Grades in Math 251:1-2-3 in spring 2010


The first exam

Problem#1 #2 #3 #4 #5 #6 #7 Total
Max grade 16 14 14 16 8 16 16 97
Min grade 0 1 0 0 0 0 1 20
Mean grade 12.38 8.21 5.87 8.60 5.29 8.78 10.43 59.57
Median grade 14 9 4 8.5 7 8 10.5 58

The two versions of the exam had similar statistical results. 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
AB+BC+ CDF
Range[85,100][80,84][70,79] [65,69][55,64][50,54][0,49]

Discussion of the grading

An answer sheet with full answers to version A (with the yellow cover sheet) is available. The sheet also contains bare answers for version B. The work needed is similar to A's. Here is a more compact version of this exam. 89 students took the exam.

Grading guidelines

Minor errors (such as a missing factor in a final answer, sign error, etc.) were penalized minimally. Students whose errors materially simplify the problem were 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.

Fantastic observations
Neither of the graders expected that people would try to get the normal vector in the second problem by differentiation of a formula for T(t). We expected people to compute the velocity and acceleration and "instantiate" these when t=0 (that is, plug in t=0!) and work with these two rather simple vectors. This is what is suggested on the answer sheet. Successful computation of the normal vector using this remembered formula would be an incredible achievement. Very few students were able to do this.
Similarly, in the fifth problem, the lecturer's expectation was that people would simply ∂/∂x the equation. Again, almost everyone memorized a formula and tried to use it. Most people did well on problem 5 but the scores on problem 2 were significantly lower than anticipated.

Problem 1 (16 points)
a) (4 points) 2 points for getting a vector in the direction of the line and then 2 points for assembling the parametric equations correctly.
b) (8 points) 2 points for getting vectors in the direction of the plane, 4 points for computing the cross product correctly, and then 2 points for assembling the equation of the plane correctly.
c) (4 points) 2 points for substitution of the parametric terms into the plane equation, and then 2 points for carrying out the arithmetic and getting the point of intersection.
Note A combination of vector notations will lose one point. For example, something like <3i,-4j,2k> doesn't make sense!

Problem 2 (14 points)
2 points for getting the velocity vector at time 0 and 2 points for getting the acceleration vector at time 0. Then writing a correct T(0) (dividing by the length) is worth 2 points. The computation of N(0) is worth the remaining 8 points. 6 points of that is to get a dot product and subtract off the tangential part. The remaining 2 points is for reporting the correct unit vector.
An equation such as N(t)=T´(t)/||t´(t)|| earns 3 points. Some further work with differentiation of this formula can earn 1 more point without a good answer.
Note A combination of vector notations will lose one point. For example, something like <3i,-4j,2k> doesn't make sense!

Problem 3 (14 points)
a) (2 points) For correct computation of ∇f(x,y).
b) (6 points) Each curve is worth 3 points. The curves must go through P (respectively, Q), and must be parabolas with the correct symmetry and "opening".
c) (6 points) 1 point each for correct values of ∇f at P and Q. 2 points each for correct sketches of these vectors. Each vector should be "based" at the correct point, should have approximately the correct length, and should be perpendicular to the correct level curve. 1 point (up to 2!) will be deducted from the full score if any of these qualities are incorrect. No points will be earned for sketching incorrect vectors.

Problem 4 (16 points)
In almost every part of this problem, a correct answer alone received 1 point. The other points for the part were earned by displaying readable relevant work. The cover sheet stated, Show your work. An answer alone may not receive full credit. Part of the student's effort should be devoted to explaining the method used, with standard notation, since the problems all use rather simple formulas and sometimes correct answers can be obtained by coincidence!
a) (1 point) For the correct value.
b) (2 point) 1 point for the Chain Rule and 1 point for the correct value.
c) (3 points) 2 point for the Chain Rule (a second differentiation is needed!) and 1 point for the correct value.
d) (5 points) 4 points for correct use of both the Chain Rule and the Product Rule and 1 point for the correct value.
e) (5 points) 4 points for correct use of both the Chain Rule and the Product Rule and 1 point for the correct value.

