Continuous function with applications.pptx

Copyright © 2020, 2016, 2012 Pearson Education, Inc. Slide - 1
Chapter 2
Exponential and
Logarithmic
Functions

A LWAY S L E A R N I N G
2.5 Applications: Exponential Decay
OBJECTIVE
• Find a function that satisfies dP/dt = – kP.
• Convert between decay rate and half-life.
• Solve applied problems involving exponential decay.

The equation
shows P to be decreasing as a function of time,
and the solution
shows it to be decreasing exponentially. This is
exponential decay. The amount present
initially at time t = 0 is P0.
, where 0,
dP
kP k
dt
 
  0
kt
P t Pe


THEOREM 10
The decay rate k and the half–life T are related by
ln2 0.693147,
or
ln2 0.693147
,
and
ln2 0.693147
.
kT
k
T T
T
k k
 
 
 

Example 1: Plutonium-239, a common product of a
functioning nuclear reactor, can be deadly to people
exposed to it. Its decay rate is about 0.0028% per
year. What is its half-life?
T 


ln2
k
ln2
0.000028
24,755 years.

Quick Check 1
a.) The decay rate of cesium-137 is 2.3% per year. What is its
half-life?
b.) The half-life of barium-140 is 13 days. What is its decay rate?
ln 2
T
k

ln 2
0.023
 30.1 years.

ln 2
k
T

ln 2
13
 5.3% per day

0.053


Example 2: The radioactive element carbon-14 has a
half-life of 5730 yr. The percentage of carbon-14
present in the remains of plants and animals can be
used to determine age. Archaeologists found that the
linen wrapping from one of the Dead Sea Scrolls had
lost 22.3% of its carbon-14. How old was the
linen wrapping?

Example 2 (continued):
First find the decay rate, k.
Then substitute the information from the problem and
k into the equation
ln2 ln2
0.00012097
5730
k
T
  
0 .
kt
N N e


Example 2 (concluded):

0
0.777 N
 



0.00012097
0
t
N e
0.777 0.00012097t
e
ln0.777 0.00012097
ln t
e
ln0.777 0.00012097t

t
t

2086
ln0.777
0.00012097

The linen wrapping from the Dead Sea Scroll is about 2086 yr old.

Quick Check 2
How old is a skeleton found at an archaeological site if tests show
that it has lost 60% of its carbon-14?
First find the decay rate. We know from Example 2 that the decay
rate
Then use the information from the problem and into
ln 2 ln 2
0.00012097, or 0.012097% per year.
5730
k
T
  
k 0 .
kt
N N e


Quick Check 2 Concluded
0.00012097
0 0
0.4 t
N N e
 
0.00012097
0.4 t
e

0.00012097
ln0.4 ln t
e

ln0.4 0.00012097t

ln0.4
0.00012097
t


7575 t

Thus the skeleton is approximately 7575 years old.

Example 3: Following the birth of their
granddaughter, two grandparents want to make an
initial investment of P0 that will grow to $10,000 by
the child’s 20th
birthday. Interest is compounded
continuously at 4%. What should the initial
investment be?
We will use the equation 0 .
kt
P Pe


Thus, the grandparents must deposit $4,493.29, which
will grow to $10,000 by the child’s 20th
birthday.
10,000
10,000





0.04 20
0
Pe 
0.8
0
Pe
0.8
10,000
e 0
P
0
P
0
P
0.8
10,000e
4493.29

Quick Check 3
Repeat Example 3 for an interest rate of 6%
We will use the equation
Thus the grandparents must deposit $3011.94, which will grow to
$10,000 by the child’s 20th
birthday.
0 :
kt
P P e

0.06(20)
0
10,000 P e

1.2
0
10,000 P e

0
1.2
10,000
P
e

0
3011.94 P


THEOREM 11
The present value P0 of an amount P due t years later,
at an interest rate k, compounded continuously, is
given by
0 .
kt
P Pe


Newton’s Law of Cooling
The temperature T of a cooling object drops at a rate
that is proportional to the difference T – C, where C is
the constant temperature of the surrounding medium.
Thus,
The function that satisfies the above equation is
( ).
dT
k T C
dt
 
( ) .
kt
T T t ae C

  

Example 4: A body is found slumped over a desk in
a study. A coroner arrives at noon, immediately takes
the temperature of the body, and finds it to be 94.6.
She waits 1 hr, takes the temperature again, and finds it
to be 93.4. She also notes that the temperature of the
room is 70.
When was the murder committed?

We let t be the time elapsed, in hours, where t = 0
represents noon, when the coroner took the first
reading. Thus,
This gives ( ) 24.6 70.
kt
T t e
 
94.6 

0
70
k
ae 

24.6 a

To find the number of hours, N, since the murder
was committed, we must first find k. Use the fact
that at t = 1, the body’s temperature was 93.4°
Then, we can solve for k.
(1)
93.4 24.6 70
23.4 24.6
k
k
e
e


 


23.4
24.6
23.4
ln
24.6
 
  
 
0.05 

 k
e
23.4 24.6 k
e

k
k
So we now have
0.05
( ) 24.6 70.
t
T t e
 

Example 4 (concluded):
Determine the time of death, assuming the individual had a
normal body temperature, 98.6, at the time of death.



28.6
24.6
0.05t
e
28.6
ln
24.6
 
 
 
0.05t

3.01
 t
0.05
98.6 24.6 70.
t
e
 
Since t = 0 represents noon, the time of death was about 3 hours
earlier, or at about 9:00 a.m.

Section Summary
• The decay rate, k, and the half-life, T, are related by
or and
• The present value of an amount P due t years later, at
an interest rate k, compounded continuously, is given by
ln2,
kT 
ln2
k
T

ln2
.
T
k

0
P
0 .
kt
P Pe


Section Summary Concluded
• According to Newton’s Law of Cooling, the temperature
T of a cooling object drops at a rate proportional to the
difference T - C, when C is the constant temperature of
the surrounding medium. Thus, we have
for
( ),
dT
k T C
dt
  0,
k  and ( ) .
kt
T t ae C

 

Continuous function with applications.pptx

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