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

Max grade  8  8  8  8  8  8  10  10  8  12  12  93 
Min grade  0  0  0  0  0  0  0  0  0  0  0  18 
Mean grade  7.08  6.36  4.09  4.31  7.48  3.22  2.81  6.45  4.18  5.88  7.00  58.87 
Median grade  7  8  4  6  8  4  1  8  6  5  7  60 
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. I recommend that students with a grade of 50 or less
seriously consider dropping this course and concentrating on other
courses. The material in the course gets much more intricate and
difficult. This exam was nearly minimal. The majority of the problems
were taken directly from homework problems and these problems were not
the most difficult which could have been asked.
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 (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.
Regrading and grading errors
13 students requested regrading of some problems on their exams. The
grading of 5 students' exams was incorrect, and a total of 15 points
were added. This is about 17/100ths of a percent of the total of 8500
points on the exams.
Problem 1 (8 points)
Differentiation of the components of r is worth 5 points: 2
points each for the first two components (1 point off for multiplier
error), and 1 for the third. Plugging in is worth 2 points, and
normalizing is worth 1 point. I took off 1 if the 2 "plugging" points
if values of sine/cosine were missing or incorrect.
Problem 2 (8 points)
Finding a vector in the direction of the line is worth 4
points. Realizing that one of the points is on the line is
worth 1 point. Writing a vector equation is worth 3 points (1 point
off for a collection of scalar equations  I copied the phrasing of
the text here).
Problem 3 (8 points)
Each limit is worth 1 point, so the values of the limits are worth a
total of 2 points. The graph is worth 6 points. It should be smooth,
and "unimodal": up then down, with limiting behavior near 0 (2
points). It should have a part which is close to .5 (1 point), and
that part should be about 6 to 7 units long (1 point). Near the origin
(s=0), the graph should be increasing from near 0 to near .5 within
about 1 unit (2 points).
Problem 4 (8 points)
Each correct vector earns 2 points. Discussion verifying that they are
orthogonal is worth 2 points and verification that they are not
multiples of one another is worth 2 points. I will not accept
(0,0,0) as either one of the vectors.
Problem 5 (8 points)
The correct answer is worth 2 points. 6 points for the process. At
least some evidence must be offered that a process has been done!
Problem 6 (8 points)
2 points each for the answers, and 2 points each for supporting
reasons. The answers must both be correct and
contextual: I mean the answers should have some relevance to the
bee. Therefore a statement about "the arclength" or "the
distance" is not acceptable as an answer (because I want to know what
arclength or what distance as it refers to the circumstances of this
problem). An answer such as "The bee is located at r(T)" will earn no
credit since it does not indicate the student has responded to the
circumstances of the question. Also, to earn full credit, some
supporting reasons for each of the answers must be presented. Many
interesting answers were presented which earned no credit.
Problem 7 (10 points)
Each level curve can earn 3 points: 2 points for the picture, and 1
point for the correct label. Students will not lose credit for
their graphs if special attention is not given to (0,0) although such
attention is indeed correct. That the limit does not exist is
worth 1 point, and the explanation is worth 3 points. The value of f
at (0,0) (or even if f(0,0) exists) is not relevant to this
question (go back and look at calc 1 examples, please, and check on
this!).
Problem 8 (10 points)
3 points for the first differentiation: 2 points for the Chain Rule,
and 1 point for further work. 5 points for the second differentiation:
2 points for the Chain Rule, 2 points for the product rule, and 1
point for further work. 2 points for numerical substitution (1 of
these is for e^{0}=1).
Problem 9 (8 points)
2 points for a correct f_{x} and 2 points for a correct
f_{y}. 4 points for assembling these into a correct
verification of the PDE. Working with a specific function f earns no
points (!). I believe that g´ must appear to give evidence
that the correct version of the Chain Rule has been used. This problem
can't be done without the Chain Rule!
Problem 10 (12 points)
a) (8 points) 2 points for correct computation of partial derivatives;
1 point for evaluation of them; 2 points for recognizing the result as
a normal vector, and then 3 points for correctly assembling the
equation.
b) (4 points) 2 points for the unit vector: recognizing the gradient
evaluated in a) and then normalizing. 2 points for the value of the
directional derivative: recognizing that the length of the gradient is
the answer. The points are for numerical answers which need not be
simplified. An error in the computation of the gradient in a), if it
does not trivialize this part of the problem, allows a student to be
eligible to get full credit here. In version B, as explained above,
either (2,1,1) or (1,2,1) may be used.
Problem 11 (12 points)
Computing f_{x} and f_{y} is worth 2 points. Locating
the critical points earns 2 points. Finding the algebraic form of the
discriminant is 2 points. Each diagnosis (identification) of the
nature of a critical point is worth 3 points.
