Calculus with Analytic Geometry II
Homework 1, Solutions and Grading Notes
The instructions on how to present homework and other written material are posted, linked to our website. For
your convenience, I am also attaching them at the end of this homework set. I will be soon rejecting homework
that violates too many of these rules. I get to decide what is “too many.”
1
Grading Notes.
• I decided that when solutions by two people were so similar that the probability that one just copied it from
the other was extremely high, then I would split the grade between the two. So if the exercise merited 8
points, each would get 4. I am willing to make a different split, say 5, 3 or 8,0. Just let me know.
• Section 6.2 is about computing volumes using calculus. If a volume was calculated without the use of calculus
(say by finding a formula in the internet or elsewhere), I gave 0 credit for the solution. This applies particularly
to Exercise #4, the frustum of a pyramid.
• I did not grade the optional exercises. I apologize to the one person who did a very thorough job on Exercise
10.
2
1. The antiderivative of e−x plays a big role in statistics and other areas of mathematics and its applications.
2
(a) Find an expression for a function F such that F (0) = 0 and F 0 (x) = e−x for all x. This expression can
(and should!) involve an integral.
Solution. By the fundamental theorem of calculus,
Z x
2
F (x) =
e−t dt.
0
(b) Use a 10 point midpoint rule to get an approximate value for F (1). If you don’t know what this rule is,
it is explained in Section 5.2 of our textbook.
Z 1
2
Solution. We have F (1) =
e−x dx. We have to partition the interval [0, 1] into 10 parts by
0
x0 = 0 < x1 = 0.1 < x2 = 0.2 < · · · < x9 = 0.9 < x1 = 10.
The midpoint rule is a Riemann sum in which we choose for each interval its midpoint as the point where
the function gets evaluated, so x∗1 = 0.05, x∗2 = 0.15, . . . , x∗10 = 0.95 and the integral is approximated by
10
X
∗ 2
e−(xi ) ∆x,
i=1
where ∆x is the length of each partition subinterval; that is ∆x = 0.1. Evaluating (and we obviously
need a calculator for this)
10
X
∗ 2
e−(xi ) ∆x =
2
2
2
2
e−0.05 + e−0.15 + e−0.25 + · · · + e−0.95
(0.1)
i=1
=
(0.9975031224 + 0.9777512372 + 0.9394130628 + 0.8847059049
+0.8166864826 + 0.7389684883 + 0.6554062543 + 0.5697828247
+0.4855368952 + 0.4055545051) ∗ 0.1 = 0.7471308778
When adding numbers given approximately, one loses one digit every ten sums, so we’ll drop the last
digit and give the answer we obtained from the ten point midpoint rule as 0.747130877.
(c) Compare your answer with the answer you would get using some computer algebra system, for example,
Wolframalpha, Maple, Matlab or Mathematica.
WolframAlpha states that
Z
1
2
e−x dx ≈ 0.746824,
0
so we got two exact digits.
2. In the land of Calculandia there is an island. Placing axes with the origin at the southern tip of the island,
one can say that the island is the region bounded by the parabolas x = 5y − y 2 and x = y 2 − 3y, where y is
measured in kilometers (it’s a big island). Aldon (known as A by his friends) and Betty (known as B) both
claim the island as theirs. To settle the dispute, the wise King Sagittarius III of Calculandia decreed that
the island would be divided equally between A and B by a line perpendicular to the S-N axis (also known as
the y-axis). Since both A and B flunked College Algebra, and get nightmares every time they hear the word
Calculus, they hire you to draw the dividing line. Your task it is to answer the following questions:
(a) How far away from the southern tip should one draw the line (i.e., what is the y coordinate where the
line has to be drawn)?
(b) How much area do A and B get?
For your convenience, here is a picture of the island.
Solution. It will be best to consider y as the independent, x as the dependent variable. Equating 5y −y 2 =
y 2 − 2y, and solving, the intersections are at y = 0, y = 4. It will also be convenient to have an expression for
the area from y = 0 to a generic point y, namely
Z y
Z y
2
A(y) =
(5y − y 2 − y 2 + 2y) dy =
(8y − 2y 2 ) dy = 4y 2 − y 3 .
