Mathematics M119
A Brief Survey of Calculus I
Fall 2012
Answers to even numbered recommended problems
Section
Problem
Answer
1.1
12
a. f (0) = 2
b. f (3) = 11
c. x = ±3
d. No, as y = x2 + 2 ≥ 2
a. f (3) = 0.14.
This means that after 3 hours, the level of nicotine is 0.14 mg
b. About 4 hours.
c. The vertical intercept is 0.4. It represents the level of nicotine
in the blood right after the cigarette is smoked
d. A horizontal intercept would represent the value of t when
N = 0, or the number of hours until all nicotine is gone from the
body.
a. The original deposit is the balance, B, whent = 0, which is the
vertical intercept. The original deposit is $1000.
b. It appears that f (10) ≈ 2200. The balance in the account after
10 years is about $2200.
c. When B = 5000, it appears that t ≈ 20. It takes about 20 years
for the balance in the account to reach $5000
a. Since N = f (h), the statement f (500) = 100, means that if
h = 500, N = 100. This tells us that at an elevation of 500 feet
above the sea level, there are 100 species of bats.
b. The vertical intercept k is on the N -axis, so it represents the
value of N when h = 0. The intercept, k, is the number of bat
species at sea level. the horizontal intercept, c, represents the
value of h when N = 0. The intercept, c, is the lowest elevation
above which no bats are found.
14
16
18
1.2
10
28
1.3
10
a. Lines l2 and l3 are parallel and thus have the same slope. Of
these two, l2 has a larger y-intercept.
b. Lines l1 and l3 have the same y-intercept. Of these two, line
l1 has a larger slope
Both formulas are linear, with slope -1, which means that in one
year there is a decrease of one beat per minute, so (c) is correct.
The function is increasing and concave up between D and E, and
between H and I. It is increasing and concave down between A
and B, and between E and F . It is decreasing and concave up
between C and D, and between G and H. Finally, it is
decreasing and concave down between B and C, and between F
and G.
1
Section
Problem
Answer
1.3
24
a. The average rate of change R is the difference in the amounts
divided by the change in time.
831−1517
≈ −98 million pounds/yr
R = 2003−1996
This means that in the yeas between 1996 and 2003, the
production of tobacco decreased at a rate of approximately
98 million pounds per year.
b. To have a positive rate of change, the production has to
increase during one of these 1 year intervals. Looking at the data,
we can see that between 1996 and 1997, production of tobacco
increased by 270 million pounds. Thus, the average rate of change
is positive between 1996 and 1997.
30
We have Average rate of change =
73, 365, 880 − 12, 168, 450
2003 − 1977
= 2, 353, 747.3 households/year.
During this 26-year period, the number of US households with
cable television increased, on average, by
2,353,747.3 households per year.
1.4
12
We know the fixed cost is the cost that the company would have
to pay if no items were produced. Thus
Fixed cost = C(0) = $5, 000.
We know that the cost function is linear and we know that the
slope of the function is exactly the marginal cost. Thus
= 20
= $4 per unit produced.
Slope = Marginal Cost = 5020−5000
5−0
5
Thus, the marginal cost is $4 per unit produced. We know that
since C(q) is linear, thus C(q) = 4 · q + 5000.
1.5
2
a. Initial amount =100; exponential growth; growth rate=7%=0.07.
b. Initial amount=5.3; exponential growth; growth rate=5.4%=0.054.
c. Initial amount=3500; exponential decay; decay rate=-7%=-0.07.
d. Initial amount=12; exponential decay; decay rate=-12%=-0.12.
a. The function is linear with initial population of 1000 and slope
of 50, so P = 1000 + 50t.
b. The function is exponential with initial population of 1000 and
growth rate of 5%, so P = 1000(1.05)t .
4
8
a. We see from the base 1.0126 that percent rate of growth is
1.26% per year.
b. The initial population(in the year 2004) is 6.4 billion people.
When t = 6, we have P = 6.4(1.0126)6 = 6.90. The predicted
world population in the year 2010 is 6.90 billion people.
c. We have
6.90−6.4
Average rate of change = 4P
= 2010−2004
= 0.083 billion people per yr.
4t
The population of the world is increasing by about 83 million people
per year.
2
Section
Problem
Answer
1.5
20
Use an exponential function of the form P = P0 at for the number
of passengers. At t = 0, the number of passengers is P0 = 190, 205,
so P = 190, 205at .
After t = 5 yrs, the number of passengers is 174,989, so
174, 989 = 190, 205a5
a5 = 0.920002
a = 0.983
Then P = 190, 205(0.983)t , which means the annual percentage
decrease over this period is 1 − 0.983 = 0.017 − 1.7%.
Let y represent the number of zebra mussels t years since 2003.
= 486 and the vertical intercept
a. The slope of the line is 3186−2700
1
is 2700. We have y = 2700 + 486t. The slope of the line, 486 zebra
mussels per year, represents the rate at which the zebra mussel
population is growing
b. When t = 0, we have y = 2700, so the exponential function is in
the form y = 2700at . When t = 1, we have y = 3186, and we
substitute to get 3186 = 2700a1 , so a = 1.18. The exponential
function is y = 2700(1.18)t and the zebra mussel population is
growing at 18% per year.
22
pp 480-484
4
6
Use a graphic calculator.
Suppose P0 is the initial deposit, we have P = P0 e0.06t .
After one year, P = P0 e0.06 = P0 (1.0618). So the effective annual
rate is 6.18%
1.6
22
a. (i) P = 1000(1.05)t ; (ii) P = 1000e0.05t
b. (i) 1629; (ii) 1649
a. Substituting t = 0.5 in P (t) gives
P (0.5) = 1000e−0.5(0.5)
≈ 1000(0.7788)
≈ 779
trout.