Problem  #1  #2  #3  #4  #5  #6  #7  #8  Total 

Max grade  10  12  12  12  12  12  12  16  97 
Min grade  0  3  0  0  0  0  0  0  25 
Mean grade  6.64  8.13  8.21  7.34  7.72  8.68  5.99  4.43  57.13 
Median grade  6  8  8  8  8.5  10  6  3  57 
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 (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.
Regrading and grading errors
3 students requested that I check the grading of their exams. One exam
had a 5 point error, due to me, which is
not good, even though it is 5 points out of a total of 7800. Sigh.
Problem 1 (12 points)
4 points for setting up the Lagrange multiplier equations. Either all
4 equations must be explicitly listed together, or they must be
identified and used in the analysis which follows.
6 points for the correct analysis of these equations. The analysis must include discussion of
the case y=0 and z=0, and this is worth 2 of the 6 points. 1 point
is lost if the signs are not correctly understood. That is, x,y,z
should be +,+,+ or +,, or ,,+ or ,+,: this is somewhat subtle! 2
of these 6 points are given if some analysis, even if resulting in
incorrect answers, is done. A third point is earned if the analysis is
successful except for the sign comments and the consideration of 0
values as previously remarked.
2 points for the answers. Specification of where the max and min occur
without the requested values loses 1 point.
Problem 2 (12 points)
Computation of the given iterated integral is worth 4 points. 3 points
for the picture and a total of 5 points for the integrals in answer to
c): 2 points for the rectangle and 3 points for the other. 1 point
penalty for not including the integrand.
In b) a student earns 1 point for sketching the correct curve but
specifying an incorrect region.
If a student supplies only one double integral in c), then at
most 3 points may be earned. If a bound is missing or incorrect, then
that's reduced. Therefore the common and incorrect answer of
one integral with the form
∫_{0}^{Constant}∫_{0}^{ycorrect power}xy dx dy
earns 2 points.
Problem 3 (12 points)
8 points for the setup. This should be a "grammatical" triple integral
(1 point: by this I mean that the limits do not refer to variables
which have already been integrated!) and 1 point for the outer
limits. 3 points for each pair of inner limits. 4 points for a correct
computation (1 of these for a correct answer).
If the triple integral begins
∫_{A}^{B}∫_{C}^{D}∫_{stuff}^{other stuff}Whatever dV
(where A, B, C, and D are constants related to the problem) then a
minimum of 3 points are deducted for the limits, and a maximum of 2
points can be earned for the computation since the computation has
been materially simplified by this error. A negative or zero volume
also loses a point!
Problem 4 (12 points)
Setup of the boundaries of the double integral is worth 4 points, and
4 more points for conversion of the integrand and the area element. 4
points for a correct computation.
3 points for the setup in rectangular coordinates (and those will be
the only 3 points awarded!).
A relevant correct picture alone earns 1 point.
Problem 5 (12 points)
6 points for the boundaries of the triple integral, and 2 more points
for the integrand and the volume element. 4 points for a correct
computation (1 of these for the correct answer).
Problem 6 (12 points)
5 points for the boundaries of the triple integral, and 3 more points
for the integrand and the volume element. 4 points for a correct
computation.
Problem 7 (12 points)
Part a) is worth 6 points. 2 points for some correct formula for the
Jacobian, and then the other 4 points for working out the result
desired.
Part b) is worth 6 points. Each semicircle is 2 points and each line
segment is 1 point. Not knowing e fairly well loses a point!
Problem 8 (16 points)
Part a) is worth 8 points. The line integral over the circular arc is
4 points and over the line segment is 4 points. 1 point is lost if
there is no correct total answer. If the student tries to use a
potential, then 2 points for getting the correct potential, 2 points
for using it correctly, and 4 points for verification that the
suggested potential is correctly related to the given vector field.
Part b) is worth 8 points. 2 points are for the correct value of A and
2 points for reasoning supporting this value. 2 points are for the
correct general potential (1 point off for no "+C") and 2 points for
some correct supporting reasoning (can be differentiation or
integration, but some reasoning is needed).
Problem  #1  #2  #3  #4  #5  #6  #7  #8  #9  #10  #11  #12  Total 

Max grade  18  20  14  16  20  20  16  20  16  19  12  8  182 
Min grade  0  7  0  0  0  0  0  0  0  0  0  0  19 
Mean grade  11.33  14.38  1.08  8.33  10.85  11.94  9.38  10.42  7.68  8.36  10.31  6.18  110.24 
Median grade  15  15  0  8  15  13  11  10  5  9  12  7  110.5 
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  [155,200]  [145,154]  [130,144]  [120,129]  [100,119]  [90,99]  [0,89] 
Some general background
No answer sheet is available (there's got to be something good about a
final exam, and in this case, the "good" is that the lecturer won't
prepare an answer sheet). Here is a more compact version
of this exam, taken by 72 students.
