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 | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|
Range | [85,100] | [80,84] | [70,79] | [65,69] | [55,64] | [50,54] | [0,49] |
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!
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 | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|
Range | [85,100] | [80,84] | [70,79] | [65,69] | [55,64] | [50,54] | [0,49] |
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 | A | B+ | B | C+ | C | D | F |
Range | [165,200] | [153,164] | [140,152] | [125,139] | [110,124] | [100,109] | [0,99] |
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.