Substituting t = 1 in P (t) gives
P (0.5) = 1000e−0.5
≈ 1000(0.6065)
≈ 607
trout.
b. Substituting t = 3 in P (t) gives
P (0.5) = 1000e−0.5(3)
≈ 1000(0.2231)
≈ 223
trout.
This tells us that after 3 years the fish population has gone down to
about 22% if the initial population.
32
3
Section
Problem
Answer
1.6
32
c. We are asked to find the value of t such that P (t) = 100. That is,
we asked to find the value of t such that
100 = 1000e−0.5t . Solving this gives
100 = 1000e−0.5t
0.1 = e−0.5t
ln(0.1) = −0.5t
t = ln(0.1)
≈ 4.6.
−0.5
Thus, after roughly 4.6 years, the trout population will have
decrease to 100.
d. A graph of the population is given below:
Looking at the graph, we see that as time goes on there are fewer
and fewer trout in the pond. A possible cause for this may be the
presence of predators, such as fishermen. The rate at which the
fish die decrease due to the fact that there are fewer fish in the
pond, which means that there are fewer fish that have to try to
survive. Notice that the survival chances of any given fish remains
the same, it is just the overall mortality that is decreasing.
1.7
2
4
Every 2 hours, the amount of nicotine remaining reduced by 12 .
Thus, after 4 hours, the amount of 12 of the amount present after 2
hours.
t(hours)
0
2
4
6
8
10
Nicotine(mg) 0.4 0.2 0.1 0.05 0.025 0.0125
From the table it appears that it will take just over 6 hours for the
amount of nicotine to reduce to 0.04mg.
We use the equation B = P ert . We want to have a balance of
B = $20, 000 in t = 6 years, with an annual interest rate 10%.
20, 000 = P e0.1(6)
P = 20,000
e0.6
≈ $10, 976.23.
4
Section
Problem
Answer
1.7
22
We have Po = 500, so P = 500ekt . We can use P = 1, 500 when
t = 2to find k:
1, 500 = 500e2k
e2k = 3
2k = ln(3)
k ≈ 0.5493.
The size of population is given by P = 500e0.5493t .
At t = 5, P = 500e0.5493(5) ≈ 7794.
a. Using the Rule of 70, we get that the doubling time of the
= 8.75.
investment is 70
8
That is, it would take about 8.75 years for the investment to double.
b. We know that the formula for the balance at the end of t years is
B(t) = P at where P is the initial investment and a is
1+(interest per year). We are asked to solve for the doubling time,
which amounts to asking for the time t at which B(t) = 2P .
Solving, we get
2P = P (1 + 0.08)t
2 = 1.08t
ln(2) = t ln(1.08)
ln(2)
≈ 9.01.
t = ln(1.08)
Thus, the actual doubling time is 9.01 years. And so our estimation
by the “rule of 70” was off by a quarter of a year.
28
1.8
12
x f (x) + 3
x f (x − 2)
0
13
2
10
1
9
3
6
6
3
a. 2
b. 4
c.
3
7
5
4
4
10
6
7
5
14
7
11
x −f (x) + 2
x g(x − 3)
0
-8
0
2
1
-4
1
3
-1
5
d. 2
e. 2
f.
3
-2
3
8
4
-5
4
12
5
-9
5
15
34
5
x 5g(x)
0
10
1
15
2
25
3
40
4
60
5
75
x f (x) + g(x)
0
12
1
9
2
8
3
12
4
19
5
26
Section
1.9
Problem
Answer
40
a.
b.
c.
d.
2
4
6
10
12
22
24
y = 3x−2 ; k = 3, p = −2.
y = 83 x−1 ; k = 38 , p = −1.
−1
y = 25 x 2 ; k = 52 , p = −1
.
2
3
3
3
y = 5 · x = 125x ; k = 125, p = 3.
y = 51 x; k = 15 , p = 1.
Let M =blood mass and B=body mass. Then M = k · B. Using the
fact that M = 150 when B = 3000, we have
150 = 3000k
150
= 0.05.
k = 3000
We have M = 0.05B. For a human with B = 70, we have
M = 0.05 × 70 = 3.5 kilograms of blood.
The specific hear is larger when the atomic weight is smaller, so
we test to see if the two are inversely proportional to each other by
calculating their product. The method works because if y is
inversely proportional to x, then y = xk for some constant k, so
xy = k. The following table shows that the product is approximately
6 in each case, so we conjecture that sw ≈ 6 or s ≈ w6 .
Element
Lithium Magnesium Aluminum
w
6.9
24.3
27.0
s(cal/deg-gm)
0.92
0.25
0.21
s·w
6.3
6.1
5.7
Element
Iron Silver Lead Mercury
w
55.8 107.9 207.2
200.6
s(cal/deg-gm) 0.11 0.056 0.031
0.33
s·w
6.1
6.0
6.4
6.6
6
Section
Problem
Answer
Ch1 review
22
a. The hottest oven results in the largest final temperature, so the
answer is (I).
b. The initial temperature is represented by the vertical intercept,
and the curve with the lowest vertical intercept is (IV).
c. The initial temperature is represented by the vertical intercept
and the two curves with the same vertical intercepts are (II) and (III).
d. The loaf represented by curve (III) heats up much faster than the
loaf represented by curve (II). One possible reason for this is that the
curve shown in (III) is for a much smaller loaf.
a. This is the graph of a linear function, which increases at a constant
rate, and thus corresponds to k(t), which increases by 0.3 over each
interval of 1.
b. This graph is concave down, so it corresponds to a function whose
increases are getting smaller, as is the case with h(t), whose increase
are 10,9,8,7,6.
