### The first exam

Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Extra credit(RREF) Total 16 18 20 20 14 12 5 101 0 0 4 2 0 0 0 41 12.31 10.45 12.61 11.79 11.34 10.38 2.76 71.65 15 12 13 13 13 12 5 72

29 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] [80,84] [70,79] [65,69] [60,64] [55,59] [0,54]

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) (10 points) This was supposed to be a straightforward computation. Certainly there is opportunity for error regarding signs and exponents. 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. Students were specifically asked to use the definition to compute the Laplace transform in this problem. Other methods did not earn any points, but a correct answer was eligible for points in part b).
b) (6 points) 2 points for explaining why L'Hopital's rule was needed. For this, displaying the answer to a) as one quotitent and then using this algebraic form is, I think, necessary. 4 points for correctly applying L'Hopital's rule.

Problem 2 (18 points)
This is a problem resembling several which were assigned for homework. which can be solved with various techniques
a) (13 points) 2 points for taking the Laplace transform of the equation. 4 points for solving for Y(s) (there's a compound fraction to deal with). 4 points for stating the partial fraction problem (1) and then solving it (3). 3 points for finding the inverse Laplace transform of the result, and giving y(t).
Common errors Not reading or using correctly the Laplace transform table which was supplied. Sutdents who did not have s2+2 in the denominator of their Laplace transform could earn at most 10 points (1,4,4,2) since I felt they had unduly simplified the problem. Students who had no Y(s) resulting from the Laplace transform of the integral expression could earn at most 3 points (1,1,1,0) since they had almost made the problem trivial.
b) (5 points) 1 point for the derivative, 2 points for the integral, and 2 for the final assembly which was suppose to check the answer. Again, students whose answers were incorrect from a) could not earn full credit in b).

Problem 3 (20 points)
a) (6 points) I gave 2 points for using linearity of the integral and breaking the problem up into 2 parts. The computation with the Delta function would then earn 1 point, while recognizing that the Heaviside part of the computation reduced to an integral from 2 to 5 earned another point. The answer from the Heaviside part of the problem earned 2 points.
Comment A number of students invented their own problem here. Their problem seemed to involve computation of a Laplace transform. That is not what was requested, and no credit was earned.
Additional comment In fact, the previous comment is not totally correct. Mr. O'Sullivan convinced me, quite correctly, that the problem could be done using Laplace transforms. Take the transform of the function, divide the result by s, find the inverse transform, and evaluate the resulting function at t=5. This gives the same answer as the computation done by the instructor.
b) (7 points) Straightforward application of one of the translation theorems and the table of formulas. 2 points for telling that the answer would be e-3s multiplied by the Lapalce transform of something else. 2 points for recognizing that the "something else" was the function which resulted when t+3 was substituted for t in the expression 4t+e7t. Finally 3 points for assembling the correct answer.
c) (7 points) 2 points for recognizing that the Laplace transform of a convolution was the product of the Laplace transforms of the convolution's "factors". I needed to be convinced that the student knew what was going on. 3 points for the successful solution of the resulting partial fraction problem, and 2 points for correctly writing the inverse Laplace transform of the result. Several students correctly used the definition of convolution to solve this problem. Students who incorrectly wrtoe 1/s as the Laplace transform of t could earn only 4 of 7 points in the problem. Again, I felt that such an error made the partial fractions and inverse Laplace transform parts significanly easier.