3
0
0
128
64
=
. With the whole area being 64/3, the half area is 32/3
3
3
2
32
and we have to find y so A(y) = 32/3. This leads to the equation 4y 2 − y 3 =
; which can be rearranged
3
3
3
2
in the form y − 6y + 16 = 0. This is a cubic equation and there are several ways of solving it. If one has
had a good high school or college algebra course and one remembers things from there one might search for
rational roots and discover that one root is y = 2. One might also just guess this by inspection. Once one
has this one root one can divide it out and factor
The area of the region is then A(4) = 64 −
y 3 − 6y 2 + 16 = (y − 2)(y 2 − 4y − 8)
and one finds the other roots by solving the quadratic equation y 2 − 4y − 8 = 0. Or one can put the equation
into some computer
(WolframAlpha or√Matlab, for example), and get all roots at once.
√ algebra program
√
√ they
are, in order, 2 − 2 3, 2, 2 + 2 3. It turns out that 2 − 2 3 < 0, so it is out being too small, while 2 + 2 3 > 4
is too big. But 2 is just right. so now we can answer easily both questions:
(a) The line should be drawn 2 km from the southern tip of the island.
(b) A and B both get an area equal to 32/3 square kilometers.
3. Textbook, Exercise 44, §6.1 (p. 428).
Solution. This seems to me a somewhat misleading problem. Too many widths are given. The pool seems
to be 16 meters in maximum left to right length. If we use all the data, the intervals are 2 meters apart and the
measurements are at the endpoints. For a midpoint rule we need midpoint data. If we think of partitioning
into intervals of length 4 meters, then half of the data is at the midpoint of an interval. That means that
we are ignoring half of the measurements and only using the first, third, fifth and seven measurement. That
gives an area of
(6.2 + 6.8 + 5.0 + 4.8) × 4 = 91.2m2 .
I might accept solutions using a left or a right Riemann sum, assuming the solvers realize that in this case
the first, respectively the last, measurement is 0.
4. Textbook, Exercise 50, §6.2 (p.440). In this exercise you are supposed to find the volume of the frustum of a
pyramid.
It might interest you that this problem was posed
and solved in one of the oldest extant mathematical documents, the Moscow Papyrus (so called
because it is in a museum in Moscow), which has
been dated to about 1850 BCE. The ancient Egyptians were, of course, obsessed with pyramids. It is
somewhat of a mystery how they solved the problem without calculus. A picture of a fragment of
the papyrus (and a transcription) is at the right,
showing the problem in question.
Solution. If we cut the frustum by a vertical plane through that divides two opposite sides of the base in
two, the cross section will look like the picture below, where the dotted lines represent the missing part of
the full pyramid.
By similarity of triangles, or otherwise, one gets that the side of the square at height y is x = b −
Thus
2
Z h
b−a
(a2 + ab + b2 )h
V =
b−
y
dy =
.
h
3
0
b−a
y.
h
If a = 0, we get the volume of the full pyramid, namely V =
cylinder of base radius b, height h, V = b2 h.
b2 h
. If a = b we get the volume of the square
3
5. Textbook, Exercise 62, §6.2 (p.440).
Another slightly misleading statement. Instead of saying ”cross sections parallel to the line of intersection of
the two planes, which is ambiguous, the textbook should have said cross-sections perpendicular to the y-axis
(as opposed to perpendicular to the x-axis). Such a cross-section, cutting
the y axis at a generic point of
p
coordinate y is a rectangle having the side along the p
base of length 2 16 − y 2 and the side perpendicular to
√
√
the base of length y/ 3. The area is thus A(y) = 2y 16 − y 2 / 3, the volume is equal to
2
V =√
3
Z
4
y
0
p
128
16 − y 2 dy = √ .
3 3
6. (Optional) Textbook, Exercise 64, §6.2 (p. 440). This exercise at first glance seems horribly difficult, but it
is easier than it seems. Notice that horizontal cross sections are squares.
