The first exam

The versions of the exam were very similar. 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:

Discussion of the grading

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) (6 points) The formulas for the partial derivatives are worth 4 points. Evaluating them to get ∇f(p) 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) (4 points) 2 points for recognizing the relevance of the numbers in ∇f(p) and p, and then 2 points for assembling the equation correctly. The last 2 points are not earned if there is no equation!
c) (2 points) Each answer is worth 1 point.
d) (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.
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)
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 points) 1 point for the Chain Rule and 1 point for the correct value.
c) (2 points) 1 point for the Chain Rule and 1 point for the correct value.
+1: 1 point is added for starting with general formulas in d), e), and f) (without "plugging in"!).
d) (4 points) 3 points for the Chain Rule again and 1 point for the correct value.
e) (5 points) 4 points for the use of the Chain Rule and the Product Rule, and 1 point for the correct value.
f) (5 points) 4 points for correct use of both the Chain Rule and the Product Rule and 1 point for the correct value.

Problem 3 (16 points)
a) (3 points) 1 point for realizing that (0,0,0) is the center of the sphere, and 2 points for writing the parametric equations. 1 point if the line given contains only 1 of the specified points.
b) (5 points) Substituting t's into the sphere equation is worth 2 points. Solving for the t values is worth 1 point. The answers are worth 2 points.
c) (2 points) The correct answer.
d) (6 points) The sphere is worth 2 points (radius and center, with the center being (0,0,0) and radius equal to the square root of an integer, with radius between 2 and 3), the line is worth 2 points (it should go through the center of the sphere and should extend beyond the sphere), and the labeled two intersection points are worth 2 points.

Problem 4 (18 points)
a) (2 points) For the answer.
b) (4 points) 2 for limits and 2 for the integrand.
c) (2 points) For the answer.
d) (2 points) 1 point for each answer. If a student's answer is incorrect but does not make part e) too easy (for example, vectors similar to <4,0,0> and <0,2,0>) then the student's answers to d) are eligible for credit in part e).
e) (8 points) 5 points for the process and 3 points for the answer.

Problem 5 (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 that the triangle's area should be half of the magnitude of the cross product. If the answer is "trivial" (as described) 1 point can be earned (but is not if the length is computed incorrectly or if the half factor is omitted).

Problem 6 (14 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 but 1 is taken off if the graph is only near 0 in that interval). 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 –) ∞ are each worth 2 points.
6 points for the explanation. What's needed is an explanation, not a description in words of the graph of the limiting behavior or the behavior near 0. There must be some reasoning or explanation given. I attempted diligently to grade consistently and carefully here. 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.

• For the limit explanations, the word circle ideally should occur, and some relationship between the radius of a circle and the curvature of that circle. Then how the circles A and B and the curve are connected should be mentioned. An acceptable explanation need not be long. If 1/r or 1/R was encountered, I searched for an explanation of what r or R might mean. I looked for the word "radius". I did not guess what was meant! Full credit can be earned with just a few well-written sentences. Such phrases as "the curvature of a point" or "the definition of curvature is 1/R" are meaningless or incorrect, and didn't earn credit. Students whose only error is somehow systematically interchanging +∞ and –∞ will be penalized 2 points only.
• A reason for the curvature being 0 near s=0 should include some connection with the idea that straight lines have curvature 0. I penalized people who discussed curvature only at s=0 and not near s=0. I penalized people who wrote "no curvature" instead of "the curvature is 0". These statements are not the same!

Problem 7 (8 points)
The idea of applying ∂/∂z to the equation is worth 2 points. The mechanics of differentiation are worth 4 points. Solving for y_z earns 2 points.
A quoted formula for ∂y/∂z such as –F_z/F_y should be correct or 3 points are lost (someone remarked: "If you want to use a shortcut be sure it is correct!").
If no formula is quoted but some correct partials of the left-hand side of the given equation are computed correctly and the answer is not asembled correctly, only 3 points are awarded.

The second exam

The versions of the exam were very similar. 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:

Discussion of the grading

On the exams as distributed, problem 1 was worth 16 points and problem 4 was worth 14 points. This was incorrect, and the point values should be interchanged. This is shown in the discussion below and in the posted versions of the exam and the answers.

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 (14 points)
a) (4 points) 2 points for the first antidifferentiation and 2 points for the second (which includes the final answer).
b) (4 points) The boundary of the region should consist of three straight line segments and a perceptibly curved segment. The "corners" should be at (0,0), (2,0), (2,2), and (0,6).
c) (6 points) There should be two iterated integrals. One, worth 2 points, should be a rectangular box (all four limits constant), and the second, worth 4 points, should have a variable upper boundary.