Problem 4 (20 points)
This is a first-order ODE.
a) (10 points) Here taking the Laplace transform earns 3 points, 2 points for the partial fractions, 1 point for the (simple) inverse Laplace transform of 2/(s+1), and 4 points for the more complicated inverse transform which results in various Heaviside functions.
If students gave a solution which was sufficiently close in complexity to the correct solution, I attempted to read the remainer of the problem using that solution. If a serious error is made in this part which "trivializes" (makes much easier!) some or all of the successive parts of the problem, then I will not give points for those parts.
b) (3 points: 1 point for each part) I was happy to get unsimplified formulas in various intervals. Of course, to me algebraic "massaging" of these formulas makes the sketch in c) possible.
c) (3 points) I accepted very rough sketches here. I certainly looked for 2e-t (exponential decay from 2) in [0,1] (1 point). 1 point was earned for a graph that went down and up and down and stayed within the box. Finally, I gave the last point for continuity.
If the algebraic formulas are written in the form C1+C2e-t for various constants, then the graph is easier to sketch, and the questions in later parts become more tractable.
d) (1 point) For the correct answer.
e) (2 points) For the correct answers.
f) (1 point) For the correct answer.

Problem 5 (14 points)
I gave 5 points for writing the symbolic linear combination. Then correct use of a RREF or other method to get a solution earns the remainder of the problem's credit, and I reserved (at least) 2 points for a clear statement of the correct solution. I do not believe that the Atlantic RREF can be used to solve this problem.
To get a significant number of points, I need to be satsified that the student knows what a linear combination is, in the context of this problem. The problem is very easy (two lines or three?) with the use of Pacific. The problem is quite irritating if the RREF is ignored.
Comment I believe that computation sometimes can be an effective teaching and learning device, I don't like to subject students to large amounts of pointless computation, especially on an exam. If you find yourself doing that on one of my exams, it may be time to think about your methods.

Problem 6 (12 points)
Students should know and "manifest" what is needed to verify linear independence. Then the linear system has to be set up and solved. 4 points for knowing about linear independence. 8 points for manipulating the system correctly and showing that the functions are linearly independent. Various valid methods were used, and I was happy to give credit for them. Students who wrote collections of symbols which I could not understand did not get credit. Such material also resulted in points being deducted from possibly valid solutions. I wanted to understand what was written.
Engineers You can indeed survive an exam in an upper-level math course where one of the questions has the word "Prove" in its statement.

Extra credit (5 points)
I gave 5 points to students who were able to take a "random" (?) matrix produced by Maple and convert it to RREF. Ample opportunities for retakes were offered.
Comment Only 16 of 29 students got 5 points this way.

### The second exam

Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Problem #7 Problem #8 Extra credit(BLOCK) Total 12 20 8 12 12 17 10 8 5 104 0 10 0 0 1 3 0 0 0 45 7.75 17.32 3.11 5.75 8.68 11.11 5.64 7.43 2.68 69.46 8 18.5 3 6.5 11 11 8 8 2 73

28 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] [80,84] [70,79] [65,69] [60,64] [55,59] [0,54]

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)
6 points for each part. I will read what was written. I can not guess what you meant to write. I required that what was written be responsive to what was asked.
a) (6 points) First I gave 2 points if I could discern ("perceive clearly") by reading what you wrote that At was a q-by-p matrix. 4 more points would be earned if I could understand what the entries of At were. A phrase unanchored by references ("the rows and columns are interchanged") would not alone earn credit. Rows and columns of what? Resulting in what?
b) (6 points) In part b), an important "If ... then ..." needed to be inferred from what you wrote (for example, "If [a certain linear combination is 0] then [all the coefficients are 0]". Or "The only way a certain linear combination is 0 is if all the coefficients are 0". I also accepted statements about none of the vectors being linear combinations of the others. But I read what you wrote, and tried to read it carefully.

