Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Problem #7 |
Total | |
---|---|---|---|---|---|---|---|---|

Max grade | 16 | 12 | 18 | 18 | 16 | 10 | 10 | 96 |

Min grade | 12 | 0 | 11 | 4 | 0 | 0 | 3 | 55 |

Mean grade | 14.76 | 9.24 | 16.57 | 13 | 10.57 | 4.48 | 7.38 | 76.00 |

Median grade | 15 | 12 | 18 | 12 | 11 | 5 | 8 | 74 |

21 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [85,100] | [75,84] | [65,74] | [55,64] |

**Problem 1** (16 points)

a) (4 points) This should be a straightforward computation.

b) (4 points) 2 points for the curves, labeled (1 point per curve). 2
points for the coordinates of the intersections, labeled.

c) (8 points) 4 points for the setup. 4 points for substituting and
then solving. Some students got lost in a swamp of algebra. I
generally gave these students 6 out of 8 points for this part.

**Problem 2** (12 points)

The standard way to solve this, used by all but one or two students,
is to slice the solid by planes perpendicular to the axis of symmetry
which is a line perpendicular to the base through the center of the
square. I gave 2 points for the definite integral ranging from 0 to
60. I gave 4 points for a correct formula for the length of the side
of a square slice. I gave 2 points for squaring that, to get the
cross-sectional area of the slice. The last 4 points were earned for
taking the antiderivative and obtaining the correct answer. Students
who made the antidifferentiation much too easy due to earlier errors
were not eligible for the last 4 points. I tried to read and grade the
solutions of the few students who chose a different method of
computing the volume in the same spirit as the previous outline.

**Problem 3** (18 points)

a) (9 points) 2 points for factoring the denominator of the rational
function, 2 points for writing the correct structure of the partial
fraction decomposition, 2 points for getting the correct values of the
unknown constants, 2 points for the integration, and 1 point for
recognizing that the answer was the same as the quoted one.

b) (9 points) The grading scheme of part a) was used here.

**Comment** Of course one can go backwards from the answers given,
and the answers surely hint at, say, the factorization. I realize
this, but according to the problem statement, I wanted a partial
fraction decomposition displayed for each integrand, the appropriate
equations solved, etc. If these things were not shown (or something
very close!) then full credit was not given. Please note that the
general form of the partial fraction pieces was given on the formula
sheet. Also the derivative of arctan is on the formula sheet.

**Problem 4** (18 points)

**Comment** Some student answers were *especially* difficult
to read in this problem. Here and in the other problems, if I can't
make sense out of what I read, then I cannot give
credit. Communication is part of the exam, and, as I wrote on several
papers, I cannot read "your" mind, and can't (or, rather, won't!)
guess at what should be on the page. I also believe that overly sloppy
writing probably hurts the student's own further work.

a) (9 points) I think integration by parts is needed somewhere in this
problem. Indication of such earns 3 points. Students who made a
mistake early in the parts may end up with an impossible integral, and
nothing can really help them. Please note that the derivative of
arcsin(x) is given on the formula sheet. The value of arcsin(1) was
worth 1 point.

b) (9 points) Several algebraic approaches are successful. Almost
surely a substitution is needed somewhere along the way. Again, I
tried to read what was written. Students who did some preliminary
algebra usually received 3 points for that effort. It is difficult to
go further without some good idea about a substitution. If algebraic
errors simplified the integrand too much, then students were
*not* able to earn full credit for this problem.

**Problem 5** (16 points)

a) (4 points) 2 points for each use of L'Hopital's rule. Eligibility
for the rule is a quotient with specific behavior. Students should
show that each time.

b) (10 points) 8 points for two uses of integration by parts resulting
in the correct answer. 2 points were reserved for some response to the
sentence "Indicate why the limits you need exist ..." This was an
effort to have students not just plug in numbers because ln(x)
*can't* be "plugged in" at x=0.

c) (2 points) Basically the student should agree that the picture does
support the computation, and indicate that the area even considered
approximately is close to the answer in b).

**Problem 6** (10 points)

This was the only problem for which the mean grade was less than half
the possible score (and the median grade was half of the score). I did
not give credit for material which was false or irrelevant to the
problem. Students could earn 5 points for estimating or evaluating an
integral which was relevant to the problem, but the other 5 points was
earned when that estimation was connected correctly to the integral in
the problem statement. I awarded 8 out of 10 points to a solution
which relied on inequalities that were "almost" correct, in the sense
that for x large enough, the specified inequalities were true.

