Transcript Slide 1

Homework
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Homework Assignment #9
Review Section 5.8
Page 365, Exercises: 1 – 49(EOO)
Quiz next time
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
1.
A bacteria population P obeys the exponential growth law
P(t) = 2,000e1.3t (t in hours).
(a) How many bacteria are present initially?
P  0  2000e
1.3 0
 2000*1  2,000 bacteria
(b) At what time will there be 10,000 bacteria?
P  t   2000e1.3t  10000  e1.3t  5  ln e1.3t  ln 5
ln 5
1.3t  ln 5  t 
 1.238
1.3
There will be 10,000 bacteria at t  1.238 hours.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
5.
The decay constant of Cobalt–60 is 0.13 years–1. What is
its half-life?
ln 2 ln 2
k  0.13  half  life 

 5.332
k
0.13
half-life  5.332 years
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
9.
Find the solution to y ′ = 3y satisfying y(2) = 4.
32
y  3 y  y  Ce3t  y  2   4  Ce    C 
4
e6
C  0.009915  y  t   0.009915e3t
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
13.
Assuming that population growth is approximately
exponential, which of the two sets of data is most likely to represent
the population (in millions) of a city over a 5-year period?
Year 2000
Data I 3.14
Data II 3.14
2001
3.36
3.24
2002
3.60
3.54
2003
3.85
4.04
2004
4.11
4.74
Data II is most likely to represent the population of the city over the
five-year period, as Data I is almost linear.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
17.
A 10-kg quantity of radioactive isotope decays to 3-kg
after 17 years. Find the decay constant of the isotope.
P  t   10e kt  P 17   3  10e
k 17 
 ln 3  ln10  ln e
k 17 
ln 3  ln10
ln 3  ln10  17 k ln e  k 
 0.0708
17
Decay constant = 0.0708
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
21.
The atmospheric pressure P(h) (in psi) at a height h (in
miles) above sea level on earth satisfies a differential equation
P′ = – kP for some positive constant k.
(a) Measurements with a barometer show that P(0) = 14.7 and
P(10) = 2.13. What is the decay constant k?
P  h   Ce  kh  P  0   14.7  Ce
 k  0
 C  14.7
P  h   14.7e  kh  P 10   2.13  14.7e
 k 10 
ln 2.13  ln14.7  ln e
 k 10 
ln 2.13  ln14.7
ln 2.13  ln14.7  10k ln e   k 
 0.19317
10
Decay constant = 0.19317
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
21.
(b) Determine the atmospheric pressure 15 miles above
sea level.
P  h   14.7e 0.19317 h  P 15   14.7e
0.1931715
 0.811
Atmospheric pressure at 15 miles = 0.811 psi
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
25.
In 1965, Gordon Moore predicted that the number of
transistors on a microchip would increase exponentially.
Year
No. Trans,
(a)
Does the table of data
1971
2,250
confirm Moore’s prediction?
1972
2,500
If so, estimate the growth constant.
The data seems to support
Moore’s prediction.
1974
5,000
1978
29,000
1982
120,000
1985
275,000
1989
1,180,000
1993
3,100,000
1997
7,500,000
1999
24,000,000
2000
42,000,000
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
25.
(b) Plot the data in the table.
Year
No. Trans,
1971
2,250
1972
2,500
1974
5,000
1978
29,000
1982
120,000
1985
275,000
1989
1,180,000
1993
3,100,000
1997
7,500,000
1999
24,000,000
2000
42,000,000
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
25.
(c) Let N(t) be the number of transistors t years after
1971. Find an approximate formula N(t) ≈ Cekt, where t is the
number of years after 1971.
N  0   2250  N  t   2250e kt  N  7   29000  2250e
k 7
ln 29000  ln 2250  ln e    ln 29000  ln 2250  7 k ln e
ln 29000  ln 2250
k
 0.365
7
k 7
N 18   11800000  2250e
k 18 
 ln1180000  ln 2250  ln e
k 18 
ln1180000  ln 2250
k
 0.348
18
N  29   420000000  2250e
k  29 
 ln 42000000  ln 2250  ln e
ln 42000000  ln 2250
k
 0.339
29
k  29 
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
25.
(d) Estimate the doubling time in Moore’s Law for the
period 1971 – 2000.
ln 2
N  t   2250e
 doubling time 
 1.98
0.35
doubling time  2 years
0.35t
(e) If Moore’s Law holds to the end of the decade, how
many transistors will a microchip hold in 2010?
N  39   2250e
0.35 39 
 1,906, 787, 060
N  39   1,910, 000, 000
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
25.
(e) Can Moore expect his prediction to hold indefinitely?
Moore cannot expect his prediction to hold indefinitely, as a
some point transistors will get as small as they can, one or more
molecules, in size. That coupled with the finite length of the
connecting conductors will limit the miniaturization we have seen
over the last several decades.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
29. A certain quantity increases quadratically: P(t) =P0t2.
(a)
Starting at time t0 = 1, how long will it take for P to double
in size? How long will it take starting at t0 = 2 or 3?
P 1  P0 1  P0  P  t   2 P0  P0  t   t 2  2  t  2  1.414
2
2
P  2   P0  2   4 P0  P  t   8P0  P0  t   t 2  8  t  2 2  2.828
2
2
P  3  P0  3  9 P0  P  t   18P0  P0  t   t 2  18  t  3 2  4.243
2
2
(b)
In general, starting at time t0, how long will it take for P to
double in size?
In general, starting at t0 , it will take until t0 2 for the size to double.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
33. A bank pays interest at the rate of 5%. What is the yearly
multiplier, if interest is compounded
(a)
annually?
1.05
(b)
three times per year?
3
(c)

 0.05 
1 
  1.016
3 

continuously?

3
e0.05
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
37.
An investment increases in value at a continuously
compounded rate of 9%. How large must the initial investment
be to build up a value of $50,000 over a seven-year period?
A  Pe rt  50000  Pe
0.09 7 
 50000  Pe0.63
50000
P  0.63  26629.590  P  $26, 659.59
e
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
41.
If a company invests $2,000,000 to upgrade a factory, it
will earn additional profits of $500,000 per year. Is the
investment worthwhile, assuming an interest rate of 6%?
A  Pe rt  A  2000000e0.06 5  $2, 699, 717.61
A  500000e0.06 4  500000e0.06 3  500000e0.06 2 
500000e0.06 1  500000
 635624.57  598608.68  5637483.42  530918.27
+500000
 $2,828,899.94
Since there will be more money after five-years, the upgrades
are the better investment.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
45.
Use equation 3 to compute PV of an income stream
paying out R(t) = $5,000/yr continuously for ten yr at r = 0.05.
PV  0 R  t  e  rt dt
T
PV  5000 e
10
0
0.05t
dt  5000  20e
0.05t

 5000  12.131  20   $39,346.93
10
0
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 366
49.
Show that PV of an investment that pays out R dollars/yr
continuously for T years is R(1 – e–rt)/r where r is the interest
rate.
Using Equation 3,
T
  1  rT   1  r  0  
 1  rt 
PV   R  t  e dt  R   e   R    e     e

r
r
r

0
 


T
0
 rt
 1  e  rT 
 R

r


Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework




Homework Assignment #10
Review Sections 5.1 – 5.8
Page 369, Exercises: 1 – 97(EOO)
Quiz next time
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company