Problem 2 (22 points)
Hand computation of a 2-by-2 problem is much easier than a 3-by-3 problem. THe last part can be done economically or with much computation and potential for error.
a) (2 points) Take the determinant of A-I. You may compute this determinant in any way you like.
b) (2 points) You may have already simplified the characteristic polynomial in part a), or you can do it here. The roots should be obvious.
c) (3 points) You need to solve two (2) homogeneous systems of linear equations. They are rather simple but errors can be made.
d) (2 points) You are merely asked to write P and C, which you certainly should be able to do after parts b) and c). Verification of your statements occurs in the next few parts of the problem. If incorrect results from b) and c) are used correctly here, I gave full credit.
e) (2 points) You may find P-1 in any way you like. I wanted to know the method you used to get C-1 so a correct answer alone received only 1 of the 2 points. For a 2-by-2 matrix, almost any computational strategy is easy.
f) (2 points) Compute the product requested.
g) (2 points) Compute the product requested. If you do not get a correct diagonal matrix, I gave no points.
h) (6 points) 2 points for setting up the requested relationship A=CDC-1. The other 4 points were for the computation. Certainly an efficient way to compute A-A2+A3 is realizing that must be C(D-D2+D3)C-1. I gave 2 of the remaining 4 points to students would showed that they understood this "reorganization" of the computation. The last 2 points were reserved for the answer. Information was given allowing some aspects of the answer to be checked.

Problem 3 (8 points)
No!
Based on what they wrote on this exam, some people seem to have strange beliefs about diagonalization. Please note that a matrix with a determinant equal to 0 may be diagonalizable. A matrix with one or more rows or columns equal to zero maybe also be diagonalizable. A matrix with low rank may be diagonalizable. A matrix with main diagonal entries all equal to 0 may be diagonalizable. One example which shows this is the matrix with all zero entries. Other, more interesting examples, can be given. Therefore all of these "reasons" are false and irrelevant in connection with this problem.

Here is a partial credit outline for this problem which I was able to follow easily in almost all cases: 3 points for showing that 0 is the only eigenvalue. 3 points for showing that the collection of eigenvectors consists of multiplies of only one vector. 2 points for concluding that, since there are not three linearly independent eigenvectors, the matrix cannot be diagonalized.

No, no!
The word "distinct" or similar words or phrases is not the same as linearly independent. I can only read what is written on the exam, and I cannot and would rather not imagine what students wanted to write. In this problem such a restriction is almost painful, because I'd like to believe that what students wanted to write was a correct explanation of non-diagonalizability. But I tried to be consistent. I read and graded and regraded several times.

Problem 4 (12 points)
a) (8 points) Students earned 3 points for writing that the rank of A was 1 or 2 because the rank(A) is at most rank(A|B) and 1 point for excluding 0 as a possible rank, since A is non-zero. 2 points were given for analysis of the case rank(A)=1: the system is inconsistent since B supplies a compatibility condition which can't be satisfied. 2 points were given for analysis of the case rank(A)=2: since rank(A|B)=2, no additional compatibility conditions exist, and the system is consistent. In each case, 1 of the 2 points was for the answer, and 1 for the reason. Very brief reasoning was accepted.
b) (4 points) 2 points for the answer and 2 points for supporting reasoning. Again, rather sketchy reasoning, usually referring to the number of columns of A compared to rank(A), was accepted.

Problem 5 (12 points)
It was difficult to assign partial credit consistently in grading this problem. I first graded and then excluded those papers which were perfect and those which had no readable or sensible work. This was actually the majority of the class. I then read and graded three times the other papers and attempted to give grades consistently.
If a student missed a sign near the end of the computation or in some place which wasn't important, then I deducted only 1 point. Students who demonstrated some knowledge of properties of determinants generally got at least 4 points, and those who, it seemed to me, showed enough knowledge to do the problem correctly earned 8 points. Some of this grading was more subjective than I like to admit (!) but it was done with the front pages concealed.
I tried diligently to grade fairly and consistently. By the way, most of the students who earned 12 out of 12 did the problem by expanding on the unique b entry. The solution on the answer sheet is what I did when I went back to the exam some time after writing it and sat and wrote solutions. I wasn't as efficient as most students!

