6.4 exponential growth and decay. the number of bighorn sheep in a population increases at a rate...
Post on 20Jan2018
214 views
Embed Size (px)
DESCRIPTION
The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interestbearing account. If the rate of change is proportional to the amount present, the change can be modeled by:TRANSCRIPT
6.4 Exponential Growth and Decay
The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.)
So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interestbearing account.
If the rate of change is proportional to the amount present, the change can be modeled by:
Rate of change is proportional to the amount present.
Divide both sides by y.
Integrate both sides.
Integrate both sides.
Exponentiate both sides.
When multiplying like bases, add exponents. So added exponents can be written as multiplication.
Exponentiate both sides.
When multiplying like bases, add exponents. So added exponents can be written as multiplication.
Since is a constant, let .
Since is a constant, let .
At , .
This is the solution to our original initial value problem.
Exponential Change:
If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay.
Note: This lecture will talk about exponential change formulas and where they come from. The problems in this section of the book mostly involve using those formulas.
Modeling Growth
At the beginning of summer, the population of a hive of hornets is growing at a rate proportional to the population. From a population of 10 on May 1, the number of hornets grows to 50 in thirty days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100?
Continuously Compounded Interest
If money is invested in a fixedinterest account where the interest is added to the account k times per year, the amount present after t years is:
If the money is added back more frequently, you will make a little more money.
The best you can do is if the interest is added continuously.
Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine.
We could calculate:
but we wont learn how to find this limit until chapter 8.
Since the interest is proportional to the amount present, the equation becomes:
(The TI89 can do it now if you would like to try it.)
Continuously Compounded Interest:
You may also use:
which is the same thing.
Radioactive Decay
The equation for the amount of a radioactive element left after time t is:
This allows the decay constant, k, to be positive.
The halflife is the time required for half the material to decay.
Halflife
Halflife:
Using Carbon14 Dating
Scientists who use carbon14 dating use 5700 years for its halflife. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed.
Newtons Law of Cooling
Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air.
(It is assumed that the air temperature is constant.)
If we solve the differential equation:
we get:
p
Newtons Law of Cooling
where is the temperature of the surrounding medium, which is a constant.
Newtons Law of Cooling
Newtons Law of Cooling
Using Newtons Law of Cooling
A hard boiled egg at 98C is put in a pan under running 18C water to cool. After 5 minutes, the eggs temperature is found to be 38C. How much longer will it take the egg to reach 20C?
1
dykdt
y
=
dy
ky
dt
=
1
dykdt
y
=
ln
yktC
=+
C
eA
=
C
e
Ckt
yee
=
kt
yAe
=
Ckt
yee
=
ln
y
ktC
ee
+
=
0
t
=
0
yy
=
0
0
k
yAe
=
0
yA
=
0
kt
yye
=
1
0
lim1
kt
k
r
A
k
+
(
)
0
1
kt
r
AtA
k
=+
0
rt
AAe
=
rt
APe
=
00
1
2
kt
yye

=
0
kt
yye

=
(
)
1
lnln
2
kt
e

=
ln1ln2
kt
=
0
ln2
kt
=
ln2
t
k
=
ln2
halflife
k
=
[
]
0
kt
ss
TTTTe

=
[
]
s
dT
kTT
dt
=
s
T
kt
kt
30k
ln5
t
30
ln5
t
0
30
1.y
y
50
ln
30
= y e
= 10 e
= 10 e
5
2.
y = 10 e
100 = 10 e
(0,10), (
=
30,5
0)
find t when y
=
10
0
k
10
ln5
t
30
30
ln
= 10 e
ln 10
t =
= 42.920 days
5
t
5700
0
t
5700
00
t
5700
1
A = A
2
1
.9 A = A
2
1
.9 =
2
t1
ln (.9) = ln
57002
ln .9
t = 5700 = 866 yrs old
ln .5
Newton's Law of Cooling: The rate at wh
ich an object's
temperature is changing at any time is p
roportional to the
difference between its temperature and t
he temperature of the
surrounding medium. If
(
)
(
)
s
s
s
s
kt + C
s
kt
s
T = temperature at time t, and T is
the surrounding temperature, then:
dT
= k T  T
dt
dT
=  k dt
T  T
ln T  T = kt + C
T  T = e
T  T = C e


(
)
(
)
kt
s0s
kt
0ss
0s
T  T = T  T e
at t = 0, C = T  T
T = T 
T
e
T


+
(
)
0
0
Newton's Law of Cooling:
For T = initial temperature of object
T = temperature of surrounding en
vironment
k = constant of variation, found
by using given
Then,
T = 
Te +

kt
s
s
T
T
s
s0
 k t
 k t
 5k
 5k
(.2 ln 4
T = 18 and T = 98, then
T  18 = (98  18) e
T = 18 + 80 e
Find k by using T = 38 when t = 5 to get
:
38 = 18 + 80 e
1
e =
4
1
5 k = ln =  ln 4
4
1
k = ln 4
5
so now, T = 18 + 80 e

) t
(.2 ln 4) t
Solve 20 = 18 + 80 e to get t = 13.3 min
13.3  5 = 8.3 seconds longer
Recommended