### Grading student work in Math 251:05-10, spring 2006

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.

• 2 points for including all relevant Maple instructions
• 3 points: 1 point EACH for identifying each of the three vectors (pq and pr and v).
• 5 points for the labeled picture distributed as follows:
• 2 points for the picture. Only 1 point if the picture does not display the perpendicularity of v to the triangle T.
• 1 point for labeling p and q and r.
• 1 point for labeling T.
• 1 one for labeling v. The vector v should have its initial point at p otherwise this point is not earned and 1 point should be deducted from the possible initial 1 or 2 points for the picture.

The first workshop problems
Here is the body of a message I sent to the recitation instructors, who read and graded two-thirds of the workshop problems. I graded one section for each lecture. The contents of the message should provide some background on the grading.

### The first exam

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:

 Letterequivalent Range A B+ B C+ C D F [80,100] [75,79] [65,74] [60,64] [50,59] [45,49] [0,44]

Some general background
Quoting from the formula sheet (even relevant formulas!) will not earn 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.
Most of the problems on this exam were taken directly from the text, and were on the list of suggested textbook homework problems.

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 V1 gets 4 points, and a correct value of V2 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 non-smooth 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 well-written 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 zx and 2 points for a correct zy. 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 non-trivial 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 R3, not in in R2. A two-dimensional 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 non-trivial. A correct geometric observation such as "The tangent vector of the curve is perpendicular to both surface normals" would earn at least a point.

### The second exam

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:

 Letterequivalent Range A B+ B C+ C D F [80,100] [75,79] [65,74] [60,64] [50,59] [45,49] [0,44]

Some general background
Quoting from the formula sheet (even relevant formulas!) will not earn 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.
All of the problems on this exam were simple variations of textbook or review problems. Each problem actually has an appropriate reference signalling its origin.

Problem 1 (8 points)
2 points for zx; 3 points for zy; 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 r2 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 Vx and Vy 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.

### The final exam

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 e-mail.

 Letterequivalent Range A B+ B C+ C D F [160,200] [150,159] [130,149] [120,129] [100,119] [90,99] [0,89]

Some general background
Problems 4 and 9 asked students to draw graphs. The results were rarely satisfactory even though the instructor believed that the curves (parabolas) and the surface (a paraboloid) should have been familiar. The parabola and line sketch in problem 10 was usually done correctly.

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.