### The first exam

Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Problem#7 Total 16 12 18 18 16 10 10 96 12 0 11 4 0 0 3 55 14.76 9.24 16.57 13 10.57 4.48 7.38 76.00 15 12 18 12 11 5 8 74

21 students took this exam. 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] [75,84] [65,74] [55,64]

### Discussion of the grading

Generally arithmetic errors will be penalized only minimally. If, however, your error makes the problem much simpler, more credit will be deducted and your answer may not be eligible for all of the credit of the problem.

Problem 1 (16 points)
a) (4 points) This should be a straightforward computation.
b) (4 points) 2 points for the curves, labeled (1 point per curve). 2 points for the coordinates of the intersections, labeled.
c) (8 points) 4 points for the setup. 4 points for substituting and then solving. Some students got lost in a swamp of algebra. I generally gave these students 6 out of 8 points for this part.

Problem 2 (12 points)
The standard way to solve this, used by all but one or two students, is to slice the solid by planes perpendicular to the axis of symmetry which is a line perpendicular to the base through the center of the square. I gave 2 points for the definite integral ranging from 0 to 60. I gave 4 points for a correct formula for the length of the side of a square slice. I gave 2 points for squaring that, to get the cross-sectional area of the slice. The last 4 points were earned for taking the antiderivative and obtaining the correct answer. Students who made the antidifferentiation much too easy due to earlier errors were not eligible for the last 4 points. I tried to read and grade the solutions of the few students who chose a different method of computing the volume in the same spirit as the previous outline.

Problem 3 (18 points)
a) (9 points) 2 points for factoring the denominator of the rational function, 2 points for writing the correct structure of the partial fraction decomposition, 2 points for getting the correct values of the unknown constants, 2 points for the integration, and 1 point for recognizing that the answer was the same as the quoted one.
b) (9 points) The grading scheme of part a) was used here.
Comment Of course one can go backwards from the answers given, and the answers surely hint at, say, the factorization. I realize this, but according to the problem statement, I wanted a partial fraction decomposition displayed for each integrand, the appropriate equations solved, etc. If these things were not shown (or something very close!) then full credit was not given. Please note that the general form of the partial fraction pieces was given on the formula sheet. Also the derivative of arctan is on the formula sheet.

Problem 4 (18 points)
Comment Some student answers were especially difficult to read in this problem. Here and in the other problems, if I can't make sense out of what I read, then I cannot give credit. Communication is part of the exam, and, as I wrote on several papers, I cannot read "your" mind, and can't (or, rather, won't!) guess at what should be on the page. I also believe that overly sloppy writing probably hurts the student's own further work.
a) (9 points) I think integration by parts is needed somewhere in this problem. Indication of such earns 3 points. Students who made a mistake early in the parts may end up with an impossible integral, and nothing can really help them. Please note that the derivative of arcsin(x) is given on the formula sheet. The value of arcsin(1) was worth 1 point.
b) (9 points) Several algebraic approaches are successful. Almost surely a substitution is needed somewhere along the way. Again, I tried to read what was written. Students who did some preliminary algebra usually received 3 points for that effort. It is difficult to go further without some good idea about a substitution. If algebraic errors simplified the integrand too much, then students were not able to earn full credit for this problem.

Problem 5 (16 points)
a) (4 points) 2 points for each use of L'Hopital's rule. Eligibility for the rule is a quotient with specific behavior. Students should show that each time.
b) (10 points) 8 points for two uses of integration by parts resulting in the correct answer. 2 points were reserved for some response to the sentence "Indicate why the limits you need exist ..." This was an effort to have students not just plug in numbers because ln(x) can't be "plugged in" at x=0.
c) (2 points) Basically the student should agree that the picture does support the computation, and indicate that the area even considered approximately is close to the answer in b).