This graph is concave up, so it corresponds to a function whose
increase are getting bigger, as is the case with g(t), whose increases
are 1,2,3,4,5.
a. The death rate from stomach cancer has decreased fairly steadily.
For lung cancer the death rate has increased fairly steadily.
Colon/rectum and prostrate cancer deaths increased until about 1945,
and then have stayed the same level. Death rates from cancer of the
esophagus have remained steady. Liver cancer deaths have decreased
gradually. Deaths from leukemia and cancer of the pancreas have
increased slightly.
b. Lung cancer deaths have had the greatest average rate of
change that line is by far the steepest on the graph.
43−2.5
Average rate of change = 1967−1930
≈ 1.1 additional deaths per 100,000
per year between 1930 and 1967.
c. Stomach cancer deaths has the most negative slope on the graph.
10−29
Average rate of change = 1967−1930
≈ −0.5 or 0.5 fewer deaths.
per 100,000 per year between 1930 and 1967.
a. The slope of the line is
28.2−22.0
Slope = 4C
= 2005−1995
= 0.62 billion tons per year.
4t
The initial value (when t = 0) is 22.0, so the linear formula is
C = 22.0 + 0.62t.
The annual rate of increase in carbon dioxide is 0.62 billion tons
per year.
24
26
40
7
Section
Problem
Answer
Ch1 review
40
b. If the function is exponential, we have C = C0 at . The initial
value (when t = 0)is 22.0, so we have C = 22.0at . We use
the fact that C = 28.2 when t = 10 to find a:
28.2 = 22.0a10
a10 = 1.28
a = 1.0251.
The exponential function is C = 22.0(1.025)t . Carbon dioxide
emission is increasing 2.5% per year. (Alternately, if we used
the form C = C0 ekt , the formula would be C = 22.0e0.0248t .)
We use logarithms:
ln(e5x ) = ln(100)
5x = ln(100)
x = 0.921.
Since e0.08 = 1.0833, and e−0.3 = 0.741, we have
P = e0.08t = (e0.08 )t = (1.0833)t and
Q = e−0.3t = (e−0.3 )t = (0.741)t .
If the pressure at sea level is P0 , the pressure P at altitude h is
h
0.4 100
given by P = P0 1 − 100
, since we want the pressure to be
0.4
multiplied by a of 1 − 100 = 0.996 for each 100 feet we go up
to make it decrease by 0.4% over that interval. At Mexico City
7340
h = 7340, so the pressure is P = P0 (0.996) 100 ≈ 0.745Po .
So the pressure is reduced from P0 to approximately 0.745P0 ,
a decrease of 25.5%.
44
46
52
56
58
8
Section
Problem
Answer
Ch1 review
60
If the function f (x) is
then 5f (x) is
2.1
2
4
6
a. The function N = f (t) is decreasing when t = 1950. Therefore,
f 0 (1950) is negative. That means that the number of farms in the
US was decreasing in 1950.
b. The function N = f (t) is decreasing in 1960 as well as in 1980
but it is decreasing faster in 1960 than in 1980. Therefore,
f 0 (1960) is more negative than f 0 (1980).
For
the interval 0 ≤t ≤ 0.8, we have
Average velocity
= 6.5
= 8.125 ft/sec.
= s(0.8)−s(0)
0.8−0
0.8
0
≤
t
≤
0.8
Average velocity
= 0.5
= 2.5 ft/sec.
= s(0.2)−s(0)
0.2−0
0.2
0
≤
t
≤
0.2
Average velocity
= 1.3
= 6.5 ft/sec.
= s(0.4)−s(0.2)
0.4−0.2
0.2
0.2 ≤ t ≤ 0.4
To find the velocity at t = 0.2, we find the average velocity to the
right of t = 0.2 and to the left of t = 0.2 and then average them.
So a reasonable estimate of the velocity at t = 0.2 is the average
of 21 (6.5 + 2.5) = 4.5 ft/sec.
a. The average rate of change of a function over an interval is
represented graphically as the slope of the secant line to its graph
over the interval. The secant line joining (0, f (0)) and (3, f (3))
has slope greater than the secant line joining (3, f (3)) and (5, f (5)).
Therefore, the average rate of change between x = 0 and x = 3 is
greater than the average rate of change between x = 3 and x = 5.
b. We can see from the graph that the function is increasing
faster at x = 1 than at x = 4. Therefore, the instantaneous rate of
change at x = 1 is greater than the instantaneous rate of change at
x = 4.
9
Section
Problem
Answer
2.1
6
c. The units of rate are obtained by dividing units of cost by units
of product: thousands of dollars/kilogram.
a. Since the values of P go up as t goes form 4 to 6 to 8, we see
that f 0 (6) appears to be positive. The percent of households with
cable television is increasing at t = 6.
b. We estimate f 0 (2) using the difference quotient for the interval
to the right of t = 2, as follows:
= 63.4−61.5
= 0.95.
f 0 (2) ≈ 4P
4t
4−2
0
The fact that f (2) = 0.95 tells us that the percent of households
with cable television in the United States was increasing at a rate
of 0.95 percentage points per year when t = 2 (that means 1992).
Similarly,
= 68.9−67.8
= 0.55.
f 0 (0) ≈ 4P
4t
12−10
0
The fact that f (10) = 0.55 tells us that the percent of households
with cable television in the United States was increasing at a rate
of 0.55 percentage points per year when t = 10 (that means 2000).
Using a difference quotient with h = 0.001, say, we find
f 0 (1) ≈ 1.001·ln(1.001)−1·ln(1)
= 1.0005
1.001−1
2.001·ln(2.001)−2·ln(2)
0
= 1.6934
f (2) ≈
2.001−2
0
The fact that f is larger at x = 2 than at x = 1 suggests that f is
concave up between x = 1 and x = 2.
