Grades in Math 152:5,6,7 and 9,10,11 in spring 2007


The first exam

Problem#1 #2 #3 #4 #5 #6 #7 #8 Total
Max grade 12 12 12 12 12 14 12 14 97
Min grade 0 0 0 0 0 0 0 0 12
Mean grade 7.14 3.68 10.22 5.78 5.01 9.38 5.93 4.92 52.04
Median grade 7 1 12 6 4 10 6 3 51

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. Indeed, students with low grades should evaluate the amount of time and effort they can devote to this course as part of their entire plan of work for the balance of the semester. Such students may wish to drop this course or drop another course in order to increase their chances of success in what remains. I strongly recommend considering such actions.
We will continue to use improper integrals, estimation techniques, integration by parts, and familiarity with functions and with algebraic manipulation a great deal in this course, and much of what we will do is important to your later technical education and practice. Please don't waste your time and effort with false expectations. I want students to be successful and will work diligently to help this occur, but I don't want to deceive them.
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 answers to version A (the light blue cover sheet) is available, and here is a more compact version of this exam. The questions of the other versions were close to those of version A. I hope that students themselves will be able to create answers for other versions after reading the version A answers. 161 students took the exam. Statistical measures of the performance of the four versions of the exam were sufficiently close that I feel the exams were reasonably functionally equivalent.

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.

Problem 1 (12 points)
a) (4 points) 2 points for the formula: Pi02 (ex)2dx. 1 point for the antidifferentiation and 1 point for the answer.
b) (8 points) 2 points for the formula: 2 Pi02x exdx. 5 points for correctly integrating by parts. 1 point for the answer. Some students attempted to do this problem with washers (dy) rather than shells. The setup using dy is more complicated and no one was successful.

Problem 2 (12 points)
6 points for correct integration by parts and getting the antiderivative (4 points for the first u,v step and then 2 points for the final antiderivative). 3 points for setting up an integral from 1 to a large number and substituting (1 of these points was earned for writing the improper integral as a limit of a proper integral with a parameter). 1 point for finding the limit correctly, and 2 points for showing where L'H is used and using it correctly. Just "plugging in" infinity in the antiderivative is not an acceptable strategy!
Note No points were earned if the student ignored the integral and applied L'H to the integrand and presented the result as the "solution". This is a totally different and much easier problem. It is not the problem students were asked to solve.

Problem 3 (12 points)
2 points for setting up the pieces in the partial fraction expansion. 2 points for getting an equation relating the top to the constants in the pieces. 3 points for correct solution of this equation. 3 points for antidifferentiation. 2 points for the answer (1 of these for using ln properties to verify the displayed result).

Problem 4 (12 points)
6 points for a useful integration by parts. 4 points for integrating the u dv term correctly (there are possible strategies other than the one displayed in the answers). 2 points for substituting and getting the answer as shown (1 point if arctan(1) is not evaluated correctly).

Problem 5 (12 points)
Stating one useful substitution gets 4 points. Correctly carrying out the substitution to get an antiderivative (including substituting back to x's!) is 6 points. 2 points for a correct answer, which does not need to be simplified. If students choose not to "go back to x", then the substitution's antidifferentiation earns 4 points. Changing the bounds in the integral correctly earns 2 points, with 2 more points given for the correct answer.
One possible strategy for computing this integral is direct manipulation of the integrand: expanding the power, then integrating each term. A few students did this tedious computation successfully. Partial attempts were awarded appropriate points.

Problem 6 (14 points)
a) (6 points) There should be seven terms, values of cosine at the appropriate x's, "weighted" appropriately (1/4/2/4/2/4/1), and these should be multiplied by (1/2)/3. The multiplier is worth 1 point, the weights are worth 2 points, the appropriate x's are worth 2 points, and cosine appearing correctly is worth 1 point. No function evaluations need to be done. If the function in the sum is only identified as "f" then the 1 point for cosine is not earned. Students should realize that cos(x3) and (cos(x))3 are very different functions.
b) (8 points) Use of the second derivative graph to get some estimate of the size of |f´´(x)| on [0,3] is worth 2 points. Use of the Trapezoidal Rule error term is worth 2 points. Getting some numbers, with a correct inequality for n is worth the remaining 4 points. No further direct approximation for n needs to be given.
If the inequality is reversed, 1 point will be deducted. The error estimate must be less than the desired accuracy for the computation to be valid! The K should not appear out of "thin air" but should be related either to the graph or to a computed and correctly estimated second derivative.

