Students should know how their work is graded. Then they can better understand the relative importance of various requests in the assignment. They also can more knowledgeably review the correctness of the grading.
The first Maple assignment
Here are the instructions the graders were given:
There will be a total of 10 points.
The first workshop problems
Here is the body of a message I sent to the recitation instructors,
who read and graded twothirds of the workshop problems. I graded one
section for each lecture. The contents of the message should
provide some background on the grading.
I just finished grading the workshop problems for section ¶ taught by
Mr. ¶¶. Here are some observations about the student work and my
grading which almost surely will be true for all of the sections.

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

Max grade  8  12  10  12  12  12  12  12  12  96 
Min grade  0  0  0  0  0  0  0  0  0  8 
Mean grade  5.69  5.31  6.03  6.49  7.43  7.44  2.16  6.25  7.54  54.35 
Median grade  6  2  6  6  8  9  0  6  8  54.5 
140 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. The different versions did not seem to have very different statistical results. Here are approximate letter grade assignments for this exam:
Letter equivalent  A  B+  B  C+  C  D  F 

Range  [80,100]  [75,79]  [65,74]  [60,64]  [50,59]  [45,49]  [0,44] 
Problem 1 (8 points)
a) (5 points) 1 point each for finding two vectors (say pq and pr, for
example). 3 points for finding the cross product.
b) (3 points) Students may use their answer (even if incorrect) to a)
in this part if the answer is not trivial (such as the 0 vector
or just <1,0,0>). 1 point for not knowing the the triangle's area
should be half of the magnitude of the cross product.
Problem 2 (12 points)
2 points for computing the dot product of V and W
correctly. 2 points for computing the magnitude of W
correctly.
A correct value of V_{1} gets 4 points, and
a correct value of V_{2} gets 4 more points.
Students certainly may use any of the formulas on the formula sheet,
provided that the answers are correct. If a student makes a mistake
early in the problem but does not trivialize the problem (for example,
makes an error in computing the dot product) then the student will be
penalized for that error but can earn all the points in the balance of
the problem, working as if the earlier answer were correct.
Problem 3 (10 points)
a) (4 points) 2 points for recognizing the correct normal vector. 2 more
points for writing the correct parametric equations.
b) (4 points) 2 points for recognizing the problem: that is, addressing
the distance question correctly. 2 more points for getting a correct
point (either correct point earns the points!).
c) (2 points) This is earned for writing the correct equation for the
plane. 1 point off for either a wrong "point" on the plane or for a
wrong normal vector. The student may use an incorrect point from b)
and still earn 2 points here.
Problem 4 (12 points)
2 points for a correct r´(t).
2 point for computing (and recognizing the "need" for) r´(t).
Students do not need to "simplify" this expression, or any other in
the problem.
2 points for a correct result for T(t) and then for T(1).
3 points for T´(t) (take off a point for each error but continue
to read for other results!).
3 points for N(t) and then N(1): any "mess" (if correct!) will earn
the points.
Quoting from the formula sheet (even relevant formulas!) will not earn
credit.
Problem 5 (12 points)
4 points for the graph: the graph should be "unimodal": down then up,
with good limits (2 points). And it should be 0 on an appropriate
interval in between. (2 points). Since I'm convinced that the curve
sketched is smooth 1 point will be deducted for graphs of
curvature which seem offensively nonsmooth to me.
4 points for the limits: the limit as s goes to + (or ) infinity are
each worth 2 points.
4 points for the explanation. What's needed is an explanation,
not a description in words of the graph of the limiting
behavior. There must be some reasoning or explanation given. Ideally,
the word circle should occur, and some relationship between the radius
of a circle and the curvature of that circle. Then the relationship
between circles A and B and the curve should be mentioned.
An acceptable explanation need not be long. Certainly some students
received full credit with just one or two wellwritten sentences. When
the word "it" was encountered, an effort was made to identify the
referent (what the "it" means). If this identification was ambiguous
or impossible, credit was reduced. I read what students wrote. Such
phrases as "the curvature of a point" or "the definition of curvature
is 1/R" are meaningless or incorrect, and don't earn credit.
Students whose only error is somehow systematically interchanging
+infinity and infinity will be penalized 2 points only.
Comment I meant to ask for an explanation of both the
graph and the limits, but did not. Then I would have wanted a
statement about why curvature is 0 for an interval around s=0 (the
curve is a straight line segment there, and straight lines have
curvature 0.) I didn't phrase the question as I should have, however!
Problem 6 (10 points)
1 point for computing f's value at (2,1).
2 points each (total: 4 points) for computing the first partial
derivatives of f correctly.
2 points (1 each) for evaluating these first partial derivatives.
3 points for the linear approximation formula. It does not need
to be explicitly stated. A numerical version is o.k. Students who
insist upon doing numerical work incorrectly should lose a
point. Realize that there is no need to do any numerical work
in this (or any!) problem  answers do not need to be simplified.
Problem 7 (12 points)
a) (6 points) 2 points for a correct z_{x} and 2 points for a
correct z_{y}. 2 points for assembling these into a correct
verification of the PDE. Working with a specific function f earns
no points (!).
b) (6 points) 4 points for a correct use of both the product rule and
the chain rule (2 for one of them not correct). Then 1 point each for
stating explicitly what the two functions (A and B) are (that
is requested!).
Comment This problem had the worst results. Therefore either
the chain rule wasn't taught well and/or it wasn't learned well. There
will be a similar question on the next exam.
Problem 8 (12 points)
a) (6 points) 1 point for F(x,y,z). 1 point for the correct value of F at p. 2 points for the maximum
directional derivative stated correctly and 2 points for the correct
unit vector.
b) (6 points) C's value earns 1 point. Realizing that F at p is a normal vector earns 2
points (so incorrect data from a) can still earn these
points). Writing the equation of the tangent plane earns 3 points.
Students may use their result from a) if incorrect and nontrivial
without penalty.
Problem 9 (12 points)
a) (4 points) 2 points for the correct computation of f (the correct computation of the
partial derivatives alone earns 1 point  assembling them into the
gradient earns the other), and then 1 point for f(2,1,2) and 1 point
for f(2,1,2). The gradient of
this function should be a vector in R^{3}, not in in
R^{2}. A twodimensional answer loses a point.
b) (4 points) 2 points for the correct computation of g (the correct computation of the
partial derivatives alone earns 1 point  assembling them into the
gradient earns the other), and then 1 point for g(2,1,2) and 1 point
for g(2,1,2).
c) (4 points) 2 points for realizing that the cross product (or
something equivalent!) is needed were rewarded, and 2 points for the
computation. Students could use their results from a) and b) even if
incorrect without penalty provided that the resulting problem was
nontrivial. A correct geometric observation such as "The tangent
vector of the curve is perpendicular to both surface normals" would
earn at least a point.
Problem  #1  #2  #3  #4  #5  #6  #7  #8  Total 