16
20
2.2
10
a. x3
2.4
20
b. The positive first derivative tells use that the temperature is
increasing; the negative second derivative tells us that the rate of
increase of the temperature is slowing.
a. IV b. III c. II d. I e. IV f. II
28
3.1
50
b. x4
c. x5
d. x3
a. We have f (2) = 8, so a point on the tangent line is (2, 8). Since
f 0 (x) = 3x2 , the slope of the tangent is given by
m = f 0 (2) = 3 · 22 = 12.
Thus, the equation is
y − 8 = 12(x − 2) or y = 12x − 16.
b. See figure below. The tangent line lies below the function f (x) = x3 ,
so estimates made using the tangent line are underestimates.
10
Section
Problem
Answer
3.2
38
a. P = 9.906 (0.997)11 = 9.584 million.
b. Differentiating, we have
dP = 9.906 (ln 0.997) (0.997)t = −0.0288 million/year.
dt t=11
Thus in 2020, Hungary’s population will be decreasing by about
28,800 people per year.
3.3
24
dy
dx
3.4
14
Divide and then differentiate
f (x) = x + x3
f 0 (x) = 1 − x32 .
f 0 (t) = 1 · e5−2t + te5−2t (−2) = e5−2t (1 − 2t).
18
Ch2 review
4
14
28
32
34
= 2 (5 + ex ) ex .
Using the interval 1 ≤ x ≤ 1.001, we estimate
f (1.001)-f(1)
f 0 (1) ≈
= 3.0033−3.0000
= 3.3
0.001
0.001
x
The graph of 3 is concave up so we expect our estimate to be
greater than f 0 (1).
a. We use the interval to the right of x = 2 to estimate the
derivative. (Alternately, we could use the interval to the left of 2, or
we could use both and average the results.) We have
(2)
= 24−18
= 62 = 3.
f 0 (2) ≈ f (4)−f
4−2
4−2
We estimate f 0 (2) ≈ 3.
b. We know that f 0 (x) is positive when f (x) is increasing and
negative when f (x) is decreasing, so it appears that f 0 (x) is positive
for 0 < x < 4 and is negative for 4 < x < 12.
We have 4h = h (6003) − h (6000) ≈ h0 (6000) 4x = (0.5) (3)
= 1.5 meters. The elevation increases approximately 1.5 meters as
the climber moves from a position 6000 meters from the start of the
trail to a position 6003 meters from the start. Thus the climber’s
elevation increases from 8000 meters to about 8001.5 meters. The
new elevation is about 8001.5 meters above sea level.
The statements f (100) = 35 and f 0 (100) = 3 tell us that at x = 100,
the value of the function is 35 and the function is increasing at a rate
of 3 units for a unit decrease in x. Since we increase x by 2 units in
going from 100 to 102, the value of the function goes up by
approximately 2 · 3 = 6 units, so
f (102) ≈ 35 + 2 · 3 = 41.
The derivative is the instantaneous rate of change. To estimate the
instantaneous rate of change in 2002 using the information given, we
estimate the average rate of change on the interval from 2000 to 2002.
11
Section
Problem
Answer
Ch2 review
34
a. We have
= −69.60 million CDs per year.
f 0 (2002) ≈ 803.3−942.5
2002−2000
and
31.1−76.0
= −22.45 million cassettes per year.
g 0 (2002) ≈ 2002−2000
In 2002, sales of music CDs were decreasing at a rate of 69.60
million CDs per year and sales of music cassettes were
decreasing at a rate of about 22.45 million cassettes per year.
b. In 2002, sales of music CDs were 803.3 million and were
decreasing at a rate of 69.60 million per year. Therefore, if we
assume this rate of decrease stays constant, we have
f (2003) ≈ 803.3 − 69.60 = 733.7 million CDs,
and (since 2010 is 8 years after 2002)
f (2010) ≈ 803.3 − 69.60 (8) = 246.5 million CDs.
We estimate sales of music CDs to be about 733.7 million in 2003
and about 246.5 million in 2010. We should view this estimate
very cautiously, given the rapid expansion of competing
technology.
a. For the three years ending in 1995 we have
4P
54.1−62.4
= 1995−1992
= −2.77%/year.
4t
For the three years ending in 1998 we have
4P
48.0−54.1
= 1998−1995
= −2.03%/year.
4t
For the three years ending in 2001 we have
4P
43.5−48.0
= 2001−1998
= −1.50%/year.
4t
For the three years ending in 2004 we have
4P
41.8−43.5
= 2004−2001
= −0.57%/year.
4t
is increasing from 1992 to 2004 suggests that
b. The fact that 4P
4t
d2 P
is positive.
dt2
c. The values of P and 4P
are troublesome because they indicate
4t
that the percent of students graduating is low, and that the number
is getting smaller each year.
2
b. Since ddtP2 is positive, the percent of students graduating is not
decreasing as fast as it once was. Also, in 2004 the magnitude of
4P
is less than 1% a year, so the level of drop-outs does in fact
4t
seem to be hitting its minimum at around 40%.
42
Ch3 review
56
a. When the coffee was first left on the counter, t = 0. Thus,
C (0) = 74 + 103e0 = 74 + 103 · 1 = 177.
The temperature was 177◦ F.
1
1
b. Since e−0.033t = e0.033t
, as t gets larger, e0.033t
gets smaller and
−0.033t
tends to zero. Thus, 103e
gets very small, so the
temperature tends to 74 + 0 = 74◦ F. This is the room temperature.
12
Section
Problem
Answer
Ch3 review
56
c. To find C (5), we substitute t = 5.
C (5) = 74 + 103e−0.033·5 = 161.333.
