t) = y0e kt ,
We have. f (t)=200 e0.02 t . Then
Where, y0 represents the initial state of the system
0.02 ( 300 )
and k is a positive constant, called the growth constant. f (300)=200 e
Notice that in an exponential growth model, we have f (300) ≈ 80,686
' kt
y =k y o e =ky (t) There are 80,686 bacteria in the population after 5 hours.
That is, the rate of growth is proportional to the current To find when the population reaches 100,000 bacteria, we
function value. This is a key feature of exponential growth, solve the equation
involves derivatives and is called a differential equation. 0.02 t
100,000=200 e
Exponential Decay Model 0.02t
500=e
Exponential functions can also be used to model
populations that shrink (from disease, for example), or ln (500)=0.02 t
chemical compounds that break down over time. We say that ln ( 500)
such systems exhibit exponential decay, rather than t= =¿ . 310.73
0.02
exponential growth. The model is nearly the same, except
there is a negative sign in the exponent. Thus, for some The population reaches 100,000 bacteria after 310.73
positive constant k, we have minutes.
y= y 0 e−kt
As with exponential growth, there is a differential 2. According to experienced baristas, the optimal
equation associated with exponential decay. We have temperature to serve coffee is between 155°F and 175°F.
Suppose coffee is poured at a temperature of 200°F, and
y ' =−k y 0 e−kt =−ky after 2 minutes in a 70°F room it has cooled to 180°F. When
is the coffee first cool enough to serve? When is the coffee
too cold to serve? Round answers to the nearest half minute.
−kt
We have, T =( T 0−T a ) e +T a
180= ( 200−70 ) e−k (2 )+70
EXAMPLES
, −2 k −¿¿
180−70=130 e 85=130 e
11 −k (2) 85
=e ln =¿
13 130
11 85
ln =−2 k ln =¿
13 130
ln ( 11 )−ln (13)=−2k 17
ln =¿
26
−ln ( 11 )−ln (13)
=k 2¿¿
2
Then, the model now is t=5.09 min
T =130 e
−¿¿ The coffee is too cold to be served about 5 minutes after it is
poured.
The coffee reaches 175֯ F when
175=130 e−¿ ¿
175−70=130 e−¿¿
105 −¿¿
=e
130
21
ln =−¿
26
21
ln =−¿
26
21
ln =¿
26
2¿¿
t=2.5 min
The coffee can be served about 2.5 minutes after it is
poured.
The coffee reaches 155°F at
−¿ ¿
155=130 e
0.02 t
Exponential Growth Model f (t)=200 e , where t is measured in minutes. How many
bacteria are present in the population after 5 hours (300
Many systems exhibit exponential growth. These minutes)? When does the population reach 100,000 bacteria?
systems follow a model of the form
y(t) = y0e kt ,
We have. f (t)=200 e0.02 t . Then
Where, y0 represents the initial state of the system
0.02 ( 300 )
and k is a positive constant, called the growth constant. f (300)=200 e
Notice that in an exponential growth model, we have f (300) ≈ 80,686
' kt
y =k y o e =ky (t) There are 80,686 bacteria in the population after 5 hours.
That is, the rate of growth is proportional to the current To find when the population reaches 100,000 bacteria, we
function value. This is a key feature of exponential growth, solve the equation
involves derivatives and is called a differential equation. 0.02 t
100,000=200 e
Exponential Decay Model 0.02t
500=e
Exponential functions can also be used to model
populations that shrink (from disease, for example), or ln (500)=0.02 t
chemical compounds that break down over time. We say that ln ( 500)
such systems exhibit exponential decay, rather than t= =¿ . 310.73
0.02
exponential growth. The model is nearly the same, except
there is a negative sign in the exponent. Thus, for some The population reaches 100,000 bacteria after 310.73
positive constant k, we have minutes.
y= y 0 e−kt
As with exponential growth, there is a differential 2. According to experienced baristas, the optimal
equation associated with exponential decay. We have temperature to serve coffee is between 155°F and 175°F.
Suppose coffee is poured at a temperature of 200°F, and
y ' =−k y 0 e−kt =−ky after 2 minutes in a 70°F room it has cooled to 180°F. When
is the coffee first cool enough to serve? When is the coffee
too cold to serve? Round answers to the nearest half minute.
−kt
We have, T =( T 0−T a ) e +T a
180= ( 200−70 ) e−k (2 )+70
EXAMPLES
, −2 k −¿¿
180−70=130 e 85=130 e
11 −k (2) 85
=e ln =¿
13 130
11 85
ln =−2 k ln =¿
13 130
ln ( 11 )−ln (13)=−2k 17
ln =¿
26
−ln ( 11 )−ln (13)
=k 2¿¿
2
Then, the model now is t=5.09 min
T =130 e
−¿¿ The coffee is too cold to be served about 5 minutes after it is
poured.
The coffee reaches 175֯ F when
175=130 e−¿ ¿
175−70=130 e−¿¿
105 −¿¿
=e
130
21
ln =−¿
26
21
ln =−¿
26
21
ln =¿
26
2¿¿
t=2.5 min
The coffee can be served about 2.5 minutes after it is
poured.
The coffee reaches 155°F at
−¿ ¿
155=130 e
