### 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 (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 relvant supporting reason.

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

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. 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 8 (16 points)
8 points each for sections a) and b), which were graded similarly. Luckily 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 correcte 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 computation 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 f) was increasing and bounded above, and therefore must converge. f) was worth 8 points. The graph is of a function one of whose roots is the limit of the sequence defined.

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 more 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 condsideration of the alternating series nature of cosine's Taylor series are good approaches.

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 appllies, 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 correcatly 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 succesful 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.
b) (4 points) 1 point for "Yes" and then 3 points for an explanation. A likley 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 vaild 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.