Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Extra credit (RREF) |
Total | |
---|---|---|---|---|---|---|---|---|

Max grade | 16 | 18 | 20 | 20 | 14 | 12 | 5 | 101 |

Min grade | 0 | 0 | 4 | 2 | 0 | 0 | 0 | 41 |

Mean grade | 12.31 | 10.45 | 12.61 | 11.79 | 11.34 | 10.38 | 2.76 | 71.65 |

Median grade | 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:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [85,100] | [80,84] | [70,79] | [65,69] | [60,64] | [55,59] | [0,54] |

**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
s^{2}+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+e^{7t}. 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
C_{1}+C_{2}e^{-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.

Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Problem #7 |
Problem #8 |
Extra credit (BLOCK) |
Total | |
---|---|---|---|---|---|---|---|---|---|---|

Max grade | 12 | 20 | 8 | 12 | 12 | 17 | 10 | 8 | 5 | 104 |

Min grade | 0 | 10 | 0 | 0 | 1 | 3 | 0 | 0 | 0 | 45 |

Mean grade | 7.75 | 17.32 | 3.11 | 5.75 | 8.68 | 11.11 | 5.64 | 7.43 | 2.68 | 69.46 |

Median grade | 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:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [85,100] | [80,84] | [70,79] | [65,69] | [60,64] | [55,59] | [0,54] |

**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 A^{t} was a q-by-p
matrix. 4 more points would be earned if I could understand what the
entries of A^{t} 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-A^{2}+A^{3} is
realizing that must be
C(D-D^{2}+D^{3})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

**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:

- Continuity on [-Pi,Pi] with matching values at -Pi and Pi
- Closeness (with some "wiggling") to the line segments inside the interval [-Pi,-Pi/2] and [0,Pi].
- Gibbs phenomena (overshoot) at the four ends of the interval.

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:

- Graph identical to f(x) except at -Pi,-Pi/2, and Pi.
- The appropriate average value at each of -Pi,-Pi/2, and Pi.

**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.

Problem #1 |
Problem #2 |
Problem #3 |
Problem #4 |
Problem #5 |
Problem #6 |
Problem #7 |
Problem #8 |
Problem #9 |
Problem #10 |
Problem #11 |
Total | |
---|---|---|---|---|---|---|---|---|---|---|---|---|

Max grade | 18 | 18 | 12 | 18 | 12 | 16 | 16 | 14 | 19 | 18 | 12 | 171 |

Min grade | 5 | 4 | 0 | 4 | 2 | 1 | 3 | 0 | 1 | 0 | 0 | 53 |

Mean grade | 12.42 | 14 | 7 | 13.46 | 9.69 | 9.85 | 12.15 | 7.73 | 8.27 | 9.92 | 8.73 | 113.23 |

Median grade | 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:

Letter equivalent | A | B+ | B | C+ | C | D | F |
---|---|---|---|---|---|---|---|

Range | [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) |

**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/s^{2}. 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 x^{3}. (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 x^{3} was not a linear combination of two of the
three given functions, or if they asserted that x^{3} 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 (a_{n}cos(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 a_{n} 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).

Small positive time | A bit later | Much later |
---|

**
Maintained by
greenfie@math.rutgers.edu and last modified 12/21/2005.
**