I also note that several students came up with ingenious
arguments quite different from the instructor's answer, and I
compliment them for this.

**Comment** Inequalities and reasoning connected with them seem
difficult when first encountered in a calculus setting. Things do get
easier and learning to cope with inequalities is necessary in many
applications.

**Problem 7** (10 points)

Computation of f´ was worth 2 points, and computation of
f´´ earned another 2 points. If differentiation errors
unduly simplified the balance of the problem, then all of the
remaining points could not be earned. Estimation of K was worth 4
points. Asserting that an overestimate of f´´(x) on [0,2] is
obtained by evaluating f(2) or f(0) and f(2) is certainly *not
valid* and at least 2 points were deducted for this error. One part
of the second derivative, (1+x^{3})^{-1/2}, is a
decreasing function of x and somewhere this needed to be used or,
again, 2 points were deducted. A valid answer for n, supported by the
reasoning and computation shown, earned the last 2 points.

Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Problem #7 |
Problem #8 |
Problem #9 |
Total | |
---|---|---|---|---|---|---|---|---|---|---|

Max grade | 12 | 14 | 15 | 14 | 12 | 12 | 12 | 10 | 12 | 104 |

Min grade | 6 | 5 | 8 | 7 | 0 | 0 | 6 | 1 | 6 | 68 |

Mean grade | 10.43 | 12.71 | 13.90 | 10.33 | 8.57 | 5.52 | 9.62 | 5.33 | 10.43 | 86.86 |

Median grade | 10 | 13 | 15 | 10 | 10 | 5 | 10 | 5 | 11 | 86 |

21 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [85,100] | [75,84] | [65,74] | [55,64] |

**Problem 1** (12 points)

a) (2 points) Each equation of a horizontal line gets a point.

b) (8 points) 1 point each for each correct and correctly labeled curve. 1 point each for each limit.

c) (2 points) 2 points for the correct answer.

**Problem 2** (15 points)

a) (7 points) I deducted 2 points for people who forgot +C at an
appropriate place. I deducted 1 point for bad algebra (incorrected
exponential manipulation, for example).

b) (8 points) I tried to follow the student's solution in a) here,
even if incorrect. 6 points for the correct particular solution, again
deducting points for bad algebra. 2 points for the domain. If the
student's version of the solution unduly simplified the domain
question, then no points could be earned. No points were earned if the
student confused the domain and the range. I took off a point if the
correct particular solution resulting in a domain answer which had
only the restriction that x be non-zero. *The* solution curve
which passes through the point (1,0) only has domain x>0.

**Problem 3** (15 points)

a) (5 points) I gave 2 points for *recognizing* the statement:
first term=5 and fourth term=3 (1 point each). Completing the problem
was worth the balance of credit. I deducted 1 point for faulty
algebra.

b) (10 points) The "setup" earned 6 points: this is the computation of
several of the (different) square side lengths and assembling them
into a sum. 2 more points were earned by the recognition that this was
a geometric series and the final 2 points for the sum of the
series. Faulty algebra lost 1 point.

**Problem 4** (12 points)

a) (8 points) The Ratio Test correctly used earned 6 points. The Root
Test could also have served. Then identification of the interior of
the interval and the radius of convergence was the other 2 points. I
deducted 1 point if there was no clear or correct statement of the
radius of convergence. I deducted 1 point for faulty algebra.

b) (4 points) 2 points for each endpoint. 1 point was earned in each
case for correct insertion of the value to get a series of
constants. 1 point was earned in each case for the correct conclusion
with some sufficient reason. Please note that for x=-1/3, the
Alternating Series Test only shows that the series converges
conditionally, but the correct answer is that it converges absolutely,
so additional reasoning must be given.

