The University of British Columbia
Midterm Examination - May 27, 2015
Closed book examination
Time: 60 minutes
Last Name
First
Student Number
Signature
MATH 200 253 (Circle one)
Special Instructions:
No memory aids, calculators, or electronic devices of any kind are allowed on the test.
Show all your work; little or no credit will be given for a numerical answer without the
correct accompanying work. Numerical answers should be left in calculator-ready form,
unless otherwise indicated. If you need more space than the space provided, use the back of
the previous page. Where boxes are provided for answers, put your final answers in them.
Rules governing examinations
• Each candidate must be prepared to produce, upon request, a
UBCcard for identification.
• Candidates are not permitted to ask questions of the invigilators,
except in cases of supposed errors or ambiguities in examination
questions.
• No candidate shall be permitted to enter the examination room
after the expiration of one-half hour from the scheduled starting
time, or to leave during the first half hour of the examination.
• Candidates suspected of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination
and shall be liable to disciplinary action.
(a) Having at the place of writing any books, papers
or memoranda, calculators, computers, sound or image players/recorders/transmitters (including telephones), or other memory aid devices, other than those authorized by the examiners.
(b) Speaking or communicating with other candidates.
(c) Purposely exposing written papers to the view of other candidates or imaging devices. The plea of accident or forgetfulness
shall not be received.
• Candidates must not destroy or mutilate any examination material; must hand in all examination papers; and must not take any
examination material from the examination room without permission of the invigilator.
Page 1 of 6 pages
1
10
2
8
3
6
4
7
5
9
Total
40
May 27, 2015
Math 200-253
Name:
Page 2 of 6 pages
[10] 1. Consider points A = (0, 1, 1), B = (2, 1, 0) and C = (3, 1, 1) in space, and call π the
plane that passes through A, B and C.
a) (3 marks) Find the area of the triangle with vertices the points A, B, C.
b) (3 marks) Find all vectors orthogonal to the plane π and of length 16.
−→
c) (2 marks) Determine all vectors ~v = hx, y, zi orthogonal to AB and such that
−→
AB × ~v = h0, 5, 0i.
d) (2 marks) Determine all values of the real parameter a for which the vector
ha3 , a2 − a, ea i
is parallel to the plane π.
Solution
−→
−→
−→ −→
a) AB = h2, 0, −1i and AC = h3, 0, 0i, so AB × AC = h0, −3, 0i. The area of the triangle is
half of the area of the parallelogram, thus it is | − 3|/2 = 3/2.
b) one orthogonal vector is −3~j because of the previous point. Thus all orthogonal vectors
to π are parallel to ~j, and those of length 16 are 16~j and −16~j.
c) we must have
−→
AB · ~v = 2x − z = 0
and
−→
AB × ~v = hy, −x − 2z, 2yi = h0, 5, 0i
so y = 0, z = 2x and −x − 2z = 5. The last two equations together give x = −1, so z = −2,
and the only solution is the vector ~v = h−1, 0, −2i
d) the given vector is parallel to π exactly if it is orthogonal to one of its normal vectors, for
example ~j = h0, 1, 0i. So we must have ha3 , a2 − a, ea i · h0, 1, 0i = a2 − a = 0, and thus a = 0
or a = 1.
May 27, 2015
Math 200-253
Name:
Page 3 of 6 pages
[8] 2. Consider the lines given by parametric equations




x = t + 1
x = −s
L1 : y = 3 + 2t
L2 : y = s + 1




z =1−t
z =4+s
a) (2 marks) Write down a direction vector for each of the two lines.
b) (2 marks) Show that the two lines are skew, i.e. they are not parallel and they do not
intersect.
c) (2 marks) There is a unique plane π that contains L1 and does not intersect L2 . Find
an equation of the plane π (a sketch of any two skew lines in space might help to find
a normal vector for the plane).
d) (2 marks) The distance between the two lines coincides with the distance between the
plane π and any point on the line L2 . Find this distance.
Solution
a) two direction vectors are ~v1 = h1, 2, −1i and ~v2 = h−1, 1, 1i
b) the direction vectors are not scalar multiples. Furthermore the system