Max grade  8  12  12  12  10  16  12  18  100 
Min grade  0  0  0  0  0  0  0  0  11 
Mean grade  3.96  5.53  9.08  8.39  5.01  10.46  8.74  5.80  56.98 
Median grade  5  5  11  9  4  10  9  6  57 
123 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. The different versions did not seem to have meaningfully different statistical results. Here are approximate letter grade assignments for this exam:
Letter equivalent  A  B+  B  C+  C  D  F 

Range  [80,100]  [75,79]  [65,74]  [60,64]  [50,59]  [45,49]  [0,44] 
Problem 1 (8 points)
2 points for z_{x}; 3 points for z_{y}; 3 points for
assembling things correctly. NO points will be earned if a
specific f is used. This includes the function f(t)=t. Some
acknowledgement of the "abstract" chain rule must be given.
Problem 2 (12 points)
Half the area of a circle (the bottom of the fraction) earns 3 points
(either computed or cited from a formula). 2 points for recognizing
the function (either in polar or rectangular coordinates). Setup of
the top double integral (that is, showing the the region has been
recognized and putting the function in) earns 2 points. Computing the
top integral: 4 points. The correct final answer: 1 point. If
r^{2} instead of r is used in the function to be averaged and
everything else is correct, the student's work earns 10 points.
Problem 3 (12 points)
Computing V_{x} and V_{y} is worth 2 points. Finding
the critical points is 2 points. 4 points for each second derivative
test for a total of 8 points.
Problem 4 (12 points)
4 points for setting up the Lagrange multiplier equations. 6 points
for the correct analysis of these equqations. The analysis must
include a reason why y=0 and
is not 0, and this is worth 2 of the 6 points. The answers earn
2 points.
Problem 5 (10 points)
Translation of the given region into one of the two correct iterated
integrals is 4 points. Computation of the correct iterated integral is
6 points. If the constant is in the correct position (but not the
other variable) the student can earn 1 more point. 1 point off if
there is a constant error in the inner antiderivative.
Problem 6 (16 points)
8 points for converting correctly to an iterated integral. 8 points
for the correct evaluation.
Problem 7 (12 points)
a) 6 points for the computation.
b) 6 points for the description. The description should discuss the
geometry of the solid, and should not be only a repeat of the
spherical coordinates of the solid.
Problem 8 (18 points)
a) 8 points: 2 points for stating a correct
parameterization; 2 points for dx and dy; 2 points for converting the
integral to t; 2 points for the computation.
b) 10 points: 6 points for getting f (only 1 point if there is no
explanation for how to get f from V which was specifically
requested); 4 points for using f to compute the line integral.
Problem  #1  #2  #3  #4  #5  #6  #7  #8  #9  #10  #11  #12  Total 

