Problem | #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | Total |
---|---|---|---|---|---|---|---|---|---|
Max grade | 16 | 9 | 13 | 12 | 13 | 13 | 12 | 12 | 99 | Min grade | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 12 | Mean grade | 10.32 | 7.29 | 7.35 | 6.40 | 2.51 | 6.82 | 7.64 | 5.20 | 53.53 | Median grade | 9 | 8 | 8 | 5 | 0 | 8 | 8 | 6 | 54 |
Numerical grades will be retained for use in computing
the final letter grade in the course. Students with grades of D or
F on this exam should be very concerned about their likely success in
this course.
Here are approximate letter grade assignments for this exam:
Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|
Range | [85,100] | [80,84] | [70,79] | [65,69] | [55,64] | [50,54] | [0,49] |
The following applies to students who
got below 40 on this exam.
Observation Most such students show very serious defects in
understanding and using first semester calculus, and many seem to have
problems with basic algebraic manipulation.
Expectation These students should not expect that their lack of
preparation can be addressed by the instructors in this course -- we
will have more than enough to do trying to help people learn new
material. Students who don't know earlier material may devote much
time to this course with very little chance of success.
Recommendation I am reluctant to write what follows but
realistically these students should drop the course, and begin
attending and doing the work in a calculus 1 course, and get more
knowledge of that material. When they are properly prepared, they
could try calc 2 again. I worry that people will spend time on calc 2,
time and effort which could be devoted to other tasks, and not be
successful. They have no reasonable chance of success in this
course.
Grading guidelines
Minor errors (such as a missing factor in a final answer, sign error,
etc.) will be penalized minimally. Students whose errors materially
simplify the problem will not be eligible for most of the
problem's credit.
The grader will read only what is written and not attempt to guess or
read the mind of the student.
The student should solve the problem given and should not invent
another problem and request credit for working on that problem.
Problem 1 (16 points)
a) (5 points) The region should be roughly three-sided with the
correct intersection point of the given line and curve inside the
first quadrant as one corner (1 point), one side should be the y-axis
(1 point), another side should be a straight line (1 point), and the
third side should be slightly curved, at least (1 point). The region
should be labeled (1 point).
b) (11 points) dx: integral and limits are worth 6 points. 3
points of these will be earned for the correct limits on the integral,
correct multiplier (2Π) and dx. No further points will be earned
if the integrand is incorrect. There should be 3 terms to
antidifferentiate. Then antidifferentiation and evaluation is worth 5
points. Students who insist on "simplifying" the answer arithmetically
lose 1 point if this is done incorrectly.
dy: integrals and limits are worth 7 points. 3 points of
these will be earned for the correct limits on the integrals, correct
multiplier (Π) and dy. 2 points are earned for writing x in terms of
y. No further points will be earned if the integrand is incorrect.
Evaluation is worth 4 points.
Problem 2 (9 points)
a) (5 points) Using the wrong formula is penalized 3 points, but 2
points can still be earned for reading the graph for function values.
The factor (1/3) earns 1 point. Then 2 points for the function values
and 2 points for the multipliers or weights used in Simpson's
Rule. Students who insist on "simplifying" the answer arithmetically
lose 1 point if this is done incorrectly.
b) (4 points) 1 point for selecting the correct error estimate, 1
point for the correct K4 value, and 2 points for writing
the correct a, b, and N. Using an incorrect error estimate
(not for Simpson's Rule) loses 2 points, and only the 2 points
for a, b, and N can be earned. Students who insist on "simplifying"
the answer arithmetically lose 1 point if this is done incorrectly.
Problem 3 (13 points)
Setting up the correct symbolic sum earns 3 points but without this no
points can be earned in this problem. Solving for the
correct values of the constants earns 4 points. Antidifferentiation of
the terms earns 4 points. Checking that the evaluation is correct
earns the last 2 points, but these points cannot be earned if
incorrect values of the constants or incorrect integration occurs.
Indeed, perhaps students who assert that verification of the
integral's value has occurred without actual evidence should lose
additional points for an ethical violation!
Problem 4 (12 points)
One integration by parts earns 4 points, and another earns 4 points. 2
points are earned by putting them together correctly. The last 2
points are reserved for evaluating ln's correctly (yes,
ln(ew) should be reported as w). Students who insist on
further "simplifying" the answer arithmetically lose 1 point if this
is done incorrectly. Students who give a totally incorrect
antiderivative are not eligible for any points in this problem,
including evaluation points.
