MTH164s07 Sample Final Exam Time: Mon 05/07, 4 - 7pm Name:

MTH164s07
Sample Final
Exam Time: Mon 05/07, 4 - 7pm
Name:
Student No.:
Instructions:
• Answer ALL questions from Section A
• Answer ALL questions from Section B
• You may use a handwritten sheet of notes. Calculators are NOT permitted.
• Read all questions carefully
• Unless explicitly told otherwise, you should explain all your answers fully.
• Do NOT seperate the pages of your exam.
Problem
Points
A1
10
A2
10
A3
10
A4
10
B1
6
B2
6
B3
8
B4
6
B5
6
B6
6
B7
8
B8
6
B9
8
Total
100
Score
1
Name:
Section A: Answer ALL questions.
Problem A1: [10 pts]
Find the equation for the plane which
• contains the point P (2, 1, 0); and
• contains the line L described as the intersection between the two planes: x + z = −1 and 2x − y = 0.
Solution:
We have to find L.
First, let us find a point on the intersection of the two planes: x + z = −1 and 2x − y = 0.
This is up to us, but setting x = 1, y = 2x = 2 and z = −1 − x = −2 works: Q(1, 2, −2).
Next we find a direction vector !v for L.
Since L is in both planes, !v must be orthogonal to both of the normal vectors: !n1 = "1, 0, 1# and !n2 = "2, −1, 0#.
So,
!
!
!!i !j !k!
!
!
!v = !n1 × !n2 = !!1 0 1!! = "1, 2, −1# .
!2 −1 0!
Then !v is a vector in the plane.
We need another vector in the plane to get !n for the plane.
We have two points in the plane Q and P .
So another vector in the plane is
−→
w
! = QP = "2, 1, 0# − "1, 2, −2# = "1, −1, 2# .
Therefore, finally, a normal vector to the plane is
!
!
!!i !j
!k !
!
!
!n = !v × w
! = !!1 2 −1!! = "3, −3, −3# .
!1 −1 2 !
We can scale !n: we prefer to take !n = "1, −1, −1#, instead.
The equation for the plane is
!n · !r = !n · !r1 ,
where !r1 is the position vector for some point on the plane.
We take that point to be P .
So
"1, −1, −1# · "x, y, z# = "1, −1, −1# · "2, 1, 0#
x − y − z = 1.
2
Name:
Problem A2: [10 pts]
(a) Find the curvature and osculating plane to the curve !r(t) = "cos t, 2 sin t, 2t# at the point (0, 2, π).
Solution:
We first note that (0, 2, π) corresponds to the parameter value t =
!r# (t) = "− sin t, 2 cos t, 2#,
so at t = π/2 we have
Therefore the curvature is
Next compute
!r## (t) = "− cos t, −2 sin t, 0#
!r# (π/2) = "−1, 2, 2#,
κ(π/2) =
π
2.
!r## (π/2) = "0, −2, 0#
√
|"−1, 2, 2# × "0, −2, 0#|
|"4, 0, 2#|
20
=
=
|"−1, 2, 2#|3
27
27
Now the osculating plane is perpendicular to the binormal and
"4, 0, 2#
1
!
B(π/2)
=
= √ "4, 0, 2#.
|"4, 0, 2#|
20
Since we can rescale the normal vector to a plane, we can use "4, 0, 2# and so the osculating plane is
(!r − "0, 2, π#) · "4, 0, 2# = 0
or
4x + 2z = 2π.
(b) Consider the following planar curve. Answer the following questions. You do not need to justify your solutions.
• Put the points P , Q, R in order of increasing
curvature. Q,P,R
• At which of the points does the normal point
to the right? P,Q
• At which of the points does the binormal point
down into the page? P
P
Q
R
• True or False: There is a point of zero curvature somewhere between R and P . True
• True or False: There is a point of zero curvature somewhere between Q and R. False
3
Name:
Problem A3: [10 pts]
(a) Prove the following result:
Let !r(t) be a curve on a level surface to the function g(x, y, z). Then ∇g(!r(t)) · !r(t) = 0.
Solution:
d
g(!r(t)) = ∇g(!r(t)) · !r(t).
By the chain rule, dt
But if !r(t) is on the level surface g(x, y, z) = K then g(!r(t)) = K.
d
d
So dt
g(!r(t)) = dt
K = 0.
