### The first exam

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:

 Letterequivalent Range A B+ B C+ C D F [85,100] [80,84] [70,79] [65,69] [55,64] [50,54] [0,49]

An answer sheet with answers to version A (the blue cover sheet) is available, and here is a more compact version of this exam. 85 students took the exam. I regret very much that there were two defects in the yellow/version B exam. My target is zero defects. I did not want the curve in the first question to be a curve in R4 (!). I read student solutions to that question, and, amazingly to me, just about everyone seemed to solve it routinely. The situation with regard to the point error on problem 10b) is more irritating and potentially more injurious to students. After 10a) is solved, if the point is the same in 10b), almost no further computation needed (if a student knows the relevant facts about the gradient, of course!). Certainly students who had to substitute values twice may feel aggrieved, and I regret this and sincerely apologize. In grading the exam, we accepted either an answer using the point (the "wrong" point!) stated in print (this was (2,1,1)), or using the point as corrected by me during the exam (this was (1,2,1)).

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.

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 e0=1).

Problem 9 (8 points)
2 points for a correct fx and 2 points for a correct fy. 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 fx and fy 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.

### The second exam

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:

 Letterequivalent Range A B+ B C+ C D F [85,100] [80,84] [70,79] [65,69] [55,64] [50,54] [0,49]

An answer sheet is available with detailed answers to version A (the yellow cover sheet) and also with brief answers to version B (the blue cover sheet). Here is a more compact version of this exam. 78 students took the exam.

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.

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 ∫0Constant0ycorrect powerxy 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 ∫ABCDstuffother stuffWhatever 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).

### The final exam

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:

 Letterequivalent Range A B+ B C+ C D F [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 e-mail.

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, ((x2/2)+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.
Comment I looked at a Maple picture of the intersection and was surprised: two points of intersection, close to one another. I found the intersection points by substituting the functions for the curve into the three-variable function and subtracting 2. Then I used fsolve to get the "other" value of t for the curve/surface intersection: t≈.93543. The pink curve to the right is a graph of this function of t. I zoomed into a picture of the surface, and "saw" the two points (the second picture to the right). The other intersection is ≈(2.12498,.97949,5.80628). The curve and surface are not orthogonal at the intersection(s). The surface picture is a closeup to show the two intersection points distinctly.

Problem 3 (14 points)
4 points for gx and 4 points for gy. 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, ρ2sin(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.
Here is a picture of the minimum near (0,0). Here is a picture of the saddle near (-1,1).

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 (e0 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=0z=1y=0y=zx=0x=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.