The University of British Columbia
Final Examination - June, 2015
Closed book examination
Time: 150 minutes
First
Last Name
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.
1
14
• Candidates are not permitted to ask questions of the invigilators,
except in cases of supposed errors or ambiguities in examination
questions.
2
12
3
12
4
16
5
10
6
9
7
9
8
9
9
9
Total
100
• 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 13 pages
[14] 1. Consider the points A = (1, 2, 3), B = (3, 2, 1), C = (2, 1, 3) and D = (1, 0, 0) in
space.
−→
−→
a) Find a vector that’s ortogonal to both AB and AC.
−→
−→
−→ −→
AB = h2, 0, −2i and AC = h1, −1, 0i so AB × AC = h−2, −2, −2i.
b) Compute the volume of the parallelepiped determined by the points A, B, C, D.
The volume is
−→ −→ −−→
|AB × AC · AD| = |h−2, −2, −2i · h0, −2, −3i| = 10
−−→
c) Is AD orthogonal to the plane determined by A, B, C?
−−→
−→ −→
The vectors AD = h0, −2, −3i and AB × AC = h−2, −2, −2i are not parallel (check:
there is no number c such that c · h−2, −2, −2i = h0, −2, −3i), so the answer is no.
Now consider the line with symmetric equations
L1 : x − 2 = 3(y − 1) = z
and the plane P with equation x + 2y + z = 0.
d) Find the point of intersection of L1 and the plane P , and find the (acute) angle that
L1 makes with the normal direction to the plane P .
[NOTE: you may leave the angle in terms of arccos]
From the symmetric equations of the line z = 3(y − 1) and x = 3(y − 1) + 2, so from
the plane equation
3(y − 1) + 2 + 2y + 3(y − 1) = 0
and that gives 8y = 4, so y = 12 . Then z = − 32 and x = 12 . The point of intersection is
1 1
, , − 32 .
2 2
A normal vector to the plane is ~n = h1, 2, 1i, and a direction vector of the line is
~v = h3, 1, 3i.
The angle is obtained through the dot product,
θ = arccos
~n · ~v
|~n||~v |
= arccos
8
√ √
6 · 19
.
e) Check that the point E = (2, 1, 0) is on the line L1 , and write parametric equations
for the line L2 that passes through E and is orthogonal to the plane P .
The point is on the line because in the point (2, 1, 0) the symmetric equations are
satisfied, 2 − 2 = 3(1 − 1) = 0. The parametric equations of the line L2 are


x = 2 + t
y = 1 + 2t


z=t
(initial point (2, 1, 0) and direction vector h1, 2, 1i).
[12] 2. Consider the function f (x, y) =
p
x(1 + y 2 ).
a) Write down an equation of the tangent plane to the graph of the function in the point
(x, y) = (1, 0).
The derivatives are fx = √1+y
2
fx (1, 0) =
1
2
2
x(1+y 2 )
and fy = √ 2xy
2
x(1+y 2 )
and in the point (1, 0) we get
and fy (1, 0) = 0. The equation of the tangent plane is
z − f (0, 1) = fx (1, 0)(x − 1) + fy (1, 0)(y − 0)
and that gives
1
z − 1 = (x − 1).
2
b) Estimate the error in computing f (x, y) if you know that x = 1±0.05 and y = 0±0.01.
Using the differential at the point (1, 0) we get
1
df = fx (1, 0)dx + fy (1, 0)dy = dx
2
so for the error we get
df =
1
· 0.05 = 0.025.
2
c) Does the function f (x, y) satisfy the partial differential equation fx = 3fy ? Justify
your answer.
Plugging the two derivatives in the equation gives
1 + y2
6xy
p
= p
2
2 x(1 + y )
2 x(1 + y 2 )
and for this to be true independently of the values of x and y we’d need 1 + y 2 = 6xy,
but those two functions are different. So the answer is no, f (x, y) does not satisfy the
differential equation.
d) Can you write an example of a partial differential
equation p(x, y) · fx = q(x, y) · fy
p
that is satisfied by the function f (x, y) = x(1 + y 2 ), where p(x, y) and q(x, y) are
two (non constantly 0) functions of x and y?
For example take p(x, y) = 2xy and q(x, y) = 1 + y 2 .