(b) Find the equation for the tangent plane to the surface 2z − xy = 0 at the point (x, y) = (2, 2).
Solution:
g(x, y, z) = 2z − xy. So the normal to the level surface at any point is
!n = ∇g(x, y, z) = "−y, −x, 2# .
At (x, y) = (2, 2), we have !n = "−2, −2, 2#. So the equation for the plane is
!n · ("x, y, z# − "x0 , y0 , z0 #) = 0
"−2, −2, 2# · ("x, y, z# − "2, 2, 2#) = 0
"−2, −2, 2# · "x − 2, y − 2, z − 2# = 0
−2(x − 2) + −2(y − 2) + 2(z − 2) = 0 .
This is simples as just x + y − z = 2.
(c) Using your solution to part (b), find an approximate solution of 12 x2 = 2.1.
Solution:
If z = 2.1 then we have x + y − 2.1 = 2 or x + y = 4.1. But for z =
x = y = 2.05. The answer is x ≈ 2.05.
4
1 2
2x
or 2z − x2 = 0, we have y = x. So
Name:
Problem A4: [10 pts]
Consider the function f (x, y) = 4xy − x4 − y 4
(a) Find and classify all the critical points of f (x, y).
Solution:
First we compute
∂f
= 4y − 4x3 ,
∂x
∂f
= 4x − 4y 3 ,
∂y
∂2f
= −12x2 ,
∂x2
∂2f
= −12y 2 ,
∂y 2
∂2f
=4
∂x∂y
Therefore to solve ∇f = !0 we must solve the system y = x3 , x = y 3 . Substituting, this reduces to x = x9 or
x(x8 − 1) = 0 which has solutions x = 0, 1, −1. Using y = x3 this means that the critical points are
(0, 0),
Now we find the discriminant
(1, 1),
(−1, −1)
D = fxx fyy − (fxy )2 = 144x2 y 2 − 16.
At (0, 0), D = −16 < 0 so (0, 0) is a SADDLE.
At (1, 1), D = 144 − 16 > 0 and fxx = −12 < 0 so (1, 1) is a LOCAL MAX.
At (−1, −1), D = 144 − 16 > 0 and fxx = −12 < 0 so (−1, −1) is a LOCAL MAX.
(b) The absolute maximum and minimum values that f (x, y) on the square 0 ≤ x ≤ 2 −2 ≤ y ≤ 2 and where they
occur.
Solution:
We must check the critical points that occur inside the rectangle and the boundary of the rectangle. Only the
critical points (0, 0) and (1, 1) occur inside. Now f (0, 0) = 0 and f (1, 1) = 2.
Along x = 2: f = 4y − 16 − y 4 . The derivative of this is 4 − 4y 3 which vanishes at y = 1. Thus we check (2, 1)
and the endpoints (2, 2) and (2, −2). Now f (2, 1) = −13, f (2, 2) = −24, f (2, −2) = −40.
Along x = 0: f = −y 4 . The derivative of this is 4y 3 which only vanishes at y = 0. Thus we check endpoints
and (0, 0). Now f (0, 2) = −16, f (0, −2) = −16 and f (0, 0) = 0.
Along y = 2: f = 4x − 16 − x4 . The derivative is 4 − 4x3 which only vanishes at x = −1. So we only check the
endpoints and (−1, 2). But f (0, 2) = −16, f (2, 2) = −24, f (−1, 2) = −21.
Along y = −2: f = −4x − 16 − 4x3 . The derivative is −4 − 4x3 which vanishes at x = −1, which is outside the
rectangle. Thus we must check endpoints only. But we’ve already done (0, −2) and (2, −2).
Comparing all these values. The absolute maximum is 2 and occurs at (1, 1). The absolute minimum is −40
and occurs at (2, −2).
5
Name:
Section B: Answer ALL questions.
Problem B1: [6 pts]
Integrate
""
xey
3/2
dA ,
D
where D is the domain
D = {(x, y) : 0 ≤ x ≤ 1 , x4 ≤ y ≤ 1} .
Solution:
D is a type I domain. But we can rewrite it as a type II domain:
D = {(x, y) : 0 ≤ y ≤ 1 , 0 ≤ x ≤ y 1/4 } .
