Transcript Applications of Exponential Change

Find the exponential function y  y0 e whose graph passes
through the two points.
Initial value: y  y 0  1.1
kt
0
 
y  1.1e
k  3

3 k
Other point: 3  1.1e
ln 3  ln 1.1e 
1
ln3  ln1.1  3k  k   ln1.1  ln 3
3
 ln1.1 ln 3t 3
0.3344 t
Equation:
Function:
y  1.1e
kt
 1.1e
Section 6.4b
Suppose that you just took a delicious hot pocket out
of the microwave. The tasty treat will gradually cool
to the temperature of the surrounding air…
As it turns out, the rate at which the hot pocket’s
temperature is changing at any given time is proportional to
the difference between its temperature and the temperature
of the surrounding medium!!!
This leads us to derive
NEWTON’S LAW OF COOLING
(which, incidentally, works for warming as well)
Let T be the temperature of the object in question at time t,
and Ts be the surrounding temperature. Then:
dT
 k T  TS 
dt
Since
dT  d T  TS  , we can rewrite the equation:
d
T  TS   k T  TS 
dt
By the law of exponential change, the solution is
T  TS  T0  TS  e
 kt
T  TS  T0  TS  e
 kt
where T 0 is the temperature at time t = 0.
A hard-boiled egg at 98 C is put in a pan under running 18 C
water to cool. After 5 minutes, the egg’s temperature is found to
be 38 C. How much longer will it take the egg to reach 20 ?
Define Variables:
Law of Cooling:
Substitute:
T0  98 TS  18
T  TS  T0  TS  e
 kt
T 18   98 18 e
 T  18  80e
 kt
 kt
A hard-boiled egg at 98 C is put in a pan under running 18 C
water to cool. After 5 minutes, the egg’s temperature is found to
be 38 C. How much longer will it take the egg to reach 20 ?
5 k
To find k, use the point (5, 38):
38  18  80e
e
5 k
 0.25
1
5k  ln 0.25   ln 4  k  ln 4
5
 0.2ln 4t
The Final Equation:
T  18  80e
 18  80  4
0.2 t
A hard-boiled egg at 98 C is put in a pan under running 18 C
water to cool. After 5 minutes, the egg’s temperature is found to
be 38 C. How much longer will it take the egg to reach 20 ?
T  18  80  4
0.2 t
0.2 t
20  18  80  4
ln 0.025
0.2 t
t
 13.305
0.025  4
0.2 ln 4
Solve analytically:
After about 13.305 minutes,
the egg will reach 20 degrees C.
In many situations, the resistance encountered by a
moving object (i.e., from friction) is proportional to the
object’s velocity…
That is, the slower the object moves, the less its forward
progress is resisted by the air through which it passes…
To model such a situation, we’ll start with Newton’s
Second Law of Motion…
The resisting force opposing the motion:
dv
Force = mass x acceleration  m
dt
If this force is proportional to the velocity, then:
dv
m   kv
dt
or
dv
k
 v
dt
m
 k  0
This is a differential equation of exponential change…
The solution:
v  v0e
  k m t
For a 50-kg ice skater, the k in the previous equation is about
2.5 kg/sec. How long will it take the skater to coast from 7 m/sec
to 1 m/sec? How far will the skater coast before coming to a
complete stop?
First, the general model:
v  7e
 2.5 50t
 7e
0.05 t
For a 50-kg ice skater, the k in the previous equation is about
2.5 kg/sec. How long will it take the skater to coast from 7 m/sec
to 1 m/sec? How far will the skater coast before coming to a
complete stop?
v  7e
0.05t
Now, we want the value of t when v = 1.
0.05 t
1  7e
0.05t
e
1 7
0.05t  ln 1 7 
ln 7
t
 38.918
0.05
The skater will reach 1 m/sec
  ln 7 from 7 m/sec after about
38.918 sec of coasting
For a 50-kg ice skater, the k in the previous equation is about
2.5 kg/sec. How long will it take the skater to coast from 7 m/sec
to 1 m/sec? How far will the skater coast before coming to a
complete stop?
v  7e
0.05t
To find distance, we need the integral of velocity:
s   7e
0.05t
0.05t
7e
dt 
C
0.05
0.05t
 140e
C
For a 50-kg ice skater, the k in the previous equation is about
2.5 kg/sec. How long will it take the skater to coast from 7 m/sec
to 1 m/sec? How far will the skater coast before coming to a
complete stop?
v  7e
s  140e
0.05t
0.05t
 C  140e
Assuming s = 0 when t = 0, we have
0  140e  C
 C  140
0
0.05t
 140
For a 50-kg ice skater, the k in the previous equation is about
2.5 kg/sec. How long will it take the skater to coast from 7 m/sec
to 1 m/sec? How far will the skater coast before coming to a
complete stop?
Finally, for distance:
s  140  140e
Find
0.05t
lim s  t   140
 140 1  e
0.05 t

t 
Mathematically, s never quite reaches 140. But
for all practical purposes, the skater comes to a
complete stop after traveling 140 m…
The distance traveled by a moving object that
encounters resistance proportional to its velocity:

v0 m
  k m t
s t  
1 e
k

The total distance traveled by this object:
v0 m
s
k
Suppose a battleship has mass around 51,000 metric tons
(51,000,000 kg) and a k value of about 59,000 kg/sec. Assume
the ship loses power when it is moving at a speed of 9 m/sec.
v  t   9e
59t 51,000
459, 000
59 t 51,000
s t  
1 e


59
(a) About how long will it take the ship’s speed to drop to 1 m/sec?
1  9e
59 t 51,000
51000 ln 1 9 
t
59
t  1899.296
The ship will reach 1 m/sec in
about 1899.296 seconds, or
in about 31.655 minutes
Suppose a battleship has mass around 51,000 metric tons
(51,000,000 kg) and a k value of about 59,000 kg/sec. Assume
the ship loses power when it is moving at a speed of 9 m/sec.
v  t   9e
59t 51,000
459, 000
59 t 51,000
s t  
1 e


59
(b) About how far will the ship coast before it is dead in the water?
459, 000
459, 000
59 t 51,000

lim s  t   lim
1 e


t 
t 
59
59
The ship will coast for a distance of about
7779.661 meters, or 7.780 kilometers