This tells us that, after the coffee sits on the counter for five
minutes, its temperature is approximately 161◦ F.
To find C 0 (5), we differentiate C (t) and substitute.
C 0 (t) = 103e−0.033t (−0.033) = −3.399e−0.033t .
C 0 (5) = −3.399e−0.033·5 = −2.882.
This tells us that, after the coffee has sat on the counter for 5
minutes, it is cooling at a rate of 2.882◦ F/minute.
d.The magnitude of C 0 (t) is the rate at which the coffee is cooling
at time t. We expect the magnitude of C 0 (50) to be less than the
magnitude of C 0 (5) because, when the coffee is first put on the
counter (at t = 5), it cools fast. When the coffee has been on the
counter for some time (at t = 50), it is cooling more slowly.
a. H 0 (2) = r0 (2) + s0 (2) = −1 + 3 = 2.
b. H 0 (2) = 5s0 (2) = 5 (3) = 15
c. H 0 (2) = r0 (2) s (2) + r (2) s0 (2) = −1 · 1 + 4 · 3 = 11.
r0 (2)
√ = −1 .
= 2−1
d. H 0 (2) = √
4
4
66
2
1.4
20
24
r(2)
a. The slope of the linear function is
Slope = 4V
= 20,000−100,000
= −4000.
4t
20−0
The vertical intercept is V = 100, 000 so the formula is
V (t) = 100, 000 − 4000t.
b. At t = 5, we have V = 100, 000 − 4000 · 5 = 80, 000.
In 2015, the bus is worth $80,000.
c. The vertical intercept is 100,000 are represents the value of the
bus in dollars in 2010. The horizontal intercept is the value of t
when V = 0. Solving 0 = 100, 000 − 4000t for t, we find t = 25.
The bus is worth nothing when t = 25 years, that is, in the year
2035.
d. The domain of the function is 0 ≤ t ≤ 25.
a. We know that as the price per unit increases, the quantity
supplied increased, while the quantity demanded decreases. So
Table 1.26 is the demand curve (since as the price increases,the
quantity decreases), while Table 1.27 is the supply curve (since
as the price increases the quantity increases.)
b. Looking at the demand curve data in Table 1.26 we see that a
price of $155 gives a quantity of roughly 14.
c. Looking at the supply curve data in Table 1.27 we see that a
price of $155 gives a quantity roughly 24.
d. Since supply exceeds demand at a price of $155, the shift would
be to a lower price.
e. Looking at the demand curve data in Table 1.26 we see that if
the price is less than or equal to $143 the consumers would buy at
least 20 items.
13
Section
Problem
Answer
1.4
24
f. Looking at the supply curve (Table 1.27) we see that if the price
is greater than or equal to $110 the supplier will produce at least
20 items.
a. The original demand equation,
us that
q = 100 − 2p, tells Amount per unit
Quantity demanded = 100 − 2
.
paid by consumers
The consumers pay p + 0.05p = 1.05p dollars per unit because
they pay the price p plus 5% tax. Thus, the new demand equation
is
q = 100 − 2 (1.05p) = 100 − 2.1p.
The supply equation remains the same.
b. To find the new equilibrium price and quantity, we find the
intersection of the demand curve and the supply curve q = 3p − 50.
100 − 2.1p = 3p − 50
150 = 5.19
p = $29.41.
The equilibrium price is $29.41.
The equilibrium quantity q = 3(29.41) − 50 = 88.23 − 50
= 38.23 units.
c. Since the pre-tax price was $30 and the suppliers new price is
$29.41 per unit,
Tax paid by supplier = $30 − $29.41 = $0.59.
The consumers’ new price is 1.05p = 1.05 (29.41) = $30.88 per
unit. and the pre-tax price was $30, so
Tax paid by consumer= 30.88 − 30 = $0.88.
The total tax paid per unit by suppliers and consumers together is
0.59 + 0.88 = $1.47 per unit.
d. The government receives $1.47 per unit on 38.23 units. The
total tax collected is (1.47) (38.23) = $56.20.
38
2.5
4
6
Drawing in the tangent line at the point (10000, C (10000)).
We see that each vertical increase of 2500 in the tangent line
gives a corresponding horizontal increase of roughly 6000.
This the marginal cost at the production level of 10, 000 units is
Slope of tangent line
2500
= 0.42.
C 0 (10000) =
= 6000
to C (q) at q = 10000
This tells us that after producing 10,000 units, it will cost roughly
$0.42 to produce one more unit.
a. For q = 500
Profit= π (500) = R (500) − C (500) = 9400 − 7200
= 2200 dollars
14
Section
Problem
Answer
2.5
6
b. As production increases from q = 500 to q = 501,
4R ≈ R0 (500)4q = 20 · 1 = 20 dollars.
4C ≈ C 0 (500)4q = 15 · 1 = 15 dollars.
Thus
Change in Profit= 4π = 4R − 4C = 20 − 15 = 5 dollars.
At q = 50, the slope of the revenue is larger than the slope of the
cost. This, at q = 50, marginal revenue is greater than marginal
cost and the 50th bus should be added. At q = 90 the slope of
revenue is less than the slope of cost. Thus, at q = 90 the
marginal revenue is less than marginal cost and the 90th bus
should not be added.
12
4.1
6
20
22
4.2
6
There was a critical point after the first eighteen hours when
temperature was at its highest point, a local maximum for the
temperature function.
a. The demand for the product is increasing when f 0 (t) is positive,
and decreasing when f 0 (t) is negative. Inspection of the table
suggests that demand is increasing during weeks 0 to 2 and weeks
6 to 10, and decreasing during weeks 3 to 5.
b. Since f 0 (t) = 4 > 0 during week 2 and f 0 (t) = −2 < 0 during
week 3, the demand for the product changes from increasing to
decreasing near the end of week2 or the beginning of week 3. Thus
the demand has a local maximum during this time period. Since
f 0 (t) = −1 < 0 during week 5 and f 0 (t) = 3 > 0 during week 6,
the demand for the product changes from decreasing to increasing
near the end of week 5 and or the beginning of week 6. Thus the
demand has a local minimum during this period.