Problem 5 (8 points)
The idea of applying ∂/∂z to the equation is worth 2 points. The mechanics of differentiation are worth 4 points. Solving for zx earns 2 points.
A quoted formula for ∂z/∂x such as –Fx/Fz should be correct or 3 points are lost (someone remarked: "If you want to use a shortcut be sure it is correct!").

Problem 6 (16 points)
a) (6 points) The formula for ∇f is worth 4 points. Evaluating it is worth 2 points. Students may give an incorrect answer here, and lose these points, but they may use their answer and earn points in the remaining part of the problem. (But the answer used shouldn't make the subsequent questions trivial -- for example, an answer of <0,0,0> would not be eligible for such consideration!)
b) i) (2 points) Each answer is worth 1 point.
ii) (4 points) The vector from p to q is worth 2 points, and a unit vector is worth 1 additional point. The value of the directional derivative is worth 1 point. To be eligible to score any points in this part of the problem, the student needs to clearly display the coordinates of the vector from p to q.
iii) (4 points) Setting the dot product of the gradient and an unknown vector equal to 0 is worth 1 point, and using it successfully to get a (non-zero!) vector is worth 2 points. A unit vector is worth 1 additional point.

Problem 7 (16 points)
a) (3 points) This should be "ellipse-shaped", centered at the origin, with the correct vertices (the intercepts with the vertical and horizontal axes).
b) (4 points) 2 points for the integrand and 2 points for the limits on the integral. Both the integrand and the bounds on the integral must be specifically related to the ellipse. A quoted general formula earns no points.
c) (9 points) 3 points for a correct formula for κ (no simplification is needed). Each specific curvature value is worth 3 points (1 point for a correct value of the parameter and 2 points for the correct numerical answer).


The second exam

Problem#1 #2 #3 #4 #5 #6 #7 Total
Max grade 12 12 16 12 16 16 16 100
Min grade 2 0 3 0 0 0 0 21
Mean grade 9.25 7.62 11.5 8.98 11.94 9.11 5.74 63.68
Median grade 10.5 8 13 10 13 9 4 65

The two versions of the exam had similar statistical results. 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
AB+BC+ CDF
Range[85,100][80,84][70,79] [65,69][55,64][50,54][0,49]

Discussion of the grading

An answer sheet with full answers to version A (with the yellow cover sheet) is available. The sheet also contains bare answers for those questions on version B where the questions were different. The work needed is similar to A's. Here is a more compact version of this exam. 84 students took the exam.

Grading guidelines

Minor errors (such as a missing factor in a final answer, sign error, etc.) were penalized minimally. Students whose errors materially simplify the problem were 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)
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.

Problem 2 (12 points)
3 points for setting up the Lagrange multiplier equations. Either all 3 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 cases where some of the variables are 0, and this is worth 3 of the 6 points.
3 points for the answers. The values must be given and labeled (max/min) and some connection with specific solutions of the Lagrange multiplier equations should be given. If this is not done, points may be lost.

Problem 3 (16 points)
a) (5 points) The inside integral is worth 3 points to get the answer in standard polynomial form (everything multiplied out, etc, ready to integrate), and then 2 points for the outside integral. An arithmetic error (such as incorrect "simplification") loses 1 point.
b) (4 points) 2 points for the parabola and 2 points for the line.
c) (7 points) There should be two iterated integrals (1 point). Each iterated integral is worth 3 points, with 1 point of this reserved for the constant outer limit, and 2 points for the inner limits. If students interchange limits but the answers are otherwise correct, 2 points will be deducted.

Problem 4 (12 points)
The top integral is worth 9 points. The setup is 6 of these points, with 3 for the polar bounds, and 1 for the integrand, r, and 2 for dA=r dr&dθ. 3 points for the computation (1 of these is for the specific value of the answer). If the only error in the solution is the wrong power of r (most frequently for using r2 instead of r), 2 points will be deducted out of 12 (wrong integrand, and certainly the wrong answer).
The area of the circular region (the bottom of the fraction) is worth 2 points obtained either by calculus or geometry.
The answer is worth 1 point.