[12] 3. Consider the following portion of the contour plot of a differentiable function z =
f (x, y).
Give a brief explanation for your answers.
a) Which ones of the points A, B, C, D, E, F, G, H, I, J are critical points of f (x, y)? What
is the nature (local min, local max or saddle) of each of those?
The critical points are E, F, G. From the contour it’s clear that E and G are local
maxima (the level curves around them are for smaller values of k) and that F is a
saddle point (from there you can go towards level curves with higher k and towards
level curves with lower k).
b) Draw the gradient ∇f at the points J and D on the picture in the previous page.
[NOTE: these two vectors should have different length]
c) Is the function growing faster around the point E or around the point G?
The function is growing faster around G, because the level curves (for the same increments) are much closer one to the other.
d) What’s the sign of the directional derivative D~u f in direction of h−1, −1i at the point
H?
It’s positive, because the direction points to a level curve with a higher k.
e) Let R be the rectangle [0, 2] × [0, 2]. Estimate the double integral
ZZ
f (x, y)dA
R
using a double Riemann sum, by subdividing R into 4 equal squares and using A, B, C, D
as sample points (you may draw the subdivision on the picture).
The area of the subrectangles is 1, so the double Riemann sum is
f (A) + f (B) + f (C) + f (D) = 1 + 2 + 2 + 3 = 8.
[16] 4. A person is walking on the surface given by the graph of the function f (x, y) =
x3 y + y 2 + x (so from any point he can only move in directions that are tangent to the
surface), and is standing in the point (1, 0, 1).
a) Find all directions (horizontal, so a vector with 2 components, of length 1) from
(1, 0, 1) along which the rate of change of the altitude of the surface is equal to 1.
The gradient of the function f (x, y) is
∇f = h3x2 y + 1, x3 + 2yi
and in the point (1, 0) we get
∇f (1, 0) = h1, 1i.
So we are looking for unit vectors ha, bi such that
D~u f (1, 0) = ∇f (1, 0) · ~u = 1
and that gives a + b = 1. Moreover a2 + b2 = 1 since ~u is a unit vector, and solving
these two equations together gives the two vectors h1, 0i and h0, 1i.
b) Determine all values of a and b such that the tangent plane to the graph of the surface
at the point (a, b, f (a, b)) is parallel to the plane x + 3y − z = 6.
A normal vector to the tangent plane to the graph of f (x, y) at (a, b) is given by
hfx (a, b), fy (a, b), −1i = h3a2 b + 1, a3 + 2b, −1i. If this has to be parallel to h1, 3, −1i,
then
h3a2 b + 1, a3 + 2b, −1i = c · h1, 3, −1i
for some real number c. Looking at the third component we get c = 1, and then
(
3a2 b + 1 = 1
a3 + 2b = 3.
The first equation gives
that
either a = 0 or b = 0, and from the second equation we
√
3
3
get the points 0, 2 and ( 3, 0), that are the only solutions to the question.
c) The air temperature of a point (x, y, z) in space is given by T (x, y, z) = x2 +yz 2 +z −3.
What’s the direction in space (i.e. vector with 3 components, not necessarily of length
1) towards which the temperature rises in the fastest possible way at (1, 0, 1)?
Can the person hiking on the surface travel in that direction?
If the answer is no, would he have to fly or to dig into the mountain in order to follow
the direction of fastest ascent of T ?
The direction is given by the gradient of T in the point (1, 0, 1).
We have ∇T = h2x, z 2 , 2yz + 1i and ∇T (1, 0, 1) = h2, 1, 1i. The person can travel in
this direction only if it lies on the tangent plane to the surface at (1, 0, 1). This happens
exactly if it is orthogonal to a normal vector of said tangent plane, for example
hfx (1, 0), fy (1, 0), −1i = h1, 1, −1i.
We have h2, 1, 1i · h1, 1, −1i = 2 + 1 − 1 = 2, so the two vectors are not orthogonal, and
the person can’t travel in the direction of fastest ascent for T .