Solution. We can assume that we have drawn axes so the axis of one cylinder coincides with the x-axis,
the other axis with the y-axis; the origin being at the center of the intersection of the two cylinders. Call an
axis through the origin perpendicular to the (x, y)-plane the z-axis. Suppose we intersect the solid at height
z with a plane parallel to the (x, y)-plane. Letus concentrate for a moment on only one of the cylinders,
say
√ the one whose axis is the y-axis. The horizontal plane intersects this cylinder in a strip (band) of width
2 r2 − z 2 . To calculate this width, I look at the intersection of the cylinder with the (x, z) plane. The red
segment in the picture is the profile of the horizontal plane at height z; its width is easily calculated using
the Theorem of Pythagoras.
Now let us bring the second cylinder into√the picture. At height z, the intersection is the intersection
of
√
two perpendicular bands. each of width 2 r2 − z 2 , resulting in a square of sides of length [2 r2 − z 2 ]2 =
4(r2 − z 2 ). Since z ranges from −r to r, the volume is
Z r
Z r
16r3
V =4
(r2 − z 2 ) dz = 8
(r2 − z 2 ) dz =
.
3
−r
0
7. Textbook, Exercise 46, §6.3, (p. 446) Find the volume of a torus by the shell method.
Solution. The torus is obtained by revolving a disk of radius r about an axis at distance R from the center
of the disc; where R > r. It may be best to place the disc centered at the origin and revolve about the line
x = R.
p
The shells go from x = −r to x = r; the radius of the shell at x is R − x; its height is 2 (r2 − x2 ). The
volume is thus equal to
Z r p
Z r p
Z r
p
r2 − x2 dx − 4π
x r2 − x2 dx.
V = 4π
(R − x) r2 − x2 dx = 4πR
−r
−r
−r
Of the two integrals on the right hand side of the second equality above, the second one is fairly easy to
compute with the substitution u = r2 − x2 . But if one realizes, as incredibly enough everybody did!, that it
is the integral of an odd function over an interval symmetric about the origin, one can see at once that it is
0. In a normal situation one does not expect EVERYBODY to figure this out, one expects some people to
use the substitution and make a mistake, and not get 0 as an answer. But, so is life.
The first integral is more interesting because at the time of the homework no method for computing had
been seen. So I expected one of two things: People would remember that it is the area of a half circle, and
mention it, or frankly confess not knowing how to do it and use, with credit given, something like Wolfram
Alpha to compute it. One person actually identified as the area of a half circle of radius r. Everybody else
just replaced it be πr2 /2. Amazing, isn’t it?
Returning to the exercise, as mentioned in the little diatribe above, the second one of the integrals of the
right hand side of the last equality is 0, the first one is πr2 /2, thus
V = 2π 2 Rr2 .
8. Since we are into pyramids, we should certainly do Exercise 50, §6.4 (p. 451).
Solution.
Here is a profile of a square pyramid of height H, base of sides of length b. A thin pyramidal slab at a height
y, of thickness ∆y will have a volume of ∆V = x2 ∆y, where the relation between x and y is easily found by
similar triangles:
x
H −y
b
=
, thus x = (H − y).
b
H
H
Such a volume had to be lifted a distance y; if the density of the material is ρ, then the amount of work done
to lift such a slab is
ρb2
∆W = 2 y(H − y)2 ∆y.
H
Replacing ∆ by d, and integrating from 0 to H gives the total work:
Z
ρb2 H
ρb2 H 2
W = 2
y(H − y)2 dy =
.
H 0
12
Entering the values given for ρ, b, H, namely ρ = 150, b = 756, H = 481, we get that the work equals
1, 652, 889, 256, 200lb · f t.
To figure out the number of laborers, according to the given data, one LE (laborer equivalent) did a total
work in a 20 year period of 13600000 lb·ft. Dividing out, it comes to about 121,536 laborers.
9. (Optional) Exercise 4 of the Problem Plus Section of Chapter 6 (p. 459).
Solution.