Max grade  15  15  15  20  15  25  20  15  20  20  12  8  196 
Min grade  0  0  0  3  0  0  0  0  0  0  0  0  20 
Mean grade  9.36  11.70  10.79  12.49  8.81  9.46  12.90  11.71  9.96  13.29  7.80  5.36  123.65 
Median grade  10  13  15  13  8  9.5  16  14  10  13  9  7  123.5 
112 students took this exam. Numerical grades will be retained for use
in computing the final letter grade in the course. The different
versions did not seem to have meaningfully different statistical
results, but the versions of problem 1 did show some difference: about
1.5 points. (The C/D versions scored lower.) I don't think that's
enormously significant in a 200 point exam. There was, overall, a
difference of less than a quarter of a standard deviation between the
A/B and C/D versions. Here
are approximate letter grade assignments for this exam:
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.
Letter equivalent  A  B+  B  C+  C  D  F 

Range  [160,200]  [150,159]  [130,149]  [120,129]  [100,119]  [90,99]  [0,89] 
Problem 1 (exams A and B) (15 points)
a) (7 points) Solve two of the equations: 5 points. Check the third: 2
points. If no valid method is shown, 0 points are earned.
b) (8 points) Normal vector: 4 points (2 for indicating the cross
product of the appropriate vectors and 2 for getting it correct), 2
points for a point on the plane (can be a point on any of the two
lines, or it can be the point found by the student in a), whether
correct or not). 2 points for the correct answer.
Problem 1 (exams C and D) (15 points)
a) (4 points) Dot product of the two normal vectors: 1 point;
magnitude of each normal vector: 1 point each; arccos of the correct
assemblage: 1 point.
b) (5 points) A correct answer is worth 1 point, and some valid method
earns 4 points.
c) (6 points) 3 points for the cross product of the two normal
vectors, 1 point for a point on the line which may be a correct point
or the student's answer to b), 2 points for assembling the answer
correctly.
Problem 2 (15 points)
a) (4 points) 1 point for each partial derivative, and 1 point for
the vector nature of the answer.
b) (6 points) 2 points for the gradient at p. 2 points for each answer.
c) (5 points on exams A and B) 2 points for a normal vector, 2 points
for a point on the line, and 1 point for the answer.
c) (5 points on exams C and D) 2 points for a normal vector, 2 points
for a point on the plane, and 1 point for the answer.
Problem 3 (15 points)
7 points for a correct setup in any order. 8 points for the
computation (2 points for the first antiderivative with limits, 3
points for the substitution in the integral, and 3 points for the
second antiderivative with limits and the answer). The student does
not earn the first 7 points if the setup is not correct.
If the student sets up an integral over a rectangle, no points
will be earned.
Problem 4 (20 points)
a) (10 points) This is half a paraboloid opening down. Equations (a
total of 7 points): the paraboloid equation gets 3 points, and each
plane gets 2 points. 3 points for the sketch, which should be
reasonable.
b) (10 points) r integral done correctly: 3 points; integral
done correctly: 3 points; z integral done correctly: 2 points; the
answer: 2 points.
Problem 5 (15 points)
a) (7 points) The partial derivatives each earn 2 points. Finding
the unique critical point: 3 points, including 1 for the solution.
b) (8 points) Computing the second partial derivatives earns 4
points. Evaluating them at the point found in a) earns 2 points. Using
the Second Derivative Test correctly: 2 points.
Problem 6 (25 points)
(12 points) Computing div T correctly: 2 points. Computing the
triple integral of this function over the lower half of
the ball, 10 points (2 points for stating that the goal is computing
the triple integral of the divergence over the half ball, 2 points for
setting up the correct triple iterated integral, 1 point for the d integral, 2 points each for the
d rho and d phi integrals, and 1 point for the answer).
(10 points) Computing T·n on the bottom and
setting z=0: 2 points. Computing the double integral correctly, 8
points.
(3 points) Put everything together and get the correct answer.
Problem 7 (20 points)
7. a) (10 points) 2 points for the answer alone, and 8 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 do not find the
function earn 5 of the 10 points.
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 8 (15 points)
7 points for applying Green's Theorem correctly. 8 points for
evaluating the resulting double integral. Students who insist on using
a different method (such as parameterization) will earn an appropriate
amount of credit (there are 4 different pieces of the boundary curve
to parameterize).
Problem 9 (20 points)
2 points for each level curve and 1 point for the label with the
function value: a total of 9 points. 1 point for computing the
gradient algebraically. 2 points for each gradient vector: a total of
10 points. The vector field should be drawn with correct directions and
magnitudes.
Problem 10 (points)
a) 5 points for a correct sketch. 2 of these points are for some
specific labeling (with numbers) of the intersection points.
b) and c): 5 points for each correct integral (so one part will be
worth 10 points and one part will be worth 5). In each case, the
integrand which should be f(x,y) earns a point and each correct
boundary of the double integral earns a point.
Problem 11 (12 points)
Finding and using a formula is 8 points. Identifying the t value for
each point is 1 point and computing the curvature is 1 point. Since
there are 2 points this is worth a total of 4 points.
Problem 12 (8 points)
8 points for the computation. Each misuse of the chain rule or the
product rule loses 2 points. Simpler errors such as a floating minus
sign lose 1 point.
All Maple labs were graded, and the returned labs can be picked up, as can earlier exams in the course.
Maintained by greenfie@math.rutgers.edu and last modified 5/12/2006.