The first exam

32 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. The grade distribution was remarkably bimodal. Please look at some comments about the results of the exam. Here are approximate letter grade assignments for this exam:

Discussion of the grading

Problem 1 (16 points)
a) (10 points) A routine computation using integration by parts. It will help in part b) if you get the correct answer! Of course, a small part of the purpose of part b) is to check your answer.
b) (6 points) Careful use of l'Hopital's rule is a major part of justifying some of the computations of the "operational calculus".

Problem 2 (14 points)
This is a standard ODE problem which can be solved with various techniques (several from 244, for example).
a) (10 points) 4 points for taking the Laplace transform of the equation. 4 points for solving the partial fractions problem and setting up the equation for the inverse Laplace transform. 2 points for "reading off" the solution correctly from the tables.
b) (4 points) The points are for checking that the initial conditions are satisfied. Of course, again this should help check your answer.

Problem 3 (12 points)
Straightforward application of one of the shifting theorems.

Problem 4 (8 points)
Easily done in at least two ways.

Problem 5 (20 points)
This problem is worth one-fifth of the exam. I don't think I could solve this ODE easily without the Laplace transform!
a) (10 points) Here taking the Laplace transform earns 4 points, 4 points for "massaging" the Laplace transform (mostly another partial fractions exercise), and 2 points for writing the inverse transform.
b) (6 points: 2 points for each part) I am happy to accept an unsimplified formula in the interval t>Pi.
c) (2 points) This is actually a very simple graph. You can get enough information by evaluating at 0, Pi/2, Pi, and 3Pi/2. Even if you didn't simplify previously, you should be "suspicious" about the graph. Note also that "geometry" of the axes supplied was a hint.
d) (2 points) Maybe this is the theoretical part of the exam, although maybe mechanical engineers should be concerned about shocks. That y(t) is not differentiable at only one value of t is subtle to me. Certainly the solution is differentiable everywhere except Pi/2 and Pi.

Problem 6 (14 points)
I'll give 6 points for taking the Laplace transform correctly and then 2 points for organizing the equations to be linear in X(s) and Y(s). The remaining 6 points will be for finding an expression for X(s). The intent of this problem was to get students thinking about solving linear equations even when the darn coefficients are "weird".
I have learned while grading this problem that I should avoid the use of 5 in Laplace transform problems. 5 and s (the Laplace transform variable) are difficult to distinguish in many students' handwriting.

Problem 7 (16 points)
Students should know and "manifest" what is needed to verify linear dependence. Then the linear system has to be set up and solved. 4 points for knowing about linear dependence. 8 points for manipulating the system correctly. 4 points for producing an explicit satisfactory solution. I will give only 8 points for giving an explicit linear combination when no process is shown.

The second exam

30 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:

Discussion of the grading

Problem 1 (12 points)
6 points for each part. I read what was written. I required that what was written be responsive to what was asked. Thus a discussion of span or subspace was irrelevant in part a) if linear combination was not correctly defined, and, once that was correctly defined, such discussions were superfluous.
I took 2 points off in part b) if the definition used AX=X and did not specify that X must be non-zero. This specification is important. If a definition using the equation det(A-I_n)=0 was given, there was no such risk.

Problem 2 (18 points)
a) 2 points for correctly computing the characteristic polynomial.
b) 5 points for the student's finding the correct roots of the characteristic polynomial (1 point) and then finding two correct eigenvectors. If two correct eigenvectors were displayed I gave credit for a basis. If an answer labeled basis was incorrect even though two correct eigenvectors were given, I took off 1 point. You can't get credit for the eigenvectors unless the correct eigenvalues are found, because with an incorrect eigenvalue, you can't find non-trivial solutions to AX=X. Therefore several mistakes had to be made, and that's too much for partial credit.
c) 2 points: only correct answers received credit here, because incorrect answers resulted from several previous mistakes.
d) 5 points: I gave credit for correctly finding an inverse for the student's P. 1 point for the answer alone.
e) 2 points for the correct product of A and the student's P.
f) 2 points for the correct product and answer. I deducted 1 point if the student gave the correct answer for the product of the student's P^-1 and the student's Z but the answer was not diagonal. There was enough time to check and recheck answers on the exam and numerous ways to discover errors.

