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** (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 relvant
supporting reason.

**Problem 2** (16 points)

Partial fractions. Full credit for the correct answer.

**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.
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 8** (16 points)

8 points each for sections a) and b), which were graded
similarly. Luckily 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 correcte 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 computation
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 f) was increasing and bounded above, and therefore must
converge. f) was worth 8 points. The graph is of a function one of
whose roots is the limit of the sequence defined.

**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 more
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 condsideration of the alternating series
nature of cosine's Taylor series are good approaches.

**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 appllies, 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 correcatly
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 succesful 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.

b) (4 points) 1 point for "Yes" and then 3 points for an
explanation. A likley 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 vaild
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.