Since the dot product is positive, the vector h2, 1, 1i is “on the same side” of the vector
hfx (1, 0), fy (1, 0), −1i = h1, 1, −1i with respect to the tangent plane to the surface, and
the vector hfx (1, 0), fy (1, 0), −1i always points inside of the “mountain”, so the person
would have to dig.
d) If the person moves on the surface and at time t is in the point with coordinates
(1, t, 1 + t + t2 ), what’s the rate of change of the temperature that he experiences
when he passes through the point (1, 0, 1)?
Use the chain rule:
∂T dx ∂T dy ∂T dz
dT
=
+
+
dt
∂x dt
∂y dt
∂z dt
and (since the point (1, 0, 1) corresponds to t = 0)
dx
= 0
dt
dy
= 1
dt
dz
(0) = 1
dt
∂T
(1, 0, 1) = 1
∂y
∂T
(1, 0, 1) = 1
∂z
so
dT
(0) = 2.
dt
[10] 5. Consider the function f (x, y) = 4x2 − xy + y 2 + 3 on the closed and bounded region
D given by
2 y2
1
D = (x, y) x +
≤
.
4
4
a) Find and classify all critical points of f (x, y) in the interior of the region D.
The system for the critical points is
(
fx = 8x − y = 0
fy = −x + 2y = 0
and the only solution is (0, 0).
We have fxx = 8, fyy = 2 and fxy = −1, so D(f ) = 16 − 1 = 15 > 0 and thus the point
is a local minimum.
b) Determine absolute maximum and minimum values of the function f (x, y) on the region
D, and find all points where each of those values occurs.
We already did the interior analysis. The function value on the critical point is f (0, 0) =
3.
Let’s analyze the boundary with Lagrange multipliers: write the boundary as g(x, y) =
4x2 + y 2 = 1, and the system to solve is


8x − y = λ8x
−x + 2y = λ2y

 2
4x + y 2 = 1.
If x = 0 then y would be also 0 by the first equation, but this doesn’t work in the
constraint. If y = 0, symmetrically, from the second equation x is also 0, and that
doesn’t work again.
So we can assume x 6= 0 and y 6= 0 and write
λ=
8x − y
−x + 2y
=
8x
2y
and by cross-multiplying we get −2y 2 = −8x2 , so y 2 = 4x2 , and then y = ±2x.
Plugging this in the constraint we find 8x2 = 1 so x = ± 2√1 2 , and then y = ± √12 . So
1
1
√
√
the solutions are the four points ± 2 2 , ± 2 .
Since the constraint is 4x2 + y 2 = 1, the function on the points on this curve becomes
f (x, y) = 4 − xy (you could’ve used this to simplify the system), and on the four points
above we get
1
1
15
f ± √ , ±√
=
4
2 2
2
when the two signs are equal and
1
1
17
f ± √ , ∓√
=
4
2 2
2
when the two signs are different.
obtained
So the absolute minimum is3 = 12
4
in (0, 0) and
the absolute maximum is
obtained in the two points 2√1 2 , − √12 and − 2√1 2 , √12 .
17
4
[9] 6. Consider the iterated double integral
Z 1Z
1
2
ex dxdy
y
0
a) Sketch the region of integration.
The bounds give y ≤ x ≤ 1 and 0 ≤ y ≤ 1.
The region of integration is the triangle with vertices (0, 0), (1, 1) and (1, 0).
b) Exchange the order of integration.
Exchanging the order of integration gives
Z 1Z x
0
2
ex dydx.
0
c) Evaluate the integral.
We get
Z
x
2
2
ex dy = ex [y]x0 = xex
2
0
so
Z
1
Z
x
x2
Z
e dydx =
0
0
0
1
1 2
xe dx = ex
2
x2
1
0
1
= (e − 1)
2
[9] 7. Consider the solid region E in space that is bounded by the planes z = 0, z = 2 and
by the surface with equation z = ln(x2 + y 2 ).
a) Sketch the region E.
The region is a piece (bounded by the two horizontal planes) of the solid of revolution
obtained from rotating the graph of z = ln y 2 = 2 ln y for y > 0 (and the area “under
it”) around the z-axis.
b) Write down a single (i.e. not a sum of two or more) iterated triple integral that
gives the volume of E using cartesian coordinates and one that does that in cylindrical
coordinates.
The best choice is to use cylindrical coordinates, since we have rotational symmetry
around
the z-axis. In those coordinates the solid is given by 0 ≤ z ≤ 2 and 0 ≤ r ≤
√
z
z
e = e 2 (θ is free, so varies from 0 to 2π).