**Problem 5** (12 points)

a) (6 points) The Alternating Series Test applies, but I insisted
that all three criteria be checked (or at least mentioned!). Students
who omitted one of the criteria (frequently that the absolute value of
the terms decreased) lost 2 of the 6 points. It is also possible in
this case to strip away the sign change and conclude (using the
Integral Test, say) that the resulting series converges and then,
using the implication {absolute convergence} implies {conditional
convergence} conclude that the series converges.

b) (6 points) If the series satisfies the conditions of the
Alternating Series Test, then the first omitted term supplies an error
estimate. Generally, even if a series converges, the first omitted
term supplies *no information* about the accuracy of the partial
sum. Students who did the problem but omitted the connection with
alternating series were penalized 2 points. Students who did not give
a partial sum lost 1 point. It is also possible, as in a), to delete
the (-1)^{n} and estimate the infinite tail using an integral.
Students who incorrectly estimated powers of log with powers of n and
then used an easier integral estimate did not receive credit. They
simplified the problem *too much*.

**Problem 6** (12 points)

a) (6 points) The Integral Test easily applies. Both the Ratio Test
and the Root Test, when the appropriate limits are correctly
computed, do not give sufficient information for any conclusions (the
limits are 1). Students who indicated a desire (?) to use the Integral
Test were generally given at least 2 points.

b) (6 points) Again, an estimate using an integral works. I deducted 1
point because of the common error in the upper bound (N instead of
N+1). Errors in inequalities were penalized at least 2 points. More
serious errors in logic (overestimates instead of underestimates, for
example, or {A>B} and {A>B} implies some relationship between B
and C) generally received no credit.

**Problem 7** (12 points)

a) (6 points) Most people compared the series to a convergent
geometric series. Other valid approaches, such as comparison to a
p-series with p=2, are also possible.

b) (6 points) Comparing the infinite tail to the infinite tail of a
geometric series was done by most people. The geometric series can be
computed exactly, and it also can be overestimated by an integral
which was done by some. Just asserting that the N^{th} or
(N+1)^{st} term is less than a desired tolerance earns no
points. I allowed 2 points if this was connected with an infinite
tail.

**Problem 8** (12 points)

a) (4 points) 1 point for "Yes" and then 3 points for explaining why.
A generally successful explanation would need to detour via the
implication {absolute convergence} implies {conditional
convergence}. The Ratio Test is not itself valid for the series, since
various terms can equal 0. I *should* have asked if the series
converged absolutely, and then asked students to explain why this
might imply convergence.

b) (4 points) 1 point for "Yes" and then 3 points for an
explanation. A likely explanation would begin with the 2Pi periodicity
of cosine, and that earned a point. No more credit was earned unless
all of the different cosine functions were somehow analyzed.

c) (4 points) 1 point for "Yes" and the other 3 points for some sort
of explanation. Please note that the question is an effort to explain
the graph, and therefore a verification can't *use* the graph,
which could be incorrect or an artifact:

**artifact**

- a product of human art and workmanship.
- [Archaeol.] a product of prehistoric or aboriginal workmanship as distinguished from a similar object naturally produced.
- [Biol. etc.] a feature not naturally present, introduced during
preparation or investigation (e.g. as in the preparation of a slide).
`Maple`could be showing things that aren't correct. It can draw misleading graphs.

**Problem 9** (12 points)

a) (6 points) 2 points for "True" and 4 points for a general
explanation. An example or two gains only 1 more point. Asserting that
the function goes to 0 so the reciprocal of the function ... without
more specification does not earn credit. Discussion of a_{n}'s
denominator or numerator (for an abstract a_{n}) similarly
earns no credit, since I don't understand what that means.

a) (6 points) 2 points for "False" and 4 points for a valid
example. Several people told me that a_{n}=x and thus
a_{n}=1/x, but the context was sufficiently unclear that I did
not understand whether the x was suppose to be n or a constant. I then
deducted 1 point, although either assertion would have led to a
correct example. Clarity should be rewarded and its lack, penalized.

Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Problem #7 |
Problem #8 |
Problem #9 |
Problem #10 |
Problem #11 |
Problem #12 |
Total | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Max grade | 20 | 16 | 16 | 16 | 16 | 16 | 16 | 16 | 18 | 20 | 20 | 16 | 195 |

Min grade | 14 | 15 | 3 | 0 | 6 | 0 | 4 | 13 | 0 | 0 | 4 | 0 | 102 |

Mean grade | 18.48 | 15.95 | 11.95 | 13.29 | 13.52 | 13.05 | 11 | 15.24 | 10.76 | 12.43 | 11.57 | 11.86 | 159.09 |

Median grade | 19 | 16 | 12 | 15 | 15 | 14 | 11 | 16 | 11 | 17 | 10 | 14 | 164 |

21 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [170,200] | [150,169] | [130,149] | [110,129] |

**Problem 1** (20 points)

a) (4 points) Graph of the line (1 point) and the parabola (3 points).
b) (9 points) Finding the intersections (3 points); setup of the
integral (3 points); computation of the integral (3 points).
c) (5 points) 2 points off for drawing too much of the parabola, and 1
point off for drawing the "line" insufficiently linear.
d) (2 points) 1 point for the correct answer and 1 point for a relevant
supporting reason.

