Transcript Ch 6.4 Exponential Growth & Decay Calculus Graphical

Ch 6.4 Exponential Growth & Decay
Calculus Graphical, Numerical, Algebraic by
Finney Demana, Waits, Kennedy
Solving by Separation of Variables
dy
Solve for y if
= x 2  y 2 and y = 3 when x = 0.
dx
Law of Exponential Change
dy
= ky,
dt
The differential equation that describes growth is
where k is the growth constant (if positive) or decay constant (if
negative). We can solve this equation by separating the
variables.
dy
= ky
dt
dy
= k dt
y
dy
 y =  k dt
ln |y| = k t + C
separate the variables
integrate both sides
| y | = e kt + C
exponentiate both sides
| y | = eC e kt
property of exponents
y = A e kt
Let A =  eC
v
Continuously Compounded Interest
If A0 dollars are invested at a fixed annual rate r (as a
decimal) and compounded k times a year, the amount of
money present after t years is:
r

A(t) = A0 1 + 
k

kt
If the interest is compounded continuously (or every
instant) then the amount of money present after t years
is:
r

A(t) = lim A0 1 + 
k 
k

kt
= A0 e
rt
Finding Half Life
Find the half-life of a radioactive substance with decay equation
y = y0 e  k t
The half-life is the solution to the equation y0 e
Solvea lg ebraically :
e
 k t
1
=
2
1
2
1
1
ln 2
t = - ln =
k
2
k
Half life of an element:
ln 2
t=
k
- k  t = ln
 k t
1
= y0 .
2
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?
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?
1. y = y 0 e kt
y = 10 e kt
50 = 10 e 30k
ln 5
=k
30
(0,10), (30,50)
find t when y = 100
2. y = 10 e
100 = 10 e
ln 5
t
30
ln 5
t
30
ln 5
t
30
10 = 10 e
30 ln 10
t=
= 42.920 days
ln 5
Using Carbon-14 Dating
Scientists who use carbon-14 dating use 5700 years for
its half-life. Find the age of a sample in which 10% of
the radioactive nuclei originally present have decayed.
Using Carbon-14 Dating
Scientists who use carbon-14 dating use 5700 years for
its half-life. Find the age of a sample in which 10% of
the radioactive nuclei originally present have decayed.
1
A = A0  
2
t
5700
1
.9 A 0 = A 0  
2
1
.9 =  
2
t
5700
t
5700
t
1
ln (.9) =
ln
5700
2
ln .9
t = 5700
= 866 yrs old
ln .5
Newton’s Law of Cooling
Newton 's Law of Cooling: The rate at which an object's
temperature is changing at any time is proportional to the
difference between its temperature and the temperature of the
surrounding medium. If T = temperature at time t, and Ts is
the surrounding temperature, then:
dT
= -k  T - Ts 
dt
dT
  T - Ts  = -  k dt
ln T - Ts = -k  t + C
T - Ts = e kt + C
T - Ts = C e  kt
at t = 0, C = T0 - Ts
T - Ts =  T0 - Ts  e  kt
T=
 T0 - Ts  e kt
 Ts
Newton’s Law of Cooling
Newton's Law of Cooling:
For T0 = initial temperature of object
Ts = temperature of surrounding environment
k = constant of variation, found by using given
Then,
T = T0 - Ts   e kt + Ts
Using Newton’s Law of Cooling
A hard boiled egg at 98ºC is put in a pan under running
18ºC water to cool. After 5 minutes, the egg’s temperature
is found to be 38ºC. How much longer will it take the egg to
reach 20ºC?
Using Newton’s Law of Cooling
A hard boiled egg at 98ºC is put in a pan under running 18ºC
water to cool. After 5 minutes, the egg’s temperature is found
to be 38ºC. How much longer will it take the egg to reach
20ºC?
T = 18 and T = 98, then
s
0
T - 18 = (98 - 18) e- k t
T = 18 + 80 e- k t
Find k by using T = 38 when t = 5 to get:
38 = 18 + 80 e- 5k
e- 5k =
1
4
5 k = ln
1
= - ln 4
4
1
ln 4
5
so now, T = 18 + 80 e(-.2 ln 4) t
k=
Solve 20 = 18 + 80 e(-.2 ln 4) t to get t = 13.3 min
13.3 - 5 = 8.3 seconds longer