Problem 6 (18 points)
a) (5 points) This was straightforward integration by parts. I penalized students 1 point when I could not clearly identify the answer to the question (!). I gave 2 points to students who gave the "correct" parts but had other troubles. Students lost 1 point for a minus sign error. One clue to the correct answer was an analogous formula for cosine given in part b).
b) (5 points) Students earned 1 point for every correct answer. The formula sheet contained the necessary formulas, and the problem statement gave a strong hint about the nature of the answers. Substitution and evaluation into the indicated form were the only tasks necessary. An exception to this grading scheme were those few students who systematically reversed the cosine and sine coefficients for n>0. I gave them 3 out of 4 points. Another exception occurred when students somehow omitted dividing by Pi. I decided to penalize them the first time, but not for later coefficients. Both of these exceptions affected in a minor way the grades of only about 5 or 6 students.
c) (8 points)
The graph of the partial sum (4 points)
I looked for the following qualitative behavior in the graph of the partial sum:

1. Continuity on [-Pi,Pi] with matching values at -Pi and Pi
2. Closeness (with some "wiggling") to the line segments inside the interval [-Pi,-Pi/2] and [0,Pi].
3. Gibbs phenomena (overshoot) at the four ends of the interval.
I took off only 1 point for lack of discernible Gibbs phenomena but severe lack of compliance with one of the other two could lose 2 points. Students who did not graph anything in [-Pi,-Pi/2] lost 2 points.
The graph of the whole sum (4 points)
This is supposed to be the graph of the sum of the whole Fourier sine series. Here I looked for this behavior:
1. Graph identical to f(x) except at -Pi,-Pi/2, and Pi.
2. The appropriate average value at each of -Pi,-Pi/2, and Pi.
Students who did not graph anything in [-Pi,-Pi/2] lost 2 points. Several students drew vertical line segments as part of the graph here. Such segments cannot be part of the graph of a function and were penalized.

Problem 7 (10 points)
a) (6 points) 2 points for recognizing that orthogonality in this problem means something about integrals, and then computing these integrals. 2 points for seeing that the problem requirements mean satisfying a 2-by-2 homogeneous system. The final 2 points were earned by giving a specific reason why this system has only the trivial solution. These points were not earned by labeling the system as inconsistent: some computational reason which was then demonstrated had to be given. The instructor's solution used the fact that the determinant of the coefficient matrix was non-zero. Many students "solved" and isolated one variable and then deduced that both variables had to be 0. That was certainly acceptable.
b) (4 points) Again, 2 points for the orthogonality conditions "translated" into integral computations. 1 point for realizing that the resulting homogeneous system consisted of one equation, and then the final point for giving a non-trivial solution to that equation.

Problem 8 (8 points)
a) (4 points) 3 points for the sketch. Some students sketched no curve in the [0,3] interval, and I deducted 1 point for this error. 1 point for the question about Fourier series.
b) (4 points) Scored as in a).

Extra credit (5 points)
I gave 5 points to students who successfully answered the two questions about block decomposition of matrices. I gave 2 points to students who successfully answered only one of the questions.
Comment Only 18 of 29 students got points here, and only 12 of them earned 5 points in this problem.

### 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 Total 18 18 12 18 12 16 16 14 19 18 12 171 5 4 0 4 2 1 3 0 1 0 0 53 12.42 14 7 13.46 9.69 9.85 12.15 7.73 8.27 9.92 8.73 113.23 13 15.5 8.5 14 12 9.5 15 8.5 7.5 11 10 109

26 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 [147.9,174] [139.2,147.9) [121.8,139.2) [113.1,121.8) [104.4,113.1) [95.7,104.4) [0,95.7)

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 (18 points)
a) (13 points) 3 points for the setup, which includes the correct description of f(t) and the correct domain of integration. 4 points for the integration by parts. 3 points for evaluation of one piece of the integral with 1/s and 3 points for the evaluation with 1/s2. If I could not easily understand the computations, I gave up. I think part of this problem is communication of the limits of integration, the methods, etc. If the setup was f(t)=2t for t<1 and 0 otherwise, a maximum of 10 out of 13 points could be earned.
b) (5 points) 1 point for demonstrating the need for L'H once, and 1 point for correct top and bottom differentiation. Another 1+1 was earned for the second L'H. The final point was earned for the correct answer. I was rather strict here. Students had to show why L'H was needed, and had to compute correctly. One consequence is that answers from a) which were far from true led to ineligibility for the points in this part. That should have been a signal to students that their process and/or answer in a) was incorrect.