Problem 5 (16 points)
Setup of the triple iterated integral is worth 10 points (1 of those are reserved for having the correct integrand!), 3 points for each pair of limits which will be also tied to the variable of integration. Since things are more complicated with 3 variables, grading this will be stricter than in the previous problem. So ungrammatical limits (similar to the "bad" integrals in the workshop problem) will be penalized appropriately and each bad limit/variable combination will lose 3 points. .
Computation of the integral is worth 5 points (worth only 4 points if the integrand is traded for 1 but then I think that the integration is more difficult!). The correct final answer earns 1 point. A student with 1 for the integrand is not eligible for the final answer point. If in the integration, squaring of "a binomial" was called for (for example, (4-x2)2 is needed if the order used is dz dy dz) and this was ignored (!), then 2 of the 5 computation points were awarded and no answer point.
z slices (with dz as the outside-most variable of integration) are difficult and no one attempting this did it correctly. Generally, 4 points were given if some initial reasonable attempt was made.

Problem 6 (16 points)
Computation in each coordinate system is worth 8 points. No points are earned for giving a formula for the volume of a sphere without supporting evidence. In cylindrical coordinates the Jacobian factor is worth 2 points, and in spherical, it is worth 3 points. The limits in the case of cylindrical coordinates are worth 3 points, and in spherical coordinates the limits are worth 2 points. The computations in each case are worth 3 points, with 1 of those points for the (correct!) answer.
Students who, in the spherical computation somehow integrate ρ2 dV with a correct dV and correct limits can, if the trig integral is correctly computed, earn 6 of the 8 points. The answer is dimensionally incorrect and offensive. If a similar integrand is offered in the cylindrical computation, at most 5 points can be earned since the integration which follows (powers) is much simpler than what's needed in the correct solution.

Problem 7 (16 points)
Attempts to verify the answer without using change of coordinates will receive appropriate credit -- usually very little! In particular, "computations" which just happen to get the suggested answer without being logical or understandable will earn no credit. There are ample suggestions to use change of coordinates.
3 points for suggesting u and v in terms of x and y which, if used correctly will lead to successful resolution. 3 points for converting the integrand to u and v. Solving for x and y in terms of u and v is worth 3 points, and successful computation of the Jacobian is worth 2 points. 2 points for setting up the whole problem correctly in u and v, and 2 points for the computation with 1 additional point for getting the answer displayed.


The final exam

Problem#1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 Total
Max grade 20 20 20 20 20 20 20 20 20 12 8 197
Min grade 2 0 0 0 0 0 0 0 0 0 0 36
Mean grade 15.62 9.04 13.90 12.85 9.51 13.66 12.97 14.61 10.63 3.67 7.29 123.75
Median grade 17 11 17 14 10 16 15 16 13 3 8 128

The two versions of the exam had similar statistical results. 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
AB+BC+ CDF
Range[160,200][150,159][130,149] [120,129][100,119][90,99][0,89]

Discussion of the grading

Here is a more compact version of this exam. 80 students took this exam. I hope that many students recognized many of the problems.
Problems 5 and 10, one more conceptual and one more computational, seemed difficult for many students.

Grading guidelines

Minor errors (such as a missing factor in a final answer, sign error, etc.) were penalized minimally. Students whose errors materially simplify the problem were 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) (4 points) 2 points for getting a vector in the direction of the line and then 2 points for assembling the parametric equations correctly.
b) (8 points) 2 points for getting vectors in the direction of the plane, 4 points for computing the cross product correctly, and then 2 points for assembling the equation of the plane correctly.
c) (4 points) 2 points for substitution of the parametric terms into the plane equation, and then 2 points for carrying out the arithmetic and getting the point of intersection.
d) (4 points) 2 points for the answer (Yes or No for "Are L and P perpendicular?" and 2 points for some reasoning.
Note A combination of vector notations will lose one point. For example, something like <3i,-4j,2k> doesn't make sense!

Problem 2 (20 points)
Setup (10 points) This can be earned in cylindrical or spherical or coordinates (they are all possible but realistically only the first can be completely computed by hand easily).

Ungrammatical limits (similar to the "bad" integrals in the workshop problem) will be penalized appropriately and each bad limit/variable combination will lose 3 points.