Problem 7 (12 points)
a) (10 points) The setup (that is, correctly instantiating the formula for average value which appeared on the formula sheet) earns 2 points. Computing the definite integral is worth 6 points (2 points for a useful substitution and 4 points for the successful computation). 1 point is deducted if the 1/A multiplier does not appear correctly in the answer. The final answer is worth 2 points, which includes 1 point for dividing by the length of the interval to get mA.
b) (2 points) The student's answer (if not trivial) should be analyzed, and 1 point was earned for a correct limiting statement about the student's answer. 1 point was earned for some valid explanation of this answer, which could be algebraic or could even refer to the graph of the integrand.

Problem 8 (14 points)
2 points for trying the correct substitution (some multiple of secant); 1 point for getting the multiple correct; 1 point for getting tangent out of the quadratic term; 1 point for the correct multiple of tangent; 1 point for getting dx correct; 2 points for assembling the integral in the substituted variable correctly; 3 points for cancelling and integrating (1 point for the last); 3 points for translating back to x's.
Several students translated back to x's using arcsec, and their answer involved the (disgusting?) composition of sine and arcsec. The question intended to require that answers not display any trig or inverse trig functions, but the instructor did not write this: my error, their free 3 points!


The second exam

Problem#1 #2 #3 #4 #5 #6 #7 #8 Total
Max grade 12 14 12 14 12 12 12 12 99
Min grade 0 0 0 0 0 0 0 0 15
Mean grade 10.77 9.32 4.14 7.90 9.74 4.74 2.99 6.42 56.02
Median grade 11 10 3 8 11 5 1 7 55

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 answers to version A (the light green cover sheet) is available, and here is a more compact version of this exam. The questions of the other versions were close to those of version A. I hope that students themselves will be able to create answers for other versions after reading the version A answers. 141 students took the exam. Statistical measures of the performance of the four versions of the exam were sufficiently close that I feel the exams were reasonably functionally equivalent.
I had anticipated the results for problems 3 and 6 (relatively low scores) since those problems involve inequalities, which beginning students frequently find difficult. I was surprised at the results for problem 7, which I regarded as completely routine (indeed, both parts of the problem were mostly copied from the review sheet).

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.

Problem 1 (12 points)
a) (3 points) 1 point for each equilibrium solution.
b) (8 points) Each graph is worth 2 points. If the label is missing on a graph, the student loses 1 of the points. The graph for A (respectively, B) should go through the point (0,1) (respectively, (0,-1)), and the curve drawn should be continuous, increases (respectively, decreasing), and between the horizontal lines x=0 and x=1 (respectively, x=-1). Substantial "violations" of this behavior could lose 1 point each, up to the total value of the graph.
The answer to each limit is worth 1 point.
c) (1 point) Identification of the specific solution.
The solutions in a) and c) should correctly be written as "y=constant". If instead each constant is identified in some manner which is not wrong (definitely wrong is "x=constant"!) , 1 point will be deducted.
Note This is essentially the first problem in section 9.2 of the textbook.

Problem 2 (14 points)
3 points for correct separation of the equation. 4 points for correct antidifferentiation of the y side (1 of these points is for knowing the derivative of ln). 2 points for correct antidifferentiation of the x side (this includes "+C", so if no "+C" is present 1 of these 2 points is lost).
3 points for using the initial condition successfully and 2 points for solving for y as a function of x. If no "+C" is present the student cannot earn the 3 points for using the initial condition, but could still "solve" for y as a function of x correctly and earn 2 points.
Note This is essentially problem 8 in section 9.3 of the textbook.

Problem 3 (12 points)
3 points for realizing or using the "key observation" that dropping the square root increases the size of the fraction (using the other part of the formula doesn't earn credit since it cannot be estimated usefully); 3 points for showing that the infinite tail must be estimated (here either a geometric series or a relevant [improper] definite integral can be given); estimating the infinite tail by finding the sum of the relevant geometric series or with a correct antiderivative; 3 points for using the sum and the tabular information (or a correct integral!) to get a correct answer. Merely asserting an answer is not sufficient to get credit here (lots of data was supplied). Asserting that an infinite tail is small because one or a few terms are small also earns no credit.