Problem 6 (10 points)
This was the only problem for which the mean grade was less than half the possible score (and the median grade was half of the score). I did not give credit for material which was false or irrelevant to the problem. Students could earn 5 points for estimating or evaluating an integral which was relevant to the problem, but the other 5 points was earned when that estimation was connected correctly to the integral in the problem statement. I awarded 8 out of 10 points to a solution which relied on inequalities that were "almost" correct, in the sense that for x large enough, the specified inequalities were true.
I also note that several students came up with ingenious arguments quite different from the instructor's answer, and I compliment them for this.
Comment Inequalities and reasoning connected with them seem difficult when first encountered in a calculus setting. Things do get easier and learning to cope with inequalities is necessary in many applications.

Problem 7 (10 points)
Computation of f´ was worth 2 points, and computation of f´´ earned another 2 points. If differentiation errors unduly simplified the balance of the problem, then all of the remaining points could not be earned. Estimation of K was worth 4 points. Asserting that an overestimate of f´´(x) on [0,2] is obtained by evaluating f(2) or f(0) and f(2) is certainly not valid and at least 2 points were deducted for this error. One part of the second derivative, (1+x3)-1/2, is a decreasing function of x and somewhere this needed to be used or, again, 2 points were deducted. A valid answer for n, supported by the reasoning and computation shown, earned the last 2 points.

### The second exam

Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Problem #7 Problem #8 Problem #9 Total 12 14 15 14 12 12 12 10 12 104 6 5 8 7 0 0 6 1 6 68 10.43 12.71 13.90 10.33 8.57 5.52 9.62 5.33 10.43 86.86 10 13 15 10 10 5 10 5 11 86

21 students took this exam. 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] [75,84] [65,74] [55,64]

### Discussion of the grading

Generally arithmetic errors will be penalized only minimally. If, however, your error makes the problem much simpler, more credit will be deducted and your answer may not be eligible for all of the credit of the problem.

Problem 1 (12 points)
a) (2 points) Each equation of a horizontal line gets a point.
b) (8 points) 1 point each for each correct and correctly labeled curve. 1 point each for each limit.
c) (2 points) 2 points for the correct answer.

Problem 2 (15 points)
a) (7 points) I deducted 2 points for people who forgot +C at an appropriate place. I deducted 1 point for bad algebra (incorrected exponential manipulation, for example).
b) (8 points) I tried to follow the student's solution in a) here, even if incorrect. 6 points for the correct particular solution, again deducting points for bad algebra. 2 points for the domain. If the student's version of the solution unduly simplified the domain question, then no points could be earned. No points were earned if the student confused the domain and the range. I took off a point if the correct particular solution resulting in a domain answer which had only the restriction that x be non-zero. The solution curve which passes through the point (1,0) only has domain x>0.

Problem 3 (15 points)
a) (5 points) I gave 2 points for recognizing the statement: first term=5 and fourth term=3 (1 point each). Completing the problem was worth the balance of credit. I deducted 1 point for faulty algebra.
b) (10 points) The "setup" earned 6 points: this is the computation of several of the (different) square side lengths and assembling them into a sum. 2 more points were earned by the recognition that this was a geometric series and the final 2 points for the sum of the series. Faulty algebra lost 1 point.

Problem 4 (12 points)
a) (8 points) The Ratio Test correctly used earned 6 points. The Root Test could also have served. Then identification of the interior of the interval and the radius of convergence was the other 2 points. I deducted 1 point if there was no clear or correct statement of the radius of convergence. I deducted 1 point for faulty algebra.
b) (4 points) 2 points for each endpoint. 1 point was earned in each case for correct insertion of the value to get a series of constants. 1 point was earned in each case for the correct conclusion with some sufficient reason. Please note that for x=-1/3, the Alternating Series Test only shows that the series converges conditionally, but the correct answer is that it converges absolutely, so additional reasoning must be given.