Computation (10 points) Cylindrical: The dθ integration earns 1 point. Each of the other two integrations earns 2 points for the antidifferentiation and 2 points for the substitution. Presenting the final answer as a constant earns 1 point.

Otherwise: Note that if r is missing there will be considerable difficulty, and at most 4 points can be earned. The triple integral in both spherical and rectangular coordinates also will be very difficult, and the most that can be earned is 4 points.

Problem 3 (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 4 (20 points)
a) (16 points) 2 points for computation of r´(t) and 2 points for computation of r´´(t); 6 points for correct computation of the cross product; 6 points for deducing the result.
b) (2 points) For some valid reason that the limit is 0.
c) (2 points) For some reasoning supporting the limit result ("Yes, this is a circle projected into the xy-plane, but the curve flattens out at t→∞ because the helix takes longer and longer to go around the z-axis. Therefore the curvature→0.")

Problem 5 (20 points)
5 points for computing ∫I directly (1 of those is for the answer). Green's Theorem mentioned gets 2 points. Implementation: the integrand is worth 4 points, and evaluating the double integral is worth 4 points. Connecting things is worth 4 points, and the final answer is worth 1 point.

Problem 6 (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 7 (20 points)
4 points for computing div F correctly.
8 points for correct conversion to a triple integral in spherical coordinates.
8 points for computation of the triple integral: 1 point for the volume term, 2 points for the z term, and 5 points for the ρ5 term.

Problem 8 (20 points)
a) (8 points) The idea of applying ∂/∂z to the equation is worth 2 points. The mechanics of differentiation are worth 4 points. Solving for zx earns 2 points. 1 point is deducted if the wrong numerical value is reported. 2 points are dedudted if no numerical value is reported.
A quoted formula for ∂z/∂x such as –Fx/Fz should be correct or 3 points are lost (someone remarked: "If you want to use a shortcut be sure it is correct!").
b) (8 points) 4 points for ∇f, 2 points for evaluating it correctly, and 1 point each for the maximum increasing and a vector in the direction of maximum increase.
c) (4 points) 3 points for the use of the correct normal vector (the gradient evaluated again) and 1 point for a point (which can be read from b) again).

Problem 9 (20 points)
Setup of the triple iterated integral is worth 12 points. 2 of these points can be earned for some sort of correct sketch and 1 point for the correct integrand. 3 points for each pair of limits which will be also tied to the variable of integration. Again, ungrammatical limits will be penalized appropriately and each bad limit/variable combination will lose 3 points.
Computation of the integral is worth 8 points (worth only 5 points if the integrand is traded for 1). The correct final answer earns 1 point. A student with 1 for the integrand is not eligible for the final answer point. If in the integration, squaring of "a binomial" was called for (for example, (1-z2)2 is needed if the order used is dx dy dz) and this was ignored (!) but otherwise the computation proceded correctly, then 4 of the 8 computation points were awarded and no answer point. Generally, 4 points will be given if some initial reasonable attempt was made.

Problem 10 (12 points)
1 point for fx, 2 points for ft, and 2 points for fxx. The algebraic work for discerning the value of k and giving the value of k is worth 4 points. Finally checking that this makes a solution (what's needed is acknowledgement that the value of k works in two parts of the formula!) is worth 3 points.

Problem 11 (8 points)
Getting fx and fy is worth 2 points. Solving to find the critical point is worth 2 points. Finding the needed three second derivatives is worth 2 points, and applying the Second Derivative Test is worth 2 points.


Course grades

I computed a number for each student. Here is how I got the number. I added the final exam and the two other exam grades. To this I added suitably scaled numbers obtained from the QotD, workshop grades, Maple labs, and grades from quizzes in recitation. I changed the weighting slightly from what I had written at the beginning of the semester to reflect the fact that we had three Maple labs rather than the four I had expected, and five workshop problems rather than the three I had expected. I sorted the resulting scores, and assigned letter grades based on boundaries proportional to the numbers used to assign letter grades in the exams.

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.


Maintained by greenfie@math.rutgers.edu and last modified 5/9/2010.