Since f 0 (x) = 4x3 − 12x2 + 8, we see that f 0 (1) = 0, as we
expected. We apply the second derivative test to
f 00 (x) = 12x2 − 24x. Since f 00 (1) = −12 < 0,the graph is concave
down at the critical point x = 1, making it a local maximum.
One possible sketch is shown below.
15
Section
Problem
Answer
4.3
6
One possible sketch is shown below.
18
a. Differentiating f (x) = x3 − 3x2 produces f 0 (x) = 3x2 − 6x. A
second differentiation produces f 00 (x) = 6x − 6.
b. f 0 (x) is defined on all x and f 0 (x) = 0 when x = 0, 2. Thus
0 and 2 are the critical points of f .
c. f 00 (x)is defined for all x and f 00 (x) when x = 1. Hence x = 1
is an inflection point.
d. f (−1) = −4, f 0) = 0, f (2) = −4, f (3) = 0.So f has a local
maximum at x = 0, a local minimum at x = 2, global maxima at
x = 0 and x = 3,and global minima at x = −1 and x = 2.
36
e.
Rewriting the expression for I using the properties of logs gives
I = 192(ln − ln 762) − S + 763.
Differentiating with respect to S gives
dI
= 192
− 1.
dS
S
At a critical point
192
− 1 = 0 ⇒ S = 192.
S
d2 I
−192
Since dS
,
2 =
S2
d2 I
we see that if S = 192, we have dS
2 < 0, so S = 192 is a local
maximum. From the figure below, we see that it is a global
maximum. The maximum possible number of infected children is
therefore
I = 192 ln 192
− 192 + 763 = 306 children.
762
16
Section
Problem
Answer
4.4
12
The company should increase production if M R > M C, since
increasing production then adds more to revenue than to the cost
→net gain for the company.
a. Since M C(25) = 17.75 and M R(25) = 30,the company
should increase the production.
b. Since M C(50) = 39 and M R(50) = 30,the company
should decrease the production.
c. Since M C(80) = 114 and M R(80) = 30,the company
should decrease the production.
a. Since Profit=Revenue - Cost, we can calculate
π(q) = R(q) − C(q) for each of the q values given:
q
0
100 200 300 400 500
R(q)
0
500 1000 1500 2000 2500
C(q) 700 900 1000 1100 1300 1900
π(q) -700 -400
0
400 700 600
We see that maximum profit is $700 and it occurs when the
production level q is 400.
b. Since revenue is $500 when q = 100, the selling price is $5
per unit.
c. Since C(0) = $700, the fixed costs are $700.
Consider the rectangle of sides x and y.
The total area is xy = 3000, so y = 3000/x. Suppose the left and
right edges and the lower edge have the shrubs and the top edge
has the fencing. The total cost is
C = 45(x + 2y) + 20(x) = 65x + 90y.
Since y = 3000/x, this reduces to
C = 65x + 90(3000/x) = 65x + 270000/x.
Therefore, C 0 (x) = 65 − 270000/x2 , we set this to 0 to find the
critical points:
= 0
65 − 270000
x2
270000
= 65
x2
2
x = 4153.85
x = 64.450 ft
so that
y = 3000/x = 46.548 ft.
Since C(x) → ∞ as x → 0+ and x → ∞, we see that
x = 64.450 is a minimum. The minimum total cost is then
C(64.450) ≈ $8378.54.
20
26
4.7
10
a. At t = 0, which corresponds to 1935, we have
1
P = 1+3e−0.0275(0)
= 0.25
showing that 25%of the land was in use in 1935.
b. This model predicts that as t gets very large, P approaches 1.
That is, the model predicts that in the long run, all the land will be
used for farming.
17
Section
Problem
Answer
4.7
10
c. Solve P = 0.5,
1
= 0.5
1+3e−0.0275(t)
−0.0275(t)
1 + 3e
= 2
−0.0275(t)
e
= 13
−0.0275(t) = ln( 13 )
t = 39.95
According to this model, the tojolobal were using half their land
in 1935+39.95≈1975.
d. The point of diminishing returns occurs when P = L/2 or at
one-half the carrying capacity. In this case, P = 1/2 in 1975, as
shown in part c.
a. We use k = 1.78 as a rough approximation. We let L = 5000
since the problem tells us that 5000 people eventually get the
virus. This means the limiting value is 5000.
We know that
5000
P (t) = 1+Ce
−1.78t and P (0) = 10
so
5000
5000
= 1+C
10 = 1+Ce
0
10(1 + C) = 5000
1 + C = 500
C = 499.
5000
c. We have P (t) = 1+499e
−1.78t . This function is graphed in figure
below.
14
16
d. The point of diminishing returns appears to be at the point
(3.5,2500); that is, after 3 and a half weeks and when 2500 people
are inflected.
a.The dose-response curve for the product C crosses the minimum
desired response line last, so it requires the largest dose to
achieve the desired response. The dose-response curve for
product B crosses the minimum desired response line first, so it
requires the smallest dose to achieve the desired response.
b. The dose-response curve for product A levels off at the highest
point, so it has the largest maximum response. The dose-response
curve for product B levels off at the lowest point, so it has the
smallest maximum response.
c. Product C is the safest to administer because its slope in the
safe and effective region is the least, so there is a broad range of
dosages for which the drug is both safe and effective.