Then
""
xey
3/2
dA =
"
#"
1
0
D
=
"
=
1
2
xey
3/2
"
#
1
$
dx dy
0
1
0
y 1/4
!x=y1/4 $
1 2 y3/2 !!
dy
x e
!
2
x=0
ey
3/2
y 1/2 dy .
0
Make the substitution
3 1/2
y dy .
2
Then, y1 = 0 and y2 = 1 correspond to u1 = 0 and u1 = 1. So
!1
"
"
1
1 1 y3/2 1/2
1 1 u2
1 !
e
y dy =
e du = eu !! = (e − 1) .
2 0
2 0
3
3 0
3
u = y 3/2 ,
du =
6
Name:
Problem B2: [6 pts]
Solution:
Find the surface area of the surface parametrized by
%
%
!r(z, θ) = " 1 − z 2 cos θ, 1 − z 2 sin θ, z#,
−1 ≤ z ≤ 1, 0 ≤ θ ≤ 2π.
First compute
!rz = "− √
and
Thus
%
%
!rθ = "− 1 − z 2 sin θ, 1 − z 2 cos θ, 0#
z
z
cos θ, − √
sin θ, 1#,
1 − z2
1 − z2
%
%
!rz × !rθ = "− 1 − z 2 cos θ, − 1 − z 2 sin θ, −z#,
"
S
dS =
"
1
−1
"
0
7
2π
|!rz × !rθ | =
dθdz = 4π.
%
1 − z2 + z2 = 1
Name:
Problem B3: [8 pts] Let F! = "xz + y 4 , x − tan z, z + y 2 # and S be the boundary of the region x2 + y 2 + z 2 ≤ 4,
z ≥ 0 oriented outwards. (i.e. the top hemisphere, together with its base).
&
!
(a) Find S F! · dS.
Solution:
The integral looks to be hard to compute directly, so let’s look for another way. Now
div(F! ) = z + 0 + 1 = z + 1
so the vector field is not a curl. However, because the divergence is so simple we can integrate the vector field across
another region with the same boundary and use the divergence theorem to compute the difference. In particular,
let’s set E to be the solid hemisphere x2 + y 2 + z 2 ≤ 4, z ≥ 0. Then ∂E = S. Now using spherical polar coordinates
"""
div(F! )dV =
E
"
"
2π
0
= 2π
π/2
0
π/2
"
2
(1 + r cos φ)r2 sin φ drdφdθ
0
' r3
3
0
"
"
sin φ +
(2
r4
sin φ cos φ
dφ
4
r=0
8
sin φ + 4 sin φ cos φ dφ
3
0
' 8
(π/2
= 2π − cos φ + 2 sin2 φ
0
) 3 *
8
28π
= 2π 2 +
=
3
3
= 2π
Now by the divergence theorem
"""
π/2
div(F! )dV =
E
Thus
(b) Now find
!
F! · dS
S
""
! = 28π
F! · dS
3
S
"
S1
Solution:
""
! for the open surface S1 which is just the hemisphere without the base.
F! · dS,
Set S2 to be the disc x2 + y 2 ≤ 4 oriented downwards. Then S = S1 + S2 . Using polar coordinates
""
""
!
!
F · dS =
"y 4 , x, y 2 # · (−!k)dxdy
S1
x2 +y 2 ≤4
=
""
x2 +y 2 ≤4
" 2π
2
=−
0
−y 2 dxdy =
sin θdθ ·
= −4π
Now
"
S1
!=
F! · dS
"
S
!−
F! · dS
"
"
2
"
0
2π
"
0
2
−r3 sin2 θdrdθ
r3 dr
0
! = 28π + 4π = 40π .
F! · dS
3
3
S2
8
Name:
4
Let F! = "y + sin x, z 2 + cos y, x3 # and C be the oriented curve parametrized by !r(t) =
3
"
"sin t, cos t, sin(2t)#. Find
F! · d!r
Problem B4: [6 pts]
C
Hint: recall that sin(2t) = 2 sin t cos t and use this to describe a surface that contains C.
