Prob 1: (From Midterm I)
A number of people did not remember that the "trick" to this problem
was to use the fact that sin2 t + cos2 t = 1.
Some seemed to assume that x2 + y2 = r2,
which is not true since the circle is not centered at the origin. One should
START with a known truth, i.e., sin2 t +
cos2 t = 1,
and then proceed to some conclusion about the relationship of x and
y. This is done by solving the two given equations in terms of
sin t and cos t.
Prob 2: (From Midterm II)
Several people forgot the correct formula for area, given a curve
in polar coordinates. Several people forgot how to integrate sin2
theta, i.e., using the correct half-angle formula.
Prob 3: (From Midterm III)
This is NOT continuous since if one approaches the origin via the
line x = y the limit is 2 and if one approaches the origin via
the line x = -y the limit is 0.
SEVEN students said that this function WAS, in fact, continuous, even though
they had seen this problem on Midterm III, and had received the answer
key which indicated it was NOT continuous.
For a continuous function, the
limit approaching the origin must always be 1, which is what the
function is defined as at that point. A number of people had very
basic problems taking the limit along various lines. Some people
gave the limit as being "0/0" which is undefined. In fact, one of
the reason for the concept of "limit" is to determine what the "real"
value should be if one gets "0/0" (cf. Math 11). "0/0" should NEVER be given as
an answer to any math question -- it should only be used as a reminder
that more work needs to be done to find the correct answer.
More than one person seemed to forget the basic truth taught in
Calculus I (Math 11) that the limit (as theta -> 0) of (sin theta)/theta is 1
(not zero).
Some people seemed to try to use a 3D version of L'Hopital's rule, which
I never covered in class and thus may not, in fact, apply to 3D cases!
Prob 4: (From 2007 Sample Final -- Solution on-line via ERES)
A number of people got the correct vector parallel to the desired
line via the cross product of the two normal vectors. However, they
failed to give the correct equation of the line that is parallel to
the vector.
Prob 5:
This was meant as an easy problem to see if you remember how to
find small determinants and do simple algebra. Unfortunately,
some people seemed to panic and not realize how simple the problem actually
was. One basic difficulty was that some people forgot how to evaluate
simple determinants and factor small polynomials.
The 3×3 determinant
evaluates to 1-2a2+a4. This
can be written as (1-a2)2 thus giving
b the value of a2 and
c the value of 2.
Life is made harder by rewriting the value of the determinant as
(a2-1)2.
In this case, the value of b is 2-a2.
NOTE: if one evaluates the 4×4 matrix that is similar to
the ones given in this problem, one get the determinant value of
(1-a2)3.
Some people had difficulty factoring the polynomial and some several
people added a2 to a2
and got a4 rather than 2a2.
Prob 6:
(This problem was parallel to problem 8 on last year's final.)
Some people left the derivative as a vector (incorrect!). It is a
DOT PRODUCT which always gives a SCALAR result. The direction vector
must be a unit vector as well.
Some people went to 3 dimensions rather than use the 2 dimension
angular orientation.
Prob 7:
A significant number gave some sort of an equation of a LINE rather than of a
tangent PLANE as the problem requested. Several people used the
gradient components as denominators (correct for a LINE) rather than
as multiples (needed for PLANE).
Several people neglected the coefficient of -1 for the z term or
ignored the z term completely in the equation of the plane. (See
solution for problem 4 on last year's final, where -1 is in the denominator
of the z term of the equation of the line.)
Prob 8:
Most people had no problems. Several, however, did not use
the product rule when computing dy/dt and dz/dt.
Some people converted the initial equation into an equation solely
in terms of t. If this is done, there is no need to use the
chain rule, which was the point of the problem.
Some people did more work than necessary when they tried to eliminate
t from the final answer or similar "simplification."
Prob 9:
The direction of maximum rate of change
is the directional derivative evaluated at a point, and the "rate of
change" which in the case of this problem is the "rate of ascent" is
the magnitude of the directional derivative.
The major problems had to do with (1) arithmetic and algebra, and
(2) remembering the mathematics for obtaining the "rate of ascent."
Prob 10:
Some people came up with a critical value of x as 0, even
though this value cannot be derived from 3-3x2=0.
When setting fy to zero, some people "cancelled
out" the "extra" y instead of factoring
to get 4y(y2-1)=0
thereby given the possible values of y as 0, -1, 1.
There are,
thus, 6 critical points: (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1).
Some people had the incorrect derivative of fyy which
should be -4 + 12 y2 (notice that y is squared).
Prob 11:
The problem asked both for the (1) locations of max/min values, and (2)
the "corresponding values." A number of people found the locations,
but did not indicate the "corresponding values."
Many people merely seemed to guess that one value of lambda was 1 without
justification and without finding the alternative possible value of -1.
Some people did not make use of the "constraint" equation to eliminate
multiple possible values for x, y, and z.
final nfinal
250 68
250 68
245 67
244 66
244 66
243 66
241 65
239 65
235 63
234 63
228 61
228 61
227 61
225 60
225 60
212 56
197 51
196 51
188 48
187 48
187 48
182 46
181 46
179 45
161 39
156 37
149 35
138 31
134 30
132 29
132 29
108 21
36 5
MAXIMUM 250 100
x
x x x
x x x
x x x
x x x x
x x x x
x x x x x x
x x x x x x x x x
20- 40- 60- 80- 100- 120- 140- 160- 180- 200- 220- 240-
39 59 79 99 119 139 159 179 199 219 239 250
(1) (1) (4) (2) (2) (7) (1) (8) (7)
This page is maintained by Dennis C. Smolarski, S.J.
Email: dsmolarski at scu.edu
© Copyright 2008 Dennis C. Smolarski, SJ, All rights reserved.
Last changed: 11 June 2008.