**Problem 2** (16 points)

Partial fractions. Full credit for the correct answer.

**Problem 3** (16 points)

a) (6 points) I hoped that these graphs wouldn't be completely unfamiliar!

b) (10 points) 3 points for reading off the antiderivatives (they were
on the formula sheet); 3 points for realizing that this was an
improper integral; 4 points for the necessary rearranging of the
antiderivative and evaluating to get the answer.

**Problem 4** (16 points)

a) (9 points) If errors were made in the parts, several points were
taken off.
b) (7 points) I took off slight amounts of points for slight
errors. This should have been a straightforward application of
a). Actually, several students redid their work in a).

**Problem 5** (16 points)

a) (8 points) Points off for partial or incorrect solutions.

b) (8 points) Points off for partial or incorrect solutions.

**Problem 6** (16 points)

a) (10 points) A correct curve.

b) (6 points) 2 points for identifying two critical points; 2 points
for discerning what type they are; 2 points for finding an algebraic
relationship. People should know what the phrase *critical point*
means.

**Problem 7** (16 points)

Separating the variables: 2 points; antidifferentiation of y: 2 point;
antidifferentiation of x: 5 points; getting the constant: 2 points;
solving for y as a function of x: 2 points; stating the domain: 3 points.

**Problem 8** (16 points)

8 points each for sections a) and b), which were graded similarly. I
happily report that no one in the class worked directly with the
algebraic fourth derivative in b), although a few people worked with
the algebraic second derivative in a). I took off 1 point for
reversing inequalities (I graded the sections independently), and 1
point for other minor errors. Most people indeed did get the bounds on
the derivatives from the given graphs, and did use the correct error
formulas. Of course, the answers given for the number of partitions
were quite different. The largest answer was n=10^{1000},
which got full credit since enough supporting evidence was given. I
just hope that the person giving that answer never has a computational
task which takes 10^{1000} steps!

**Problem 9** (20 points)

Yes, this was a *proof* and verification for one or two values
for the sequence of initial steps is not logically definitive.

Parts a), b) c) d), e), and f) each were worth 2 points. I then hoped
that students would use this information to conclude that the sequence
defined in g) was increasing and bounded above, and therefore must
converge. g) was worth 8 points. The limit of the sequence turns out
to be a root of the function whose graph is displayed. The instructor
thought that this might help students. I did write that "These numbers
may help you emotionally but they should *not* be used in the
proof." But some people did try to use them, and information about the
first 10 terms of the sequence is not enough to conclude anything
about the limit. Please see here.

**Problem 10** (20 points)

Each part was worth 10 points. Again, a reversed inequality lost a
post. More costly were errors in logic which I penalized
strictly. Part a) had been done in front of most students several
times. Part b) was a different logical take, but again, things like it
had been done in class a number of times. In b) either Taylor's
Theorem with n=4 or 5 or consideration of the alternating series
nature of cosine's Taylor series are good approaches.

The instructor was disappointed by the grades on this problem. On the
other hand, perhaps the grades indicate that the topic was not taught
as well as it should have been.

**Problem 11** (20 points)

a) (8 points) Here the best strategy is to use the Ration Test (almost
every student did this).

b) (2 points) I looked for "Absolute convergence implies convergence."
(*Any* kind of convergence!) Read the question, which asks
"... why the result of a) implies that the series converges ..."

c) (10 points) 5 points each for discussions of J(1) and J(4). I
wanted specific reasons, not that "something" was "smaller" than
"something".

**Problem 11** (16 points)

a) (8 points) 4 points for each curve. A two-lobed "circle" lost 2 points.
b) (8 points) 4 points for the setup and 4 points for the
computation. The setup is slightly subtle, as I warned on the exam.

**
Maintained by
greenfie@math.rutgers.edu and last modified 12/27/2005.
**