Problem 3 (12 points)
Many people got this right. "A matrix is invertible if its determinant is not 0." Very few students attempted to do row operations to simplify the matrix before evaluating the determinant which significantly reduces the work.
I took off 1 point if the strategy and the evaluation proceeded well, except for a "silly" arithmetic error. And another 1 point if I spotted a distinctly different error. I took off more if I thought there was an error more serious than simple arithmetic.
Some students attempted to find A^-1 and to discover which values of x make this possible. I consider this sa difficult strategy. Attempting to do this would earn 8 points (no one completed it successfully!). I took points off from there.
There is probably an optimal (?) sequence of row/column operations applied to the original matrix which would make the answer "clear" but I don't know it. Again, one or two row/comn operations make evaluating the determinant much easier. Very few students did this!

Problem 4 (14 points)
Student performance implies that this was the most difficult problem on the exam. I wanted correct answers and supporting reasoning in each part of the problem. I tried diligently to grade the problem consistently, so I actually had to regrade some students' work several times.
Part a) will get 6 points: 3 for remarking that there are many solutions to AX=0 (that X=0 is a solution will earn only 1 point), and 3 for analyzing the dimension of S. Again, especially in this problem, correct assertions will earn minimmal credit if unsupported by verfication. Somewhere there should be a mention that the two solutions to AX=0 supplied are linearly independent.
Part b) is worth 4 points. I seem not to have taught effectively the content of the following sentence. The larger the rank, the smaller the dimension of S will be, and the larger the dimension of S, the smaller the rank will be.
Part c) is again worth 4 points. If the rank is not 5, then there are Y's for which AX=Y has no solution.

Problem 5 (16 points)
Here I looked for correct implementation of the method which takes advantage of diagonalization: A⁵=PD^5-1 where P is the matrix of eigenvectors written as column vectors, and D is the corresponding diagonal matrix of eigenvalues. A direct computation of A⁵ earned little.
The correct outline will earn 8 of the 16 points. One error which lost 3 points was interchanging the roles of P and P^-1 in the formula for A^-1.
Implementation counts for the other 8 points. Creating a candidate for D is not difficult, and checking a candidate for P^-1, given the information in the note, can also be done. The wrong P^-1 lost 3 points. Several people took advantage of the symmetry of A to get a P which was orthogonal (by normalizing the given eigenvectors) and then computation of P^-1 was easy: P^-1=P^t. This was nice. A wrong answer loses at least a point.

Problem 6 (16 points)
a) 8 points, which must include a specific citation. Useful birds here are Goldfinch and Nuthatch, and their rank (3) should be used. Inadequate verification will earn only 4 of 8 points.
b) This was worth 8 points. Here I believe there's only one relevant RREF, Nuthatch. A citation would include verifying that the ocordinates of the student's candidate do not satisfy the compatibility/consistency conditions in the last two rows of the RREF of Nuthatch. I'll only give 4 of 8 points for inadequate verification. The suggestion that a vector in R³ is an answer cannot be correct!

Problem 7 (12 points)
I agree: knowledge of what a polynomial of degree 3 is is necessary in this problem. I do not apologize.
I do regret not making the last column of the augmented matrices in the BIRD list have entries written as y₁, y₂, etc. I used A+Bx+Cx²+Dx³ when I did the problem, but I still got very confused, with A and a, B and b, etc.
A common error seemed to be writing the coefficients in reverse order. I took off 2 points for this, since the error can be detected (although not with p(1)=2, as several students pointed out).
Otherwise, 4 points for realizing that the "general" cubic polynomial gives some linear equations with the four conditions specified, 4 points for translating the 4 equations correctly with a reference to Helicopter, and 4 more points for solving the equations correctly.

The third exam

29 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Overall, the grades were a bit lower than I had hoped, perhaps because this exam came so late in the exam period and people were tired. Here are approximate letter grade assignments for this exam:

Discussion of the grading

Problem 2. (12 points)
This problem's results were disappointing. Perhaps I should have written the question differently. I expected the following mental connections: the Fourier sine series converges to the odd extension of f(x), and the Fourier cosine series converges to the even extension of f(x). With these observations in mind, the problem is also straightforward, requiring practically no computation.

Each correct answer in part a) received 1 point. Correct answers in parts b) and c) were each awarded 2 points.