Please note that Rutgers regulations require that I keep the exams for a year. Students may look at their exams and check the grading. If you want to do this, please send email.
With several specific exceptions noted below, quoting from the formula sheet (even if the formulas are relevant!) will earn no credit. Arithmetic errors and "small" errors (such as minor errors in differentiation) will be penalized minimally. The resulting work can be considered for full credit in the remainder of the problem unless the error makes the problem much easier.
The numbers tell me ...
I did a terrible job discussing the several variable chain rule
(problem #3 above). An earlier exam problem with a different chain
rule calculation also revealed student incapacity to deal with
composition and differentiation. Composition and differentiation are
used to understand and create solutions of differential equations so
this is an important topic. If/when I teach this material again, I
should do a better job.
The solutions tell me ...
Some students tried to use the BIG theorems of vector calculus
(Green's/Stokes/Divergence) in what seem inappropriate situations to
me (such as problems 1, 4, and 5). These problems can be solved with
routine manipulations and don't need complicated tools. These students
seem to have spent much time and effort on their work, generally with
little success. The attempts surprised me.
Problem 1 (18 points)
8 points for a correct setup, in any order (each piece is worth 4
points, split up 2 and 2 for each integral). Then 10 points for the
computation (each piece is worth 5, and of that 5, 2 points for the
first integration and 3 points for the second).
Comment This was supposed to be an easy introductory
problem, designed to get students "warmed up" for the exam. Instead,
there were many details to check in the computation, and hardly any
student got the correct answer (which is 5/3, by the way)! Also a
handful of students attempted to compute the requested value of the
double integral using Green's Theorem. This was imaginative and even
valiant, and the approach certainly was not anticipated by the exam
writer. So the line integral of, say, ((x^{2}/2)+xy)dy around
the boundary of the region would equal the answer needed. I don't
think anyone was successful at this, but it is certainly an
interesting approach. Maybe I should apologize  but I really did
think this was a straightforward problem and could be computed
straightforwardly.
Problem 2 (20 points)
a) (4 points) 2 points for the surface and 2 points for
the curve.
b) (8 points) 5 points for computing ∇f and 3 points
for computing C´(t).
c) (6 points) 3 points for a correct equation of the plane; 3 points
for correct parametric equations for the line.
d) (2 points) 1 point for the answer and 1 point for a reason. Various
reasons are possible, but taking the dot product of a normal vector to
the tangent plane
and a vector in the direction of the line and noting that the result
is not zero is not sufficient! If the objects in this problem
were perpendicular, then the dot product of these vectors would not be
zero.
Problem 3 (14 points)
4 points for g_{x} and 4 points for g_{y}. Then
combining this information successfully earns the other 6 points.
No points are earned if a "general" G is not exhibited.
Problem 4 (16 points)
The information that the unit circle is a circle of radius 1
centered at (0,0) will be given if requested.
Use of polar coordinates is worth 2 points. Successful conversion of
the integrand is 2 points. Writing the problem correctly as an
improper iterated polar integral is worth 5 points. 4 of the remaining
points are earned by correct computation of the dr part, and handling
dθ correctly earns 2 points. 1 point is for the answer.
Students attempting this problem in rectangular coordinates will earn
some credit, but no more than 5 points for a totally correct
setup. Computation in rectangular coordinates is not feasible by hand.
Problem 5 (20 points)
Signaling a decision to compute in spherical or cylindrical
coordinates earns 2 points (for example, ρ^{2}sin(φ)
observed). Limits of 0 to Π and 0 to Π/2 earn 2 points. Writing
δ correctly (with a symbolic constant of proportionality) in the
selected coordinate system earns 4 points. Omission of a constant of
proportionality loses 2 points, and the wrong power of the distance
loses 2 points. The correct setup of the mass integral, with a
symbolic R, will earn 6 points. Computation with the correct answer
(with the symbolic constants) earns the remaining 8 points. Omitting
the R in the computation loses 2 points. Dividing by volume (why do
this?) loses 2 points.
Problem 6 (20 points)
a) (10 points) 3 points for the answer alone, and 7 points for a
valid process. If only constants are shown in the antiderivatives with
no variables, 2 points are deducted. If the constant "functions" have
the same names, 1 point is deducted. If no constants are shown, then 4
points are deducted. Students who compute the curl of F, get 0,
and conclude that a potential function exists but don't find the
function earn 4 of the 10 points. Those who supply an answer along
with the curl computation earn 8 points (I want some process shown to
get the answer, or direct verification of the answer).
b) (10 points) 2 points for the answer alone, and 8 points for a valid
process: 3 points for stating or using
P(END)P(START), 2 points each for start and end,
and 1 point for the correct answer. It is also possible to earn full
credit for a direct computation: parameterization, integration, and
evaluation.