Upon request.
Scoring: The 8 non-optional exercises are worth 100/8 points each. Each of the two optional exercise can give
you an extra 10 points.
Some rules to be followed in exams and homework
First of all, a thing to remember: Mathematics is a very powerful but also very delicate tool. If used improperly,
it breaks easy.
The rules.
• Homework should be well presented. No torn pages, very little if any crossed out material. If ruled paper is
used, obey the lines. If the paper isn’t ruled, write straight. Avoid double columns. Don’t try to cram in
the maximum amount of writing into the minimum amount of space.
• You are university students, on the way to becoming university graduates. One expects a university graduate
to be able to write clearly, to express him or herself correctly. I will expect you to be articulate in exams and
homework, and to write in complete sentences. A sentence, among other things, must contain a verb. Here
is an example:
1
q
16 − x12
is NOT a sentence. There is no verb. Do not just plunk down expressions. On the other hand,
1
f ◦ g(x) = q
16 −
1
x2
is a sentence. The verb is implicit in the symbol “=”, which reads “is equal to.”
• The work you present should be your own. Ideally, you did it all by yourself. Discussing homework with
a fellow student or friend is OK, as long as it reduces to a discussion, and not to you copying verbatim what
someone else did. Going to the internet and consulting with some anonymous gremlin (Clegg ex Cramster,
for example) should be an absolute no-no. But our world is not ideal and since I do not have the resources
of the NSA at my disposal, I cannot spy on you 24 hours a day, nor would I want to do it. So if you get
your answers from somewhere else, be it a friend or an internet location, at the very least make sure you
understand the answer and try not to be too obvious. If I suspect that you wrote down something
you did not understand, I reserve the right to give you 0 points for that part of the exercise. If
so, I will make you aware that the reason for the 0 is my suspicion and invite you to come to my office during
office hours and prove to me that I was wrong.
•
SHOW ALL WORK! If in an exercise the answer is obtained apparently by divine inspiration, I
may give you 0 points for that exercise. There are some exercise that can be answered at once, but if you
cannot obtain the answer in a single step, make sure you write down all the steps that led you to the answer.
Otherwise, I may suspect you copied the answer from somewhere.
• Equal signs should be used properly. Here is an example of improper use:
f ◦g = q
1
16 −
1
x2
Why is it improper? On the left hand side you have a symbol for a function, on the right hand side a symbol
or expression for a value of a function. The two sides cannot be equal. The correct form is
1
f ◦ g(x) = q
16 −
1
x2
Briefly, the two sides separated by an equal sign should be equal. Or (in the case of an equation, as opposed
to an identity) something on one side that one wishes to be equal to the other side.
• Punctuation should be used reasonably well. For example, you should know the difference between “Call me
Ishmael” (the first line of the classic American novel Moby Dick ) and “Call me, Ishmael.” Or, to give another
example, the difference between “Sharks don’t swim here,” and “Sharks; don’t swim here.”
• Your final answers in exams or homework should be easy to spot. For example, let us suppose the question is:
What is the maximum value assumed by the function y = x2 in the interval [0, 2]. Well, this is an increasing
function and it assumes its maximum at the right end of the interval, for x = 2. At that point the value is
4, so the maximum value is 4. You might get away with scribbling 4, and nothing else. Or you might frame
the 4: 4 . But we are not in grade, middle, or high school anymore. You really should write “The maximum
value is 4.” A full sentence answer.
• High school, or even college algebra, mistakes will be treated with great severity. For example, if in an
exercise you distribute a square root with respect to a sum, I might stop reading right there and√consider
your whole
exercise
as being wrong. (“I might” doesn’t mean “I will;” merely that I might do it.) a + b is
√
√
NOT a + b. We all agree, I am sure,√that√7 6= 5 (if you disagree,
I have $5 which I would be happy to
√
exchange for 7 of your dollars), but 7 = 9 + 16, while 5 = 9 + 16.
Similarly, I expect you to be able to handle fractions correctly, know how to add, subtract, multiply and
divide fractions.
• There could be more to come.