Problem 2 (18 points)
a) (12 points) 3 points for taking the correct Laplace transform of the equation and solving for Y(s). 3 points for the partial fraction decomposition. 3 points for the inverse Laplace transform of the rational function. 3 points for the inverse Laplace transform of that part involving the exponential function (one of the Shifting Theorems is needed). Students who recognized the Laplace transforms of sinh and cosh and used them correctly got full credit. I would have put these on the table if anyone had asked (but then I would not have given this precise problem!). Students who made errors which trivialized the problem were penalized here, and might not have been eligible for full credit in the parts below.
b) (3 points) 1 point for the first expression, and 2 points for the second.
c) (3 points) 1 point for the first answer, 1 point for the second, and 1 point for the third. Students needed to use their answer, and compute correctly.

Problem 3 (12 points)
a) (6 points) The most common error was omitting the condition that v not be 0, which lost 2 points. I gave credit only when I could understand the statement and it was correct.
b) (6 points) An answer alone does not get full credit. A maximum of 2 points was given to an answer supported by very little. I looked for assurance that the student had checked that every coordinate of the vector gave the same eigenvalue.

Problem 4 (12 points)
a) (10 points) The resulting polynomial should have degree 4, or the student loses 2 points. Most students "expanded" along a row or column, and I subtracted 2points if one of the resulting 3-by-3 determinants was incorrect. Other minor errors were handled appropriately.
b) (2 points) For displaying the correct roots of the student's characteristic polynomial. Almost all students got polynomials whose roots could easily be found by hand (thank goodness!).
c) (4 points) To score points in this part of the problem, statements needed to be supported by previous student work, and by valid reasons for the conclusion. Therefore I did not give points to someone who might have guessed the correct conclusion but was unable to give some valid supporting evidence.
d) (2 points) The correct answer (1 point) and some correct reason (1 point).

Problem 5 (12 points)
a) (6 points) I gave 2 points for considering a linear combination with "arbitrary" coefficients and then setting it equal to 0. The resulting homogeneous system was easy to analyze and show that it had only the trivial solution. I did not give credit if I could not understand what students wrote.
b) (6 points) The part of this question certainly is the worst constructed from the point of view of checking if a student knows the subject matter of 421. That is because it is clear (at least, I think, to me!) that a sum of polynomials whose degree is at most 2 cannot be x3. (It is clear, however, because all of us have done so much computation in the "native" basis of the polynomials, that is, the usual monomials.) Therefore any student who stated such a belief in a way I could understand got full credit. Thus those students did not have their 421 linear algebra knowledge tested. Otherwise, I deduced points if students didn't describe the problem clearly. In particular, I gave only 4 points if students asserted that x3 was not a linear combination of two of the three given functions, or if they asserted that x3 and two of the given functions were linearly independent. Either of those is not enough to conclude the desired result.

Problem 6 (16 points)
a) (8 points) I gave this problem so that people could show me they knew what Fourier coefficients and Fourier terms were. There were rather low scores on a similar problem on the second exam. Here I wanted the terms, and not just the coefficients, as I indicated. Any student who gave a linear combination of sin(x), sin(2x), and sin(3x) with other than constant coefficients got 0 for this part. By now, students should know what Fourier series look like. The 2/Pi normalization got 1 point. Each Fourier coefficient got 1 point (3 points total). Each appropriate sine function got 1 point (3 points total). 1 more point was given for assembling the information and presenting it correctly.
b) (8 points) 4 points for the graph of g(x): 1 point each for sticking to the continuous pieces "inside" the intervals of continuity; 1 point for being continuous in the whole interval and "leaping down" from the top to the bottom; 1 point for the Gibbs wiggling in both places. 4 points for the graph of h(x): again, 1 point each for being identical to f(x) on the two intervals of continuity; 1 point for having a value in the middle of the jump, and 1 point for noting in some way that there are holes on the bottom and the top of the jump. I deduced points for extra information which was contrary to the correct pictures.