The triple integral is
Z
2π
Z
2
z
Z
e2
rdrdzdθ.
0
0
0
In cartesian coordinates, if we try to see the region as type 1 we have to distinguish
between the disk D1 where x2 + y 2 ≤ 1 (where the bottom of the surface is z = 0
and is flat), and the annulus D2 where 1 ≤ x2 + y 2 ≤ e2 , which corresponds to the
“sides” of the solid, and that doesn’t come out too well (writing an annulus in cartesian
coordinates is bad). Besides we are required to write down a single integral.
Let’s let z vary first, from 0 to 2. For a fixed z, we’ll have x2 + y 2 = ez , so y p
will vary
z
z
2
2
between
−e and e , and for fixed y and z, the variable x will vary from − ez − y 2
p
to ez − y 2 . The triple integral is
Z Z z Z √
2
0
ez −y 2
e2
z
−e 2
−
√
ez −y 2
dxdydz.
Switching the role of x and y also gives a correct answer.
If you really wanted to see the thing as a type 1 region, to write it as a single integral
you’d have to define a function h(x, y) to be ln(x2 + y 2 ) if 1 ≤ x2 + y 2 ≤ e2 and 0 if
0 ≤ x2 + y 2 ≤ 1, and then the volume is
Z e Z √e2 −x2 Z 2
dzdydx.
√
−e
− e2 −x2
h(x,y)
Technically this would also be correct (but not overly helpful for computations).
c) Compute the volume of E using one of the integrals that you wrote down for b).
Let’s use cylindrical coordinates. The triple integral is
2π
Z
0
Z
0
2
z
Z
0
e2
2
Z
rdrdzdθ = 2π
Z
=π
r2
2
0
2
e z2
dz
0
ez dz
0
= π(e2 − 1).
[9] 8.
a) The solid region E whose volume is given by the sum
Z
√1
2
− √1
2
Z
−x
√
− 1−x2
Z √1−x2 −y2
−
√
1−x2 −y 2
Z
dzdydx +
√1
2
− √1
2
Z
− √1
−
2
√
1−y 2
Z √1−x2 −y2
−
√
dzdxdy
1−x2 −y 2
is a type 1 solid region over a region D in the xy-plane. Sketch the region D, and
describe the solid E.
[NOTE: pay attention to the (different!) order of integration in the two integrals]
The bound for z in both integrals gives x2 + y 2 + z 2 ≤ 1, so we are inside the sphere
of radius 1.
√
The first integral gives − √12 ≤ x ≤ √12 and − 1 − x2 ≤ y ≤ −x (D1 in the sketch),
p
and the second integral gives − √12 ≤ y ≤ √12 and − 1 − y 2 ≤ x ≤ − √12 (D2 in the
sketch).
It’s easy to see from the sketch that the union of these two regions of the plane is the
portion of the unit disk x2 + y 2 ≤ 1 that is below the line y = −x (so it’s half of the
disk).
Thus the solid E is half of a solid sphere of radius 1 centered in the origin, cut by the
vertical plane y = −x.
b) Write the volume of a) as a single integral in spherical coordinates and evaluate it.
[NOTE: don’t use formulas you might know from solid geometry]
The description of the region in spherical coordinates is 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π and
3
π ≤ θ ≤ 74 π.
4
The triple integral turns out to be
Z
7
π
4
3
π
4
Z
0
π
Z
1
0
ρ3
ρ sin φ dρdφdθ = π ·
3
2
1
2
· [− cos φ]π0 = π.
3
0
This is of course half of the volume of a sphere of radius 1.
[9] 9. Consider a lamina in the region D of the xy-plane given by
D = {(x, y) | x2 + y 2 ≤ R2 }
where R is a positive real number. Assume that the lamina has density ρ(x, y) = f (x2 + y 2 ),
where f (t) is a continuous function of one variable, such that f (t) ≥ 0 for t ≥ 0.
Recall that the coordinates of the center of mass of the lamina (x, y) are
ZZ
1
x=
x · ρ(x, y)dA
m D
ZZ
1
y=
y · ρ(x, y)dA
m D
where
ZZ
m=
ρ(x, y)dA
D
is the total mass of the lamina.
a) Verify that, independently from the value of R and from the form of f (t) as above, the
center of mass of the lamina is in the origin (0, 0), by checking with a computation (in
either cartesian or polar coordinates) that
ZZ
x · ρ(x, y)dA = 0.