Problem 4 (14 points)
a) (8 points) 4 points for computing the ratio and correctly obtaining a simple fraction (1 point is lost if the absolute value is missing); 3 points for obtaining the limit of the ratio; 1 point for the answer.
b) (6 points) 3 points for considering the case x=1. 1 of these 3 points is for the answer, and 2 points for correctly supporting the answer.
3 points for considering the case x=-1. 1 of these 3 points is for the answer, and 2 points for correctly supporting the answer.

Problem 5 (12 points)
a) (6 points) 2 points for the answer (as a rational number, not an infinite series), and 4 points for some supporting evidence similar to what the answer sheet has.
b) (6 points) 2 points for the answer (as a rational number, not an infinite series), and 4 points for some supporting evidence similar to what the answer sheet has.

Problem 6 (12 points)
4 points for some evidence connecting the sum to the integral. One acceptable item would be a picture similar to that displayed on the answer sheet. An explicit inequality connecting the Nth partial sum with a definite integral would also be acceptable. Also useful would be mention of a relevant function decreasing. But some evidence should be given.
4 points for evaluating a relevant definite integral. 4 points for a correct answer with evidence showing that specific N is valid. Students who use N in place of N+1 in an otherwise correct solution will be penalized 2 points.

Problem 7 (12 points)
a) (6 points) 1 point for taking the ln of an and pulling out the exponent. 1 point for rewriting this as a quotient. 2 points for using l'Hopital's Rule and getting correct derivatives on the top and the bottom. 1 point for doing the algebra correctly and getting the limit of the resulting quotient. 1 point for exponentiating and getting the answer.
b) (6 points) 2 points for getting an equation for the limit. The equation should be something like "L=square root of a linear function evaluated at L." 2 points for solving the resulting quadratic equation. 1 point for explaining why the negative sign is dropped, and 1 point for the answer.

Problem 8 (12 points)
a) (6 points) 2 points for the initial conversion to "a/(1-r)" and 2 points for recognizing a and r (this may be done implicitly during the solution of the problem). 2 more points for the correct answer, with errors in the answer counted against this total.
b) (6 points) Any useful idea will earn 2 points, but I hope that the useful idea will be multiplying the answer to a) by the top of the fraction. But no further points will be earned if this is not carried out to get the answer to the question. The answer then gets 4 points, with 1 point deducted for each error in the answer. 1 point off for too many terms.

Criticisms of this exam There were probably darn many problems involving geometric series, and no problems using factorials! (Well, if we had Taylor's Theorem there would be plenty of factorial problems. So wait for the final exam!)


The final exam

Almost all of the final exam was the product of the course coordinator. O.k., I did see it before it was reproduced, and one result was that the proposed length decreased by a third (really!). So maybe I am not a total monster.
I believe almost all of the exam was straightforward and that successfully answering the questions represented some degree of competence in the subject matter of Math 152. More than half of the students (perhaps almost two-thirds!) left the final by the end of two hours (of the three hour exam period).

Numerical grades on the final exam were retained for use in computing the final letter grade in the course. The lecturers in the course agreed on the following scale for converting numerical grades to letter grades for the final exam grades. This assignment would help various lecturers decide on appropriate letter grades for the numerical information for students in various sections.

Letter grades for final exam scores
Letter
equivalent
AB+BC+ CDF
Range[165,200][153,164][140,152] [125,139][110,124][100,109][0,99]

Discussion of the grading


Assignment of course grades

All of the numerical evidence I have about student scores in the course was used to assign final grades. It is important to realize, as a colleague has told me, that I report student performance, and I don't "give" grades. Thus the numbers were used (as originally described here) to create a cumulative "score". Letter grades were then assigned based closely on a scale which reproduces the ranges for letter grades in the previous exams. Notice that the final exam number to letter conversion is very close to what I used for the two in-class exams. Of course the range was appropriately extended so that the scale went from 0 to 575. You may be amused (?) to know that, for students who took all three exams, the lowest cumulative grade was 148.955 and the highest was 547.276 (out of 575). I will admit that grading was not entirely "mechanical". If students' numbers were near borderlines, I looked, almost always, for reasons to give a better grade. So students who attended often or whose exam grades steadily improved may have gotten a bit of a break. That happened rarely, and I carefully questioned each such case to be sure that the grade assignments were credible, correct, and consistent.


Maintained by greenfie@math.rutgers.edu and last modified 5/8/2007.