Problem 7 (16 points)
a) (3 points) 1 point for the graph=1 away from the interval from 0 to 2, and 2 points for the parabolic bump up inside the interval.
b) (3 points) Just quoting the D'Alembert solution (with no g term!) earns the points, with the implication being that the function f is the one defined in the problem. Students who went on to state an incorrect formula (that is, substituting the quadratic formula with no indication of the valdity of the formula in terms of the values of the variables) lost 1 point.
c) (4 points) 1 point each for the traveling wave to the left and the right; 1 point for the correct height and location; 1 point for the wave being 0 outside of intervals of the traveling waves.
d) (1 point) For identifying the correct time.
e) (4 points) 2 points for indicating that there are 2 maximums. 1 point each for the correct location of the maximums.
f) (1 point) For identifying the correct velocity.

Problem 8 (14 points)
a) (4 points) I looked for a solution which satisfied the initial conditions (2 points) and which had the desired exponentials (2 points). The solution could just be written and not justified.
b) (8 points) I gave 2 points for some evidence of orthogonality displayed. Otherwise, I looked for the double integral, I looked for orthogonality, I looked for the appearance of the normalization constants, and I looked for the t functions not being integrated. Points were taken off appropriately for algebraic errors. etc. In this part, an answer alone got little credit: I looked for some supporting evidence.
c) (2 points) The answer, supported by some evidence derived from the answer to b). I gave only 1 point to thos students who insisted that the temperature oscillated, even if supported by b)'s answer. That's just too unphysical.

Problem 9 (20 points)
a) (9 points) 2 points for separating variables correctly. 2 points for handling the x function and coming up with sin(nx), where n is a positive integer. 2 points for getting the differential equation for Y(y) correct, and realizing that solutions are exponentials or hyperbolic functions. 3 points for explicitly listing the solutions (the case n=1 requires separate comment).
b) (11 points) 2 points for excluding the sinh term. 3 points for writing the correct series. 2 points for explicitly writing how u(x,0) is related to the initial condition. 2 points for writing u(x,0) as a sum of a Fourier sine seres. 2 points for writing the formula for the coefficients and 2 more points for writing the three initial terms.

Problem 10 (18 points)
I gave a total of 10 points for the separation of variables portion of this problem. The t and x portions were graded in the same way: 2 points for getting the correct ordinary differential equation, 2 points for getting cos(n{x|t}) (not a general ) and 1 point for dropping sin(n{x|t}). 8 points for assemblling the general answer. Of this, 4 points for a sum of the solutions with coefficients (ancos(nx)cos(nt), summed from n=0 to infinity) and then 4 more points for connecting with the initial conditions: first, for using u(x,0) in connection with f(x), and then writing the coefficients an in terms of cosine Fourier coefficients of f(x).
Comment This problem discussed what's called the free boundary condition for the wave equation. Some further information is here.

Problem 11 (12 points)
a) (6 points) The curve should be smooth and higher than the global min, and lower than 3. It should have value 2 at x=0 and should have a horizontal tangent at x=Pi. I gave 2 points for the smoothness, 2 points for the horizontal tangent, and 2 points for going through (0,2). Sometimes the horizontal tangent at x=Pi was difficult to score.
b) (6 points) A horizontal line earned at least 3 points. 2 points were earned for a graph going through (0,2). The solution should be a straight line (the only steady-state one dimensional heat equation solutions), which accounted for the other points.
Here are some pictures of an approximate solution (with 100 Fourier terms). The correct eigen{function|value}s are given as the third example here. I used them to create these graphs, after first subtracting the appropriate steady-state solution (y=2).