Problem 5 (12 points)
a) (6 points) The Alternating Series Test applies, but I insisted that all three criteria be checked (or at least mentioned!). Students who omitted one of the criteria (frequently that the absolute value of the terms decreased) lost 2 of the 6 points. It is also possible in this case to strip away the sign change and conclude (using the Integral Test, say) that the resulting series converges and then, using the implication {absolute convergence} implies {conditional convergence} conclude that the series converges.
b) (6 points) If the series satisfies the conditions of the Alternating Series Test, then the first omitted term supplies an error estimate. Generally, even if a series converges, the first omitted term supplies no information about the accuracy of the partial sum. Students who did the problem but omitted the connection with alternating series were penalized 2 points. Students who did not give a partial sum lost 1 point. It is also possible, as in a), to delete the (-1)n and estimate the infinite tail using an integral. Students who incorrectly estimated powers of log with powers of n and then used an easier integral estimate did not receive credit. They simplified the problem too much.

Problem 6 (12 points)
a) (6 points) The Integral Test easily applies. Both the Ratio Test and the Root Test, when the appropriate limits are correctly computed, do not give sufficient information for any conclusions (the limits are 1). Students who indicated a desire (?) to use the Integral Test were generally given at least 2 points.
b) (6 points) Again, an estimate using an integral works. I deducted 1 point because of the common error in the upper bound (N instead of N+1). Errors in inequalities were penalized at least 2 points. More serious errors in logic (overestimates instead of underestimates, for example, or {A>B} and {A>B} implies some relationship between B and C) generally received no credit.

Problem 7 (12 points)
a) (6 points) Most people compared the series to a convergent geometric series. Other valid approaches, such as comparison to a p-series with p=2, are also possible.
b) (6 points) Comparing the infinite tail to the infinite tail of a geometric series was done by most people. The geometric series can be computed exactly, and it also can be overestimated by an integral which was done by some. Just asserting that the Nth or (N+1)st term is less than a desired tolerance earns no points. I allowed 2 points if this was connected with an infinite tail.

Problem 8 (12 points)
a) (4 points) 1 point for "Yes" and then 3 points for explaining why. A generally successful explanation would need to detour via the implication {absolute convergence} implies {conditional convergence}. The Ratio Test is not itself valid for the series, since various terms can equal 0. I should have asked if the series converged absolutely, and then asked students to explain why this might imply convergence.
b) (4 points) 1 point for "Yes" and then 3 points for an explanation. A likely explanation would begin with the 2Pi periodicity of cosine, and that earned a point. No more credit was earned unless all of the different cosine functions were somehow analyzed.
c) (4 points) 1 point for "Yes" and the other 3 points for some sort of explanation. Please note that the question is an effort to explain the graph, and therefore a verification can't use the graph, which could be incorrect or an artifact:
artifact

1. a product of human art and workmanship.
2. [Archaeol.] a product of prehistoric or aboriginal workmanship as distinguished from a similar object naturally produced.
3. [Biol. etc.] a feature not naturally present, introduced during preparation or investigation (e.g. as in the preparation of a slide). Maple could be showing things that aren't correct. It can draw misleading graphs.

Problem 9 (12 points)
a) (6 points) 2 points for "True" and 4 points for a general explanation. An example or two gains only 1 more point. Asserting that the function goes to 0 so the reciprocal of the function ... without more specification does not earn credit. Discussion of an's denominator or numerator (for an abstract an) similarly earns no credit, since I don't understand what that means.
a) (6 points) 2 points for "False" and 4 points for a valid example. Several people told me that an=x and thus an=1/x, but the context was sufficiently unclear that I did not understand whether the x was suppose to be n or a constant. I then deducted 1 point, although either assertion would have led to a correct example. Clarity should be rewarded and its lack, penalized.

### The final exam

Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Problem #7 Problem #8 Problem #9 Problem #10 Problem #11 Problem #12 Total 20 16 16 16 16 16 16 16 18 20 20 16 195 14 15 3 0 6 0 4 13 0 0 4 0 102 18.48 15.95 11.95 13.29 13.52 13.05 11 15.24 10.76 12.43 11.57 11.86 159.09 19 16 12 15 15 14 11 16 11 17 10 14 164

21 students took this exam. 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 [170,200] [150,169] [130,149] [110,129]

### Discussion of the grading

Generally arithmetic errors will be penalized only minimally. If, however, your error makes the problem much simpler, more credit will be deducted and your answer may not be eligible for all of the credit of the problem.