Problem 7 (16 points)
The first derivative level (8 points) The partial derivatives
each earn 2 points. Finding the critical points: 4 points, including 1
point each for the solutions. Students who find the first derivative
and then assert that (0,0) is a critical point, without other
information, will earn 4+1=5 points. Therefore algebraic investigation
of the critical point equations is worth 2 points.
The second derivative level (8 points) Computing the second
partial derivatives earns 4 points. Evaluating them at the points
found in a) and correctly using the Second Derivative Test on each
point earns 2 points each. Students who only investigate (0,0) may
earn 4+2=6 points.
The pictures I tried to check by asking Maple to graph the function in this problem. I had to look very carefully at the graphs to see the local nature of the critical point. The graphs shown (I didn't include the axes or the scales or the labels) are all "windows" within +/.05 of the critical points. 

Problem 8 (20 points)
a) (8 points) For the process and answer. Students who do not put the
2's in the correct place when integrating the exponential function may
earn only 5 points (this makes the problem simpler!). Similarly,
students who refuse to handle the lower bound of the evaluation
correctly (e^{0} is 1, after all!) also may earn only 5
points, since this ignorance certainly simplifies the computation.
b) (12 points) For the answer (6 points) with some indication of how
it was gotten (6 points). Students who merely interchange the bounds
on the integrals (so give as answer
∫_{z=0}^{z=1}∫_{y=0}^{y=z}∫_{x=0}^{x=y})
earn 2 of the 6 points, and that's generous! Such people are ignoring
the essential difficulties of what's asked. Some of the 6 points of
process credit may be earned if there are comprehensible diagrams
present which are relevant to the triple integral.
Problem 9 (16 points)
a) (13 points) Finding and using a formula to get some sort of correct
answer is 8 points. Manipulating the formula is worth 5 points.
Students whose second derivatives ignore the product and chain rules
earn 2 points in this part. This major error simplifies the problem
materially, and most of the algebraic manipulation is lost.
b) (3 points) Something like "It's a circle" earns 1 point. Just a
diagram of a circle alone earns 1 point. To earn full credit, a more
complete explanation is needed, including connecting this parametric
curve to a circle and mentioning that a circle has constant positive
curvature and that this circle has radius 1.
Problem 10 (20 points)
Indicating "Stokes' Theorem" will earn 2 points. Computing the curl of
the given vector field earns 7 points. Computing the correct normal
vector for the parallelogram will earn 3 points. Completing the
computation is the last 8 points.
People who parameterize will earn at most 5 points.
I may have mislabeled the picture, but I am not sure, since the darn
coordinate axes are not labeled, and the viewer's angle of the
parallelogram can really change what is seen. In any case, the picture
supplies some help in determining a surface normal and this is needed
for successful computation of the answer. Solution of the problem is
computing two cross products. One is ∇xF, and the other
is a cross product of two correctly chosen sides of the
parallelogram. The direction of the second product is the direction of
he surface normal and its magnitude is the area of the parallelogram:
a sneaky problem indeed.
In the posted version, I inserted "the" before the word
"parellelogram" in the first sentence, since I believe that makes more
grammatical sense!
Problem 11 (12 points)
Just writing "Divergence Theorem" or indicating even implicitly that
it should be used earns 2 points. Computing the divergence of the
given vector field earns 4 points, and with 6 points earned for the
correct computation of the triple integral.
Problem 12 (8 points)
2 points for the level curve and 1 point for the label with the
function value: a total of 3 points. 2 points for computing the
gradient algebraically and 1 point for evaluating it correctly. 2
points for drawing the gradient vector which should be drawn with
correct directions and magnitudes.
As mentioned previously, Rutgers regulations require that I keep the exams for a year. Students may look at their exams and check the grading. If you want to do this, please send email.
I regret very much that the evening format of the course led me to cut back on some practices which I have found useful in other instantiations of Math 251. For example, I did not have the Question of the Day, which leads to better communication between instructor and students, and which also encourages attendance. A number of students did not attend regularly (during exams I would ask a few people, "Who are you?", a rather strange question). As far as I know, all students who did not attend regularly did disastrously poorly on the exams and failed the course. I also did not have workshops and workshop writeups. This activity encourages deeper thinking about intricate problems and its omission may have diminished learning, compared to other semesters.
Maintained by greenfie@math.rutgers.edu and last modified 12/18/2008.