18
Section
Problem
Answer
5.1
16
a. Since car B starts at t = 2, the tick marks on the horizontal axis
(which we assume are equally spaced) are 2 hours apart. Thus
car B stops at t = 6 and travels for 4 hours.
Car A starts at t = 0 and stops at t = 8, so it travels for 8 hours.
b. Car A0 s maximum velocity is approximately twice that of car B,
that is 100 km/hr.
c. The distance traveled is given by the area of under the velocity
graph. Using the formula for the area of a triangle, the distance are
give approximately by
Car A travels= 12 · Base · Height = 21 · 8 · 100 = 400 km
Car A travels= 12 · Base · Height = 21 · 4 · 50 = 100 km.
5.3
2
14
16
18
5.4
12
´8
Area= 0 100(0.6)t dt ≈ 192.47
IV
I
a. The area between the graph of f and the x-axis between x = a
´b
and x = b is 13, so a f (x)dx = 13.
b.
´ c Since the graph of f (x) is below the x-axis b < x < c,
f (x)dx = −2.
b
c. Since the graph of f (x) is above the x-axis a < x < b and
is
´ cbelow the x-axis b < x < c,
f (x)dx = 13 − 2 = 11.
a
d. The graph of |f (x)|is the same as the graph of f (x) except that
the part below the x-axis is reflected to be above it. Thus
´c
´b
´c
|f (x)|dx = a |f (x)|dx + b |f (x)|dx
a
´b
´c
= a f (x)dx + b −f (x)dx
´b
´c
= a f (x)dx − b f (x)dx
= 13 − (−2)
= 15.
a. The distance traveled in first 3 hours (from t = 0 to t = 3) is
´3
given by 0 (40t − 10t2 )dt.
b. The shaded area in figure below represents the distance
travel.
c. Using a calculator, we get
´3
(40t − 10t2 )dt = 90.
0
So the total distance traveled is 90 miles.
19
Section
Problem
Answer
Ch4 review
8
There are several possibilities. The price could have been
increasing during the last few days of June, reaching a high point
on July 1, then going back down during the first few days of July.
In this case there was a local maximum in the price on July 1.
The price could have been decreasing during the last few days of
June, reaching a low point on July 1, then going back up during
the first few days of July. In this case there was a local minimum
in the price on July 1.
It is also possible that there was neither a local maximum nor a
local minimum in the price on July 1. This could have happened
two ways. On the one hand, the price could have been rising in
late June, then held steady with no change around July 1, after
which the price increased some more. On the other hand, the price
could have been falling in late June, then held steady with no
change around July 1, after which the price fell some more. The
key feature in these critical point scenarios is that there was no
appreciable change in the price of stock around July 1.
a. Increasing for x < 0, decreasing for x > 0.
b. f (0) is a local and global minimum, and f has no global
maximum.
a. Decreasing for x < 0, increasing for 0 < x < 4, decreasing
for x > 4.
b. f (0) is a local minimum, f (4) is a local maximum.
Local maximum for some θ, with 1.1 < θ < 1.2, since
f 0 (1.1) > 0 and f 0 (1.2) < 0.
Local minimum for some θ, with 1.5 < θ < 1.6, since
f 0 (1.5) < 0 and f 0 (1.6) > 0.
Local maximum for some θ, with 2.0 < θ < 2.1, since
f 0 (2.0) > 0 and f 0 (2.1) < 0.
a. If we graph the data, we see that it looks like logistic growth.
But logistic growth also makes sense from a common-sense
viewpoint. As cable television “catches on.” the percentage of
households which have it will at first grow exponentially but then
slow down after more and more people have it. Eventually, nearly
everyone who will subscribe already has and the percentage
levels off.
b. The point of diminishing returns happens around 35%. This
predicts a carrying capacity of 70% which is pretty close to the
68.9% we see in 2002 and 68.0% we see in 2003.
c. 68.8%
d. The limiting value predicts the percentage of households that
will eventually have cable. This model predicts that there will
never be a time when more than 68.8% of households have cable.
10
12
20
42
20
Section
Problem
Answer
Ch5 review
36
The fixed cost is C(0) = 1, 000, 000.
´ 500
´ 500
Total variable cost= 0 C 0 (x)dx = 0 (4000 + 10x)dx
= 3, 250, 000.
Therefore,
Total cost=Fixed Cost+Total variable cost
= 4, 250, 000 riyals.
a. The mouse changes direction(when its velocity is zero) at about
times 17, 23, 27.
b. The mouse is moving most rapidly to the right at time 10 and
most rapidly to the left at time 40.
c. The mouse is farthest to the right when the integral of the
´t
velocity, 0 v(t)dt, is the most positive. Since the integral is the
sum of areas above the axis minus the areas below the axis, the
integral is largest when the velocity is zero at about 17 seconds.
The mouse is farthest to the left of center when the integral is
most negative at 40 seconds.
d. The mouse’s speed decrease during seconds 10 to 17, from 20
to 23 seconds, and from 24 seconds to 27 seconds.
e. The mouse is at the center of the tunnel at any time t for which
the integral from 0 to t of the velocity is zero. This is true at time
0 and again somewhere around 35 seconds.
40
6.1
2
16
6.3
a. Counting the squares yields an estimate of 25 squares, each
with area=1, so we conclude that
´5
f (x)dx ≈ 25.
0
b. The average height appears to be around 5.
c. Using the formula, we get
Average value,
which is consistent with part b.
a. When t = 10 we have P = 6.1e0.0125(10) = 6.9. The predicted
population of the world in the year 2010 is 6.9 billion people.
b. We find the average value of P on the interval t = 0 to t = 10:
´ 10
1
1
6.1e0.0125t dt = 10
(64.976) = 6.5.