D
[NOTE: this will show that x = 0; don’t worry about computing y, the procedure is
the same]
You might need to use the primitive of the function f (t), let’s call it F (t); so F 0 (t) =
f (t) for all t.
The idea is that the density ρ(x, y) = f (x2 +y 2 ) only depends on r2 = x2 +y 2 (through
the function f (t)), so only on the distance of the point from the origin, and hence it’s
constant on circles centered in the origin. Therefore the “opposite” point of any point
on the lamina balances its “infinitesimal weight”, and the center of mass should be at
the origin.
Both ways are possible to formalize this idea and compute the integral.
• Cartesian: the double integral is
ZZ
x · f (x2 + y 2 )dA =
Z
Z √R2 −y2
R
−
−R
D
√
xf (x2 + y 2 )dxdy.
R2 −y 2
Call F (t) a primitive of f (t). Then for a fixed y the function of x given by
1
F (x2 + y 2 ) is a primitive of xf (x2 + y 2 ), so
2
Z √R2 −y2
√
1
1
R2 −y 2
2
2
2
2
xf (x + y )dx = [F (x + y )] √ 2 2 = (F (R2 ) − F (R2 )) = 0.
√
− R −y
2
2
− R2 −y 2
So the double integral that we have to compute is also 0, and then x = 0.
• Cartesian (alternative method): you can exploit the fact that
ρ(−x, y) = f ((−x)2 + y 2 ) = f (x2 + y 2 ) = ρ(x, y)
And
p reduce topthe example that I showed you
pin class: by splitting
p the interval
2
2
2
2
2
2
[− R − y , R − y ] into the union of [− R − y , 0] and [0, R2 − y 2 ] we
get
Z √
Z
Z √
R2 −y 2
−
√
R2 −y 2
0
xρ(x, y)dx =
R2 −y 2
√
−
xρ(x, y)dx +
R2 −y 2
R0
xρ(x, y)dx
0
R √R2 −y2
√
and to conclude, let’s check that − R2 −y2 xρ(x, y)dx = − 0
xρ(x, y)dx, so
that their sum is 0 (and then the integral in y of the function 0 will be 0 as well).
By using the substitution u = −x we get
Z 0
Z 0
xρ(x, y)dx = √
(−u)ρ(−u, y)(−du)
√
−
R2 −y 2
R2 −y 2
Z
0
= √
uρ(−u, y)du
R2 −y 2
Z √R2 −y2
=−
uρ(−u, y)du
0
and from ρ(−u, y) = ρ(u, y) we conclude
Z √
Z √R2 −y2
R2 −y 2
uρ(−u, y)du = −
−
uρ(u, y)du.
0
0
This is of course the same number as −
“dummy” variables.
R √R2 −y2
0
xρ(x, y)dx since u and x are
• Polar: by switching to polar coordinates the integral becomes
ZZ
Z 2π Z R
2
2
x · f (x + y )dA =
r cos θ · f (r2 )rdrdθ.
D
0
0
Now since we are in a “polar rectangle” (all bounds for r and θ are numbers) we
can integrate in θ first, and noting that r2 f (r2 ) doesn’t depend on θ, the integral
above becomes
Z
Z
R
2π
r2 f (r2 )
cos θdθdr
0
and
Z
0
2π
cos θdθ = [sin θ]2π
0 = 0
0
hence, again, the whole integral gives 0, and so x = 0.
b) Find an example of a non-constant function f (t) for which the total mass of the lamina
ZZ
ρ(x, y)dA
m=
D
is equal to 1.
[NOTE: the region D and the density ρ(x, y) are as in the previous page]
Let’s take f (t) = a · t, with a a number to determine. In this case the mass is
ZZ
ZZ
m=
2
ρ(x, y)dA =
D
Z
2π
Z
f (r )dA =
D
0
R
a · r2 · rdrdθ = 2πa ·
0
If we want to make this equal to 1, it suffices to take a =
The End
2
.
πR4
R4
πR4
=a·
.
4
2