Problem 5 (13 points)
A totally incorrect antidifferentiation showing no chance of success
makes the student ineligible for earning any points in this
problem, including points for evaluation.
A substitution which moves things forward earns 3 points. Changing the
entire indefinite integral correctly into the substituted variable
earns 2 more points. Performing the antidifferentiation correctly
earns 3 more points (using partial fractions followed by
antidiffererentiation). Then 2 more points for change back to x's.
1 point for evaluation, and then 2 points for showing that the value
of the integral is less than 2/3. There must be an explicit reason
given.
I mention that some students were successful (or very nearly
successful) in this problem using strategies which I had not
anticipated: weird (to me!) but successful strategies. They received
full credit (or very nearly full credit) for solution of the problem.
Problem 6 (13 points)
Performing the antidifferentiation is worth 5 points. Then evaluating
the definite integral is worth 3 points. 3 more points are earned for
dividing by Π/3. 2 points for answering the question about the size
of the answer (students who do not divide by Π/3 can earn only 1 of
these points). 1 of these points is for the answer and 1 of these
points is for an explicit reason relying on the size of
Π. Students who give a totally incorrect antiderivative or show no
useful calculus work are not eligible for any points in this
problem, including evaluation points.
Problem 7 (12 points)
Students who attempt some weird strategy which cannot be successful
are not eligible for any points in this problem. The substitution
x=sin(θ) itself earns 3 points. Changing the integral into the
correct θ integral earns 3 points. Integrating
(sin(θ))2 correctly earns 3 points. Changing back to
x's earns the last 3 points.
It is possible that other substitutions may be used, and the grader
will attempt to understand the likely final success of the
strategy. If the substitution is unlikely to be successful, then no
points will be earned.
Problem 8 (12 points)
3 points for some kind of proportional sides equation. Assembling the
slice as volume·density earns 3 points (area of a square
multiplied by height multiplied by density). Writing a definite
integral for the work earns 3 points. Computation of the definite
integral, including evaluation, earns 3 points. Students who insist on
"simplifying" the answer arithmetically lose 1 point if this is done
incorrectly.
Problem | #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | Total |
---|---|---|---|---|---|---|---|---|---|---|
Max grade | 12 | 9 | 11 | 12 | 12 | 12 | 10 | 10 | 12 | 100 | Min grade | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 14 | Mean grade | 8.49 | 2.89 | 9.22 | 6.54 | 4.96 | 4.80 | 7.65 | 5.77 | 5.11 | 55.42 | Median grade | 9 | 2 | 11 | 6 | 2 | 4 | 9 | 6 | 5 | 56 |
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] | [80,84] | [70,79] | [65,69] | [55,64] | [50,54] | [0,49] |
Grading guidelines
Minor errors (such as a missing factor in a final answer, sign error,
etc.) will be penalized minimally. Students whose errors materially
simplify the problem will not be eligible for most of the
problem's credit.
The grader will read only what is written and not attempt to guess or
read the mind of the student.
The student should solve the problem given and should not invent
another problem and request credit for working on that problem.
Problem 1 (12 points)
a) (6 points) Computation of derivatives earns 2 points. Evaluation of
the derivatives and the original function earns 1 point. 3 points for
assembling and writing the Taylor polynomial, which should be a
polynomial (!), have the correct degree and "center", and have the
correct coefficients (derivative values divided by the appropriate
factorials). No simplifying evaluations of the function or the
derivatives must be done (so 35 or 95/2 as the
constant term in T3(x) for version B is acceptable).
Students whose answer for T3(x) is not a polynomial earn at
most 3 points.
b) (6 points) Computation of the 4th derivative earns 1
point. Discussion of the behavior of this derivative on the interval
({in|de}creasing) earns 2 points, and 1 more point for obtaining a
correct overestimate of this function. Then 2 more points for
assembling the other parts of the error estimate (the radius of the
interval raised to the appropriate power divided by 4!). The method
of solution here must use the Error Estimate -- that's part of
the problem statement.
Problem 2 (9 points)
a) (5 points) 2 points for computing f´(x), and 3 points for
using this to verify that f(x) is a solution of the differential
equation.
b) (4 points) Use the initial condition to write the requested
solution. Some work must be shown. While the requested f(x) is the
"best" answer, full credit is earned if only the specific value of C
is given. 2 of the points are lost if x and y are interchanged. Only 2
points are earned if the correct C is given with no work.
Problem 3 (11 points)
a) (3 points) 1 point for each equilibrium solution. The solutions
should correctly be written as "y=constant". If instead each constant
is identified in some manner which is not wrong (definitely wrong is
"x=constant"!), 1 point will be deducted.
b) (8 points) Each graph is worth 2 points. If the label is missing on
a graph, the student loses 1 of the points. The graph for A
(respectively, B) should go through the point (0,1)
(respectively, (0,–1)), and the curve drawn should be continuous,
increasing (respectively, decreasing), and between the horizontal lines
x=0 and x=1 (respectively, x=–1). Substantial "violations" of this
behavior could lose 1 point each, up to the total value of the
graph.
The answer to each limit is worth 1 point.
Problem 4 (12 points)
3 points for realizing or using the "key observation" that dropping
the square root increases the size of the fraction (using the other
part of the formula doesn't earn credit since it cannot be estimated
usefully); 3 points for showing that the infinite tail must be
estimated (here either a geometric series or a relevant [improper]
definite integral can be given); 3 points for estimating the infinite
tail by finding the sum of the relevant geometric series or with a
correct antiderivative; 3 points for using the sum and the
tabular information (or a correct integral!) to get a correct
answer. Merely asserting an answer is not sufficient to get credit
here (lots of data was supplied). Asserting that an infinite tail is
small because one or a few terms are small also earns no credit.
Students who base their answer on the size of the Nth term
rather than analyzing the whole sum are eligible only for the first 3
points, the comparison points.
No credit will be given for a purely arithmetic approach to this
problem because this is a calculus course.
Note that N=346,578 is acceptable if supporting reasoning is given!
An approach using comparison to an improper integral can certainly be
done here. The grading will follow an outline similar to that of
problem 5.
Problem 5 (12 points)
4 points for some evidence connecting the sum to the integral. One
acceptable item would be a picture similar to that displayed on the
answer sheet. An explicit inequality connecting the Nth
partial sum with a definite integral would also be acceptable. Also
useful would be mention of a relevant function decreasing. But
some evidence should be given to earn these 4 points. Students who
only compare the term to a simpler term will earn 2 of these
points.
4 points for evaluating a relevant definite integral. 2 of those
points are awarded only if there is a limit expression for this
improper integral.
4 points for a correct answer with evidence showing that specific N is
valid. Students who use N+1 in place of N in an otherwise correct
solution will be penalized 1 point.
Students who base their answer on the size of the Nth term
rather than analyzing the whole sum are eligible only for the first 2
points, the comparison points.
No credit will be given for a purely arithmetic approach to this
problem because this is a calculus course.
Note that N=346,578 is acceptable if supporting reasoning is given!
Problem 6 (12 points)
a) (6 points) Transforming the sequence to a form where
L'Hôpital's Rule could be used is worth 2 points (that is, using
ln); 1 point for remarking on eligibility; 2 points for using
L'Hôpital's Rule (differentiation of the top and bottom); 1
point for the answer. For that point, the student should recognize
that e0=1.
b) (6 points) 2 points for an equation relating a term of the sequence
to the next term; 2 points for taking limits in this equation
resulting in a correct algebraic condition for the limit; 2 points for
the solution of the algebraic condition (1 point is lost if, in
solving the quadratic, no reason is given for selecting the positive
root).
Problem 7 (10 points)
a) (5 points) The area of a circle earns 2 points and writing or
considering the total area of all the circles earns 2 more points. The
answer gets 1 point. A correct answer without substantiation gets 1
point. An explanation which fails to distinguish clearly between the
{con|di}vergence of a series and its associated nth term
loses 2 of the points.
b) (5 points) The circumference of a circle earns 2 points and writing
or considering the total circumference of all the circles earns 2 more
points. The answer gets 1 point. A correct answer without
substantiation gets 1 point. An explanation which fails to distinguish
clearly between the {con|di}vergence of a series and its associated
nth term loses 2 of the points.
Problem 8 (10 points)
Several approaches seem likely. One is to use the Alternating Series
Test, which does apply. Stating this strategy would earn 2
points. Then verifying or mentioning the alternating nature of the
terms earns 2 points. 6 points remain for realizing that the terms
decrease with limit 0. Direct verification of this may be difficult
and only assertions earn no credit.
Some mention of absolute convergence implies convergence earns 2
points. The Ratio Test can be used for convergence (without the sign
variation, this was a textbook homework problem assigned to be handed
in). Actual detailed implementation of the Ratio Test is worth 6 more
points: 2 points for substituting n+1 for n correctly, 2 points for
going from a compound fraction to a simple fraction, and 2 points for
cancellation. Then the final 2 points for the limit and
conclusion. The conclusion officially, in the text and in the
lectures, is absolute convergence. So the first 2 points for solution
of this problem can't be earned without some mention of the connection
between absolute convergence and convergence.
Problem 9 (12 points)
3 points for the correct equation of the straight line. 2 of these
points are earned for f´(x)=b/a. 3 points for the correction
"instantiation" of the surface area formula (not quoting the formula
-- the f(x) and f´(x) should be replaced by what's relevant
here). 2 points for antidifferentiation, 2 points for evaluation, and
2 points for realizing that the result is what's needed.
Students who have ∫ab together with the
correct instantiation of the surface area formula and the correct
antiderivative earn 7 points.
The final exam and course grades
Initial grading
Grading of the final exam was done on Friday, May 8, the day following
the exam. Each lecturer and recitation instructor graded the same
problems for all students. Here are the letter grade assignments for
this exam:
Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|
Range | [160,200] | [145,159] | [130,144] | [115,129] | [100,114] | [90,99] | [0,89] |
You can see that these "bins" are similar to those used in the first two exams given to students in these sections. 554 students in Math 152 took the exam on May 7. Additional students who need to take the exam at other times (conflicts, etc.) will take the exam soon. This whole group of students had grades ranging from 8 to 192 with a median of 115. The students in sections 1, 2, 3, 6, 7, 8, and 9 had grades ranging from 28 to 185 with a median of 124 and a mean of 118.74. About 150 students in these sections took this exam.
Checking and regrading
After the exams were graded, I spent more or less the entire weekend
checking the grading of "our" exams. This was painful, boring, and
annoying. I found some grading errors (including a few of mine, of
course) and some arithmetic errors. I believe that I discovered almost
all of the errors! I hope I discovered all of them but that may
be unreasonable. I spent most of two days doing this because exams
should be accurately graded.
About the exam
Certainly the exam was somewhat different from one I would give, but
also certainly more than three-quarters of the problems were standard
and straightforward. Many people would have higher grades if they read
the problems carefully and tried to answer the questions which were
asked -- this should be a familiar statement. Enough time was
given. Generally final exam grades were consistent with earlier
performance in the course.
Use of these grades
The final exam letter grades are given to allow individual lecturers,
who are responsible for reporting course grades, to align the
performance of their specific groups of students with the overall
performance of students in the course. Random chance may give one
lecturer a group of students with better or worse performance than the
overall population, and, indeed, specific scheduling requirements may
force specific subpopulations with different math preparation to
enroll in some lectures. The common final exam and grading are part of
the faculty's effort to assign appropriate course grades, with equal
grades for equal achievement.
Course grading
The information I had included the following: grades for the three
exams (two during the semester and the final), attendance information
(for the lecture, from the QotD, and for the recitation, as reported
by the peer mentors), textbook homework scores (reported by the peer
mentors), and workshop grades (reported by Mr. Jin and Mr. Yin,
although I did some of the grading also). This is a large mess of
numbers but my electronic friends handled the arithmetic with little
strain. It is slightly humorous that six different methods of
recording the homework grades were used by the seven peer mentors,
even though I had requested a specific format at the beginning
of the semester. My electronic friends were able to cope, but this was
an irritation.
I computed a number for each student essentially weighted as previously described. Since
there were no quizzes, I reduced the weight given to "Quizzes and
attendance" from 70 points to 50 points. I then assigned a tentative
letter grade based on breakpoints proportionately derived from the
"bins" shown above for the exams. I examined each student's record to
make sure that this process had not distorted or misrepresented
student achievement. I entered the course grades into the Registrar's
computer system late on Monday, May 11. I hope students will be able
to see them soon.
If you have questions ...
Rutgers requests that I retain the final exams. Students may ask to
look at their exams and check the exam grading. These students should
send me e-mail so that a mutually satisfactory meeting time can be
arranged. Students may also ask how their course grades were
determined using the process I described. Probably e-mail will be
sufficient to handle most such inquiries.
Warning Students who officially passed the course are allowed to go on to Math 251. Those students with D's should think carefully about their progress in cumulative math courses and associated work.
Maintained by greenfie@math.rutgers.edu and last modified 5/11/2009.