Problem 3. (16 points)
Certainly the result of a) is needed in b), and the result of b) is needed in c). If an error was made somewhere in the process, I tried to "read with" the student's work so that ideally the answer would lose an appropriate amount of credit only for the initial error (if the result did not simplify the problem ridiculously).

a) 5 points for all aspects of a). The diversity of errors was interesting. I did not subtract for an omitted "+C" although it was an indefinite integral!

b) (3 points) I wanted people to recognize with appropriate evaluations cos(nPi) and sin(nPi). We have done these over and over again in this course.

c) (4 points) Exact values were requested.

d) (4 points) Here recognition of g(x) as a (high) partial sum of the sine series for f(x) (that is, the Fourier series of the odd extension of f(x) to [-Pi,Pi]) is needed. To the right is a Maple graph of g(x), which entirely obscures f(x) on this scale. Note that the drawing jumps down to 0 at x=Pi and has a Gibbs overshoot compared to x² near and to the left of x=Pi. I forgot the jump all the way down to 0, so I would have lost a point in my solution to this problem.

Problem 4. (16 points)
Students should be acquainted with e^-x2. This function is very important in statistics, and presumably your labs have told you how to analyze data statistically. This is the "Gaussian" bump. The function y(x,0) specified in the problem is 4 times as high, and shifted to center at 3.

a) (4 points) I hoped you would use the D'Alembert solution. You must state what f is, since there is no f(x) specified in the statement of the problem. I deducted 2 points if no f(x) was specified here, but I would give a point back if I could reasonably deduce later in the problem that you "meant" the correct f(x).

I graded b) (4 points) and c) (4 points) together. 2 points were deducted for each error: bumps not symmetric, bumps not rounded, bumps not the proper amplitude, and bumps not properly centered. Also each of the four bumps should have more or less the same shape (no diffusion).

d) (4 points) If an incorrect answer alone was given, the grade was 0 for this part. I gave part credit even for an incorrect answer if some correct aspect of the process was shown.

Problem 5. (14 points)
a) (6 points) A correct answer alone would get full credit, since no integration computation is necessary. Writing the formula from the formula sheet and doing no further work earns no credit. If that approach is taken, computations to get the answer must be performed.

b) (4 points) The graph must be the same as what is shown. Some explanation depending on periodicity or on cos((even integer)Pi)=1 must be given.

c) (4 points) Please note that any explanation based on the graph displayed must be incorrect. Consider the following much simpler example: y(x,t)=A sin(x)cos(t) is a solution of y_tt=y_xx for any constant A. A graph of y(x,Pi/2) will show a horizontal line with 0 displacement. The maximum displacemen t is A, which I can choose to be any number. Therefore you can't make deductions based on the displacement alone at any specific time. Perhaps the problem statement should have read, "Use your answer to a) to estimate the maximum displacement at any time ..." Answers I considered correct used information about the max of |sine| and |cosine| and the triangle inequality, even if only implicitly.

Problem 6. (16 points)
a) (5 points) The formula to be used is on the formula sheet (just copying this does not earn credit!). One computation is for c₀. Then the other coefficients must be gotten through a simple integration.

b) (5 points) The trig functions should be evaluated where possible, and the first 5 non-zero terms written out.

Below are Maple pictures of the 250^th (!) partial sums at t=0 and t=1/100 and t=100. The first picture shows the usual Gibbs stuff. The second picture is nice and smooth, and the third picture seems to be like the graph of a constant function.

Graph of the partial sums up to the term involving cos(250x)
t=0	t=1/100	t=100

c) (4 points) The graph should be rounded and smooth, decreasing and less than the original graph, and positive. The corners should be replaced by curves.

d) (2 points) A horizontal line at height 1/4.

Problem 7. (16 points)
a) (6 points) 3 points for writing T'/T=(X''/X)+1, and then 3 more points for successfully separating (the constants in the two ODE's must be linked).

b) (7 points) The deductions X(0)=0 and X(Pi)=0 earn 2 points. Solving the boundary value problem for X(x) earns 3 points, and then solving for T(t) earns 2 points. At times the work for this part and part a) were intertwined and I tried to recognize this with appropriate grades.

c) (3 points) The correct infinite series with coefficients earns 3 points.

Additional exam 1 and exam 2 problems

Adjusting the grades in exam 1 and exam 2

G(40,12);
                                  52.
G(60,12);
                                  70.
G(80,12);
                                  86.
G(90,12);
                                  94.

Course grades

Maintained by greenfie@math.rutgers.edu and last modified 3/14/2004.