Average value= 10−0
0
The average population of the world over this decade is 6.5
billion people.
2
21
Section
Problem
Answer
6.3
2
The graph reaches a peak each summer, and a trough each winter.
The graph shows sunscreen sales increasing from cycle to cycle.
This gradual increase may be due in part to inflation and to
population growth.
We compute the present value of the company’s earnings over the
next 8 years:
´8
Present value of earnings= 0 50, 000e−0.07t dt = $306, 279.24.
If you buy the rights to the earnings of the company for $350,000,
you expect that the earnings will be worth more than $350,000.
Since the present value of the earnings is less than this amount,
you should not buy.
10
7.1
66
An antiderivative is F (x) = 2e3x + C. Since F (0) = 5, we have
5 = 2e0 + C, so C = 3. The answer is F (x) = 2e3x + 3.
7.3
38
(since
a. An antiderivative of F 0 (x) = x12 is F (x) = −1
x
d
−1
1
= x2 ). So by the Fundamental Theorem we have:
´dxb 1 x
−1 b
dx
=
= −1
+ 1.
2
x 1
b
1 x
b. Taking a limit, we have
lim −1
+ 1 = 0 + 1 = 1.
b
b→∞
Since the limit is 1, we know that
´b
lim 1 x12 dx = 1.
b→∞
40
7.5
2
So
improper integral converges to 1:
´ ∞the
1
dx
= 1.
1 x2
a. Evaluating the integrals with a calculator gives
´ 10 −x/10
xe
dx = 26.42
´050 −x/10
xe
dx = 95.96
´0100 −x/10
xe
dx = 99.95
´0200 −x/10
xe
dx = 100.00
0
b.
The
results
of part a. suggest that
´ ∞ −x/10
xe
dx ≈ 100.
0
First, we observe that g is increasing when g 0 is positive, which
is when 0 < x < 4.
g is decreasing when g 0 is negative, when 4 < x < 6.
Therefore, x = 4 is a local minimum.
Table below shows the area between the curve and the x-axis
for the intervals 0-1, 1-2 etc. It also shows the corresponding
change in the value of g.
22
Section
7.5
Problem
2
14
Answer
´b
Interval Area Total change in g = a g 0 (x)dx
0-1
1/2
1/2
1-2
1
1
2-3
1
1
3-4
1/2
1/2
4-5
1/2
-1/2
5-6
1/2
-1/2
These changes are used to compute the value of g using the
Fundamental Theorem of Calculus:
´1
g(1) − g(0) = 0 g 0 (x)dx = 21 .
Since g(0) = 0, g(1) = 12 .
´2
Similarly, g(2) − g(1) = 1 g 0 (x)dx = 1
g(2) = g(1) + 1 = 23 .
Computing in this way gives the value of g in the following table.
x g(x)
0
0
1 1/2
2 3/2
3 5/2
4
3
5 5/2
6
2
Notice: the graph of g will be a straight line from 1 to 3 because
g 0 is horizontal there. Furthermore, the tangent line will be
horizontal at x = 4, x = 0 and x = 6. The maximum is at (4, 3).
See the following graph
For every number b, the Fundamental Theorem tells us that
´b 0
F (x) = F (b) − F (0) = F (b) − 0 = F (b).
0
Therefore, the value of F (1), F (2), F (3), F (4) are values of
definite integrals. The definite integral is equal to the area of
regions under the graph above the x-axis minus the area of the
regions below the x-axis above the graph. Let A1, A2, A3, A4 be
the areas shown in the following figure. The region between
x = 0 and x = 1 lies above the x-axis, so F (1) is positive and
´1
we have F (1) = 0 F 0 (x)dx = A1.
The region between x = 0 and x = 2 also lies entirely above the
x-axis, so F (2) is positive, and we have
23
Section
Problem
Answer
7.5
14
Ch6 review
14
The present value is given by
´M
Present value= 0 Se−rt dt,
with S = 1000, r = 0.09 and M = 5. Hence,
´5
Present value= 0 1000e−0.09t dt = $4026.35.
Ch7 review
44
1
If f (x) = x+1
, the average value of f on the interval 0 ≤ x ≤ 2 is
defined to be
´2
´
1 2 dx
1
f
(x)dx
=
.
2−0 0
2 0 x+1
We’ll integrate by substitution. We let w = x + 1 and dw = dx,
and we have
´ w=3 dw
´ x=2 dx
=
= ln w|31 = ln 3 − ln 1 = ln 3.
x=0 x+1
w=1 w
Thus, the average value of f (x) on 0 ≤ x ≤ 2 is 12 ln 3 ≈ 0.5493.
See the following figure.
´2
F (2) = 0 F 0 (x)dx = A1 + A2.
We see that F (2) > F (1). The region between x = 0 and x = 3
includes parts above and below the x-axis. We have
´3
F (3) = 0 F 0 (x)dx = (A1 + A2) − A3.
Since the area A3 is approximately the same as the area A2 , we
have F (3) ≈ F (1). Finally, we see that
´4
F (4) = 0 F 0 (x)dx = (A1 + A2) − (A3 + A4).
Since the area A1+A2 appears to be larger than the area of
A3+A4, we see that F (4) is positive, but smaller than the others.
The largest value is F (2) and the smallest value is F (4). None of
the numbers is negative.
24
Section
Problem
Answer
Ch7 review
56
that ´ln (x2 ) = 2 ln x. Therefore,
´Remember
ln (x2 ) dx = 2 ln xdx = 2x ln 2 − 2x + C.
Check:
d
(2x ln x − 2x + C) = 2 ln x + 2x
− 2 = 2 ln x = ln (x2 ).
dx
x
25