Problem 2 (10 points)
Setting correct first partials equal to 0 is worth 2 points. Getting the one correct solution to this system of equations is worth 2 points. Computing second partials correctly is worth 3 points, and using the second derivative test correctly earns the last 3 points.

Problem 3 (12 points)
Assembling the Lagrange multiplier equations (with the constraint equation) is worth 4 points. Solving this system is worth 6 points. Correctly reporting the minimum and maximum values is worth the last 2 points.

Problem 4 (16 points)
3 points for a valid equation for the tilted side of the tetrahedron (these can be earned by implication if such an equation is used in the problem); 6 points for the limits of the triple integral (1 for each limit) but no points are earned for limits which are incorrect but incorrect limits can be used for the computations to come, if they are syntactically correct. Then 2 points for the innermost evaluation, 3 points for the middle evaluation, and 2 points for the last evaluation (this includes 1 point for the final answer). The middle evaluation is apt to be the most troublesome.

Problem 5 (14 points)
4 points for the sketch, which should have a straight line and a circular arc meeting in approximately the correct places.
Setup of the iterated integral in polar coordinates is worth 6 points. 2 of those are for the integrand, 2 for the inner limits (the limits on r), and 2 for the other limits. Omitting the r from dA in polar is penalized 1 point.
Also, if the r limits are 1/2 and 1, 3 points will be deducted from the total score, and if the r limits are 0 and 1, 4 points will be deducted from the total score. These limits make subsequent computation much easier.
Computation of the iterated integral is worth 4 points, 1 for the inner integral, and 3 for the outer. Points may be taken off the latter 3 for not reporting values of traditional functions or for arithmetic errors affecting the final answer.

Problem 6 (12 points)
Setup of the triple iterated integral is worth 8 points. This includes the boundary (5 points) and the integrand (3 points). The integrand points include 1 for the r factor of the Jacobian.
Computation of the integral is worth 4 points. Incorrect "simplification" loses points!

Problem 7 (12 points)
Setting up the limits for the iterated integral in spherical coordinates is worth 6 points (1 point for each). Bad limits may lose points here, but if the limits don't trivialize the problem further points may be earned.
The integrand is worth 2 points, the spherical Jacobian is worth 2 points, and the computation of the answer is worth 2 points.

Problem 8 (10 points)
a) (4 points) The computation (3 points) and the result (1 point).
b) (2 points) The requested formula.
c) (4 points) 2 points for some (explained) estimate of the Jacobian, and 2 points for the desired (explained) estimate of the area. Useful and relevant explanations, even if not identical to what's on the answer sheet, will receive full credit.

The final exam

The versions of the exam were very similar. 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:

Discussion of the grading

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.
Almost all of the problems were taken from previous recent exams whose solutions are available on the web. One problem, assigned for homework, was quoted directly from the text. Most students should have recognized most problems.

Problem 1 (20 points)
This is the first problem in the spring 2010 final exam. The results then were: max=20, min=2, mean=15.62, and median=17.
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. 1 point is lost if an equation is not given.
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 (10 points)
This resembles the second problem in our second exam. The results then were: max=10, min=0, mean=6.32, and median=7.5.
Setting correct first partials equal to 0 is worth 2 points. Getting the one correct solution to this system of equations is worth 2 points. Computing second partials correctly is worth 3 points, and using the second derivative test correctly earns the last 3 points.
If there's any doubt about the answer, a picture of the graph of the surface near the critical point is shown to the right.

Problem 3 (12 points)
This is the second problem of the second exam in spring 2010. The results then were: max=12, min=0, mean=7.62, and median=8.
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.
I did ask people to check all solutions of the LM equations. The function is not constant on the constraint!

Problem 4 (12 points)
This resembles the fifth problem on our second exam, but no sketch was requested here and the integral is improper. In our past problem 5, max=14, min=0, mean=8.95, median=9.
Setup of the iterated integral in polar coordinates is worth 6 points. 2 of those are for the integrand, 2 for the limits on r (that's where ∞ should appear!), and 2 for the other limits. Omitting the r from dA in polar is penalized 1 point.
Computation of the iterated integral is worth 6 points, 2 for the θ integral and 3 for the r integral. 2 points are lost if ∞ was not given as a bound on the r integral. The final answer is worth 1 point. Points may be lost for not reporting values of traditional functions or for arithmetic errors affecting the final 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.
A negative answer is not possible here.

Problem 5 (14 points)
Variants of this question has been asked many times in previous exams; this specific version is copied from the third problem on the first exam given in spring 2010. The results then were: max=14, min=0, mean=5.87, and median=4.
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.
Sketching two parabolas and two vectors with small integer coefficients shouldn't be difficult.

Problem 6 (12 points)
I hoped that this would be a simple Chain Rule problem, but please see my comments after the grading rubric for this problem.
The first requested answer is worth 2 points. The second requested answer is worth 4 points. The third requested answer is worth 4 points. The fourth requested answer is worth 2 points. A misuse of the Chain or Product Rules loses 2 points. Notation errors (an x or y partial derivative applied to f, for example) lose 1 point each.
Maybe I should give up: I can't teach the Chain Rule in several variables, or people won't learn it, or both.

Problem 7 (20 points)
This is the fifth problem of the fall 2008 final exam. The results then were: max=20, min=0, mean=10.85, and median=15.
Signaling a decision to compute in spherical or cylindrical coordinates earns 2 points (for example, ρ²sin(φ) observed). Limits of 0 to 2Π 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 6 points, 1 of which is for the answer. Omitting the R in the computation loses 2 points.
Negative mass is not possible here.

Problem 8 (20 points)
Variants of this problem appear on essentially every final exam I've given in the last decade in Math 251. This version is from the fall 2008 final exam. The results then were: max=20, min=0. mean=11.94, and median=14.
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 6 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 9 (20 points)
This is a Divergence Theorem version of the fifth problem given on the final exam in spring 2010. The results then were: max=20, min=0, mean=9.51, and median=10.
5 points for computing the flux out the bottom directly (1 of those is for the answer). Mentioning the Divergence Theorem or flux gets 2 points. Implementation: the integrand is worth 4 points, and evaluating the triple integral is worth 4 points. "Connecting" these computations is worth 4 points, and the final answer is worth 1 point.

Problem 10 (14 points)
This is an assigned homework problem from section 13.4 done by students in my office at least a half dozen times during the semester.
The curvature computation is worth 6 points. 2 of those are for the answer. 2 points can be earned for using an applicable curvature formula (not just quoting the formula).
Analysis and explanation earn the other 8 points. Differentiation of κ is worth 2 points. Setting dκ/dα=0 is worth 2 points. Finding the critical values of α is worth 2 points (only the non-zero values need to be discussed). Explaining why these values give a maximum is worth the last 2 points.
After several semesters of calculus, you should be able to give some specific reason supporting the assertion of a maximum -- not just repeating "This is a maximum because it is."

Problem 11 (12 points)
This is "new" and asks students to invent something.
2 points for mentioning Green's Theorem; 5 points for some correct and specific answer; 5 points for a (more or less!) systematic verification of the suggested answer.
People did better than I expected on this problem.

Problem 12 (14 points)
This is a version of the first problem on our first exam. That problem's part d) was more extended. That problem could earn 16 points. The results then were: max=16, min=0, mean=9.88, and median=10.
a) (6 points) The formulas for the partial derivatives are worth 4 points. Evaluating them to get ∇f(p) 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) (4 points) 2 points for recognizing the relevance of the numbers in ∇f(p) and p, and then 2 points for assembling the equation correctly. 1 point is lost if only the left-hand side of the equation is given (no "=0").
c) (2 points) Each answer is worth 1 point.
d) (2 points) The answer is worth 2 points. 1 point will be earned if the answer is not correctly normalized. 1 point is earned if a correct and relevant dot product is indicated.
Most interesting were those students whose answers in a) were not integers. Did they think I was intentionally misleading them? I wasn't, really.

Problem 13 (20 points)
This is a version of the eighth problem on the final exam in fall 2008 (here the integrand is simpler but the shape is more curved). The results then were: max=20, min=0, mean=10.42, and the median=10.
a) (8 points) For the process and answer: 2 for the inner integral, 3 for the middle integral, and 3 for the final integral (1 of those points for the answer).
b) (12 points) For the answer (6 points -- one for each correct limit) with some indication of how it was gotten (6 points). Some of the 6 points of process credit may be earned if there are comprehensible diagrams present which are relevant to the triple integral considered.
In this problem and problem 10 (and also in the preceding problem) some students displayed serious difficulties with elementary algebra and first semester calculus.

Course grades

Rutgers regulations require that I keep the exams for a year. Students may look at their exams and check the grading which was done carefully. If you want to do this, please send e-mail.

Maintained by greenfie@math.rutgers.edu and last modified 12/27/2010.