Problem 1 (20 points)
a) (4 points) Graph of the line (1 point) and the parabola (3 points). b) (9 points) Finding the intersections (3 points); setup of the integral (3 points); computation of the integral (3 points). c) (5 points) 2 points off for drawing too much of the parabola, and 1 point off for drawing the "line" insufficiently linear. d) (2 points) 1 point for the correct answer and 1 point for a relevant supporting reason.

Problem 2 (16 points)
Partial fractions. Full credit for the correct answer.

Problem 3 (16 points)
a) (6 points) I hoped that these graphs wouldn't be completely unfamiliar!
b) (10 points) 3 points for reading off the antiderivatives (they were on the formula sheet); 3 points for realizing that this was an improper integral; 4 points for the necessary rearranging of the antiderivative and evaluating to get the answer.

Problem 4 (16 points)
a) (9 points) If errors were made in the parts, several points were taken off. b) (7 points) I took off slight amounts of points for slight errors. This should have been a straightforward application of a). Actually, several students redid their work in a).

Problem 5 (16 points)
a) (8 points) Points off for partial or incorrect solutions.
b) (8 points) Points off for partial or incorrect solutions.

Problem 6 (16 points)
a) (10 points) A correct curve.
b) (6 points) 2 points for identifying two critical points; 2 points for discerning what type they are; 2 points for finding an algebraic relationship. People should know what the phrase critical point means.

Problem 7 (16 points)
Separating the variables: 2 points; antidifferentiation of y: 2 point; antidifferentiation of x: 5 points; getting the constant: 2 points; solving for y as a function of x: 2 points; stating the domain: 3 points.

Problem 8 (16 points)
8 points each for sections a) and b), which were graded similarly. I happily report that no one in the class worked directly with the algebraic fourth derivative in b), although a few people worked with the algebraic second derivative in a). I took off 1 point for reversing inequalities (I graded the sections independently), and 1 point for other minor errors. Most people indeed did get the bounds on the derivatives from the given graphs, and did use the correct error formulas. Of course, the answers given for the number of partitions were quite different. The largest answer was n=101000, which got full credit since enough supporting evidence was given. I just hope that the person giving that answer never has a computational task which takes 101000 steps!

Problem 9 (20 points)
Yes, this was a proof and verification for one or two values for the sequence of initial steps is not logically definitive.
Parts a), b) c) d), e), and f) each were worth 2 points. I then hoped that students would use this information to conclude that the sequence defined in g) was increasing and bounded above, and therefore must converge. g) was worth 8 points. The limit of the sequence turns out to be a root of the function whose graph is displayed. The instructor thought that this might help students. I did write that "These numbers may help you emotionally but they should not be used in the proof." But some people did try to use them, and information about the first 10 terms of the sequence is not enough to conclude anything about the limit. Please see here.

Problem 10 (20 points)
Each part was worth 10 points. Again, a reversed inequality lost a post. More costly were errors in logic which I penalized strictly. Part a) had been done in front of most students several times. Part b) was a different logical take, but again, things like it had been done in class a number of times. In b) either Taylor's Theorem with n=4 or 5 or consideration of the alternating series nature of cosine's Taylor series are good approaches.
The instructor was disappointed by the grades on this problem. On the other hand, perhaps the grades indicate that the topic was not taught as well as it should have been.

Problem 11 (20 points)
a) (8 points) Here the best strategy is to use the Ration Test (almost every student did this).
b) (2 points) I looked for "Absolute convergence implies convergence." (Any kind of convergence!) Read the question, which asks "... why the result of a) implies that the series converges ..."
c) (10 points) 5 points each for discussions of J(1) and J(4). I wanted specific reasons, not that "something" was "smaller" than "something".

Problem 11 (16 points)
a) (8 points) 4 points for each curve. A two-lobed "circle" lost 2 points. b) (8 points) 4 points for the setup and 4 points for the computation. The setup is slightly subtle, as I warned on the exam.