t + 1 = −s
3 + 2t = s + 1


1−t=4+s
has no solutions in t, s as the first equation can be rewritten as t + s = −1 and the third as
t + s = −3, so the two lines don’t intersect.
c) a normal vector of π is obtained by taking ~v1 × ~v2 (this should be clear from the sketch).
We get
~n = h1, 2, −1i × h−1, 1, 1i = h3, 0, 3i
and we can take any point on L1 , for example for t = 0 we get P0 = (1, 3, 1). The plane π
has equation
3x + 3z = 6
that we can simplify to
x + z = 2.
One can check that in fact this doesn’t intersect L2 .
d) let us pick any point on L2 , for example P1 = (0, 1, 4) for s = 0, and compute the distance
between this point and the plane π using the formula
|ax1 + by1 + cz1 − d|
√
.
a2 + b 2 + c 2
We get
|0 + 4 − 2| √
√
= 2.
2
May 27, 2015
Math 200-253
Name:
Page 4 of 6 pages
[6] 3.
a) (2 marks) Describe and sketch the surface in space given by the equation x = ey , and
write down the coordinates of one point on the surface with x-coordinate equal to 5.
b) (2 marks) Given the quadric surface with equation x2 + y 2 + z = 5, find the equation
of one plane whose intersection with the surface is a circle of radius 1.
c) (2 marks) Sketch the quadric surface of point b).
Solution
a) cylinder with base the curve x = ey on the xy-plane. A point with x = 5 is (5, ln 5, 3275)
(any value of z works).
b) the plane with equation z = 4 gives x2 + y 2 = 1, as required.
c) horizontal sections are all circles and vertical sections are all concave-down parabolas.
The surface is a “circular paraboloid”.
May 27, 2015
Math 200-253
Name:
Page 5 of 6 pages
[7] 4. Consider the function f (x, y) = ln(x2 + y − 4).
a) (3 marks) Determine and sketch the domain of f .
b) (2 marks) Determine and sketch the k-level curve of f (x, y) for k = ln 4.
c) (2 marks) For what values of a does the point (a, 1, ln 100) lie on the graph of f (x, y)?
Solution
a) the domain is {x2 + y − 4 > 0}, so the region above the parabola y = 4 − x2 , excluding
the parabola.
b) the level curve is the concave-down parabola y = −x2 + 8.
c) the √point lies on the graph exactly if ln(a2 + 1 − 4) = ln 100, so a2 = 103, and finally
a = ± 103.
May 27, 2015
Math 200-253
Name:
Page 6 of 6 pages
[9] 5.
a) (3 marks) Find the first partial derivatives fx and fy of the function f (x, y) = x2 yex+y .
b) (3 marks) Find all second partial derivatives and check that fxy = fyx .
2y
c) (3 marks) For what values for the constant a does the 2 variable function f (x, y) = eax
satisfy the partial differential equation x3 · fx = 4y · fyy ?
Solution
Note that f (x, y) = x2 ex · yey .
a) we have fx = yey (2xex + x2 ex ) = yex+y (2x + x2 ) and fy = x2 ex (ey + yey ) = x2 ex+y (1 + y).
b) from a) we find
fxx = yey (2ex + 2xex + 2xex + x2 ex ) = yey (2ex + 4xex + x2 ex ) = yex+y (2 + 4x + x2 )
fyy = x2 ex (ey + ey + yey ) = x2 ex (2ey + yey ) = x2 ex+y (2 + y)
fxy = (2xex + x2 ex )(ey + yey ) = ex+y (2x + x2 )(1 + y)
fyx = (ey + yey )(2xex + x2 ex ) = ex+y (1 + y)(2x + x2 )
so in particular fxy = fyx .
2
2
2
c) we have fx = 2axyeax y , fy = ax2 eax y and fyy = a2 x4 eax y . So
2
2y
x3 · fx − 4y · fyy = 2ax4 yeax y − 4a2 x4 yeax
and this is zero exactly if 2a − 4a2 = 0, so a = 0 or a = 1/2.
The End
2y
= (2a − 4a2 )x4 yeax