Solution:
The path-integral looks hard to compute directly and
curl(F! ) = "−2z, −4x2 , −1#
so the vector field is not conservative. However we notice that if we set
S : z = 2xy,
then ∂S = C. Therefore by Stokes’ theorem
"
x2 + y 2 ≤ 1, oriented upwards
F! · d!r =
C
""
S
!
curl(F! ) · dS
!
Parametrize the surface by R(x,
y) = "x, y, 2xy# so
! x = "1, 0, 2y#,
R
! y = "0, 1, 2x#,
R
!x × R
! y = "−2y, −2x, 1#.
R
!x × R
! y is positive so this matches the orientation of S. Therefore
Note that the !k-component of R
"
""
!
!
F · d!r =
curl(F! ) · dS
C
S
""
=
"−4xy, −4x2 − 1# · "−2y, −2x, 1#dxdy
x2 +y 2 ≤1
""
=
x2 +y 2 ≤1
=
"
2π
0
=
"
=
0
,
8xy 2 + 8x3 − 1 dxdy
1
0
2π
0
"
"
+
"
0
2π
1
+
,
8r3 cos θ sin2 θ + 8r3 cos3 θ − 1 rdrdθ
8r4 cos θ − rdrdθ
1
8
cos θ − = −π
5
2
9
Name:
Problem B5: [6 pts]
Consider the vector field F! = "y 2 + 2xz, 2xy − z, x2 − y + 2z#
(a) Is F! conservative? If it is, find a potential function. If it isn’t, justify why.
Solution:
Now
curl(F! ) = "−1 − (−1), 2x − (2x), 2y − (2y)# = !0
everywhere, so the vector field is conservative. To find a potential function, note
"
y 2 + 2xzdx = y 2 x + x2 z + c1 (y, z)
"
2xy − zdy = xy 2 − zy + c2 (x, z)
"
x2 − y + 2zdz = x2 z − yz + z 2 + c3 (x, y)
Thus if we pick c1 (y, z) = z 2 − zy, c2 (x, z) = z 2 + x2 z and c3 (x, y) = y 2 x then these all match and a potential
function is
f (x, y, z) = xy 2 − zy + z 2 + x2 z
(b) Let C be the spiral path, parametrized by !r(t) = "cos(t), sin(t), t#, 0 ≤ t ≤ π. Find
Solution:
"
C
F! · d!r.
The start point of C is then (1, 0, 0) and the finish point (−1, 0, π). Since F! is conservative and we have an explicit
potential function f , by the fundamental theorem of line integrals
"
F! · d!r = f (−1, 0, π) − f (1, 0, 0) = (π 2 + π) − 0 = π(π − 1).
C
10
Name:
Problem B6: [6 pts]
The closed loop C parametrized by
!r(t) = "−1 cos(t) + cos(2π sin t), sin t#,
0 ≤ t ≤ 2π
is shown below. Find the area enclosed by the loop.
1.5
1
0.5
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
-0.5
-1
-1.5
Solution:
Using a Green’s theorem based method, we can use any of the following:
"
"
"
1
Area =
x dy − y dx =
x dy = −
y dx
2 C
C
C
Now
d!r = "dx, dy# = "− sin t − 2 sin(2π sin t) cos t, cos t# dt
so we probably don’t want to use dx if at all possible.
"
" 2π
Area =
xdy =
(−1 + cos(t) + cos(2π sin t)) cos t dt
=
C
2π
"
0
=π+
= π.
0
− cos t + cos2 t + cos(2π sin t) cos tdt
' 1
(2π
sin(2π sin t)
2π
0
11
Name:
Problem B7: [8 pts]
You are given the following information about a vector field F! = "P, Q#
(a) The vector field is defined and has continuous first order partial derivatives everywhere except (0, 0).
∂Q
∂P
=
everywhere except at (0, 0).
∂x
∂y
"
(c)
F! · d!r = 2 where C1 is the unit circle x2 + y 2 = 1 positively oriented.
(b)
C1
(d)
"
C2
F! · d!r = 4 where C2 is the straight line segment connecting (1, 0) and (3, 0).
Answer the following questions, justifying your solutions.
(a) Is F! conservative on R2 \(0, 0)?
Solution:
Item (2) means there is a closed loop such that the path integral of F! around that loop is not zero. This means the
vector field is not conservative everywhere.
[item (1) implies F! is conservative on any simply connected domain not including the origin, but this does not
imply conservