Math 103. Homework 6. Solutions
1. [Exercise 55, page 195] An economy's consumer price index
(CPI) is described by the function
|
I(t) = -(0.2) t3+ 3t2 +100, (0 £ t £ 10) |
|
where t=0 corresponds to 1989.
(a) At what rate was the CPI changing in 1994? In 1996? In 1999?
(b) What was the average rate of increase in the CPI over the
period from 1994 to 1999?
Solution.
(a) To compute rate of change of CPI you need the derivative,
Year 1994 corresponds to t=5, 1996 to t=7 and 1999 to
t=10. The CPI was changing at a rate of
I¢(5)=(-.6)(25)+6(5)=15, or 15 points/year, in 1994. In 1996 was
I¢(7) = 12.6 points/year. In 1999 was I¢(10)=0 points/year.
(b) The average rate of increase of the CPI over the period from 1994
to 1999 was
|
|
I(10)-I(5)
10-5
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= |
[-.2(1000)+3(100)+100]-[-.2(125)+3(25)+100]
5
|
= |
200-150
5
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=10, |
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or 10 points/year.
2. [Exercise 66, page 196] The percentage of U.S. population with
portable phones is projected to be
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P(t) = (24.4) t0.34 (1 £ t £ 10) |
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where t is measures in years and t=1 corresponds to 1998.
(a) What percentage of the U.S. population is expected to have
portable phones by the beginning of 2006?
(b) How fast is the percentage of the U.S. population with
portable phones expected to be changing at the beginning of 2006?
Solution. (a) Year 2006 corresponds to t=9, and
P(9)=(24.4)(9)0.34 = 51.5, or approximately 51.5%.
(b) By the Power Rule for derivatives,
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P¢(t) = (24.4)(0.34) t-0.66 |
|
Then P¢(9) = (24.4)(0.34) 9-0.66=1.946, or 1.95%
3. [Exercise 55, page 207]
From experience, the Emory Secretarial School knows that the average
student taking Advanced Typing will progress according to the rule
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N(t)= |
60 t+180
t+6
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(t ³ 0) |
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where N(t) measures the number of words per minute the student can
type after t weeks in the course.
(a) Find the derivative N¢(t).
(b) Compute N¢(t) for t=1,3,4 and 7 and interpret your
results.
(c) Sketch the graph of N(t). Does it confirm the results
obtained in part (b)?
(d) What will be the average student's typing speed at the end
of the 12 week course?
Solution.
(a) The derivative
|
N¢(t) = |
60(t+6) - (60t +180)
(t+6)2
|
= |
180
(t+6)2
|
|
|
(b) N¢(1)=180/49=3.7, N¢(3)=180/81=2.2, N¢(4)=180/100=1.8,
N¢(7)=180/169=1.1. The rate at which the average student is
increasing his or her speed one week, three weeks, four weeks, and
seven weeks is 3.7, 2.2, 1.8, and 1.1 words per minute, respectively
(c)
![[Graphics:Images/Mat1_gr_2.gif]](Images/Mat1_gr_2.gif)
(d)
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N(12)= |
60(12)+180
12+6
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=50 |
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or 50 words/minute.
4. The registrar of Kellogg University estimates that the total
number of students enrollment in the Continuing Education division
will be given by
|
N(t) = - |
20,000
(1+0.2t)1/2
|
+21,000 |
|
where N(t) denotes the number of students enrolled in the division t
years from now.
(a) Find an expression for N¢(t)
(b) How fast is the student enrollment increasing currently? 5
years from now?
(c) How many students are currently enrolled? How many will be
enrolled 5 years from now?
Solution.
(a) The derivative uses the chain rule. As you know, you can write
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N(t)=-20000(1+0.2 t)-1/2 +210000 |
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and so
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N¢(t) = -(1/2)(-20000)(1+0.2 t)-3/2 (0.2) = 2000(1+0.2 t)-3/2 |
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(b) ``Now corresponds to t=0. Thus N¢(0) = 2000, or the enrollment is currently increasing by 2000 students/year.
``5 years from now'' corresponds to t=5. Thus
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N¢(5) = 2000(1+0.2(5))-3/2 ~ 707 |
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or 707 students/year.
(c) The number of students currently enrolled is
|
N(0) = -20000(1+0.2(0))-1/2 +21000 = 1000 students |
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Five years from now
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N(5) = -20000(1+0.2(5))-1/2 +21000 = 6857.86 |
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or approximately 6856 students.
5. [Exercise 76, page 222] The quantity demanded per month, x,
of a certain make of PC is related to the average unit price, p (in
dollars), of PCs by the equation
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x=f(p) = |
100
9
|
(810,000-p2)1/2 |
|
It is estimated that t months
from now, the average price of a PC will be given by
|
p(t) = |
3200
8+t1/2
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+200 (0 £ t £ 60) |
|
dollars.
(a) What is the quantity demanded per month of the PCs 16 months from now?
25 months?
(b)
Find the rate at which the quantity demanded per month of the PCs will
be changing 16 months from now and 25 months from now.
Solution.
(a) The average price 16 mo from now is
|
p(16) = |
3200
8+(16)1/2
|
+200 = |
1400
3
|
|
|
Thus the quantity demanded 16 mo from now is
|
s=f(1400/3) = |
100
9
|
(810,000-(1400/3)2)1/2 = |
27
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=8550.66 |
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or approximately 8.550 PCs.
The average price 25 months from now is
|
p(16) = |
3200
8+(25)1/2
|
+200 = |
5800
13
|
|
|
and the quantity demanded 25 mo from now is
|
f(5800/13) = |
100
9
|
(810,000-(5800/13)2)1/2 = |
117
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= 8684.79 |
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or approximately 8,685 PCs.
(b) This is about the chain rule. We have x=f(p) and p=p(t). We want
But
and
|
|
dp
dt
|
= - |
1600
(8+t1/2)2 t1/2
|
|
|
Hence
|
|
dx
dt
|
= |
100p
|
|
1600
(8+t1/2)2 t1/2
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= |
160000p
|
9 |
| _________
Ö810000-p2
|
(8+Öt)2 Öt |
|
|
|
Thus, when
t=16, p=1400/3 and
|
|
dx
dt
|
= |
160000(1400/3)
|
9 |
| _______________
Ö810000-(1400/3)2
|
(8+ |
| __
Ö16
|
)2 |
| __
Ö16
|
|
= |
35000
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@ 18.71 |
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so the quantity
demanded 16 months from now will be changing at a rate of
approximately 19 PCs/month
When t=25, p(25)= 5800/13, and so
|
|
dx
dt
|
= |
160000p
|
9 |
| _______________
Ö810000-(5300/132
|
(8+ |
| __
Ö25
|
)2 |
| __
Ö25
|
|
= |
371200
|
@ 12.00 |
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so the quantity
demanded 25 months from now will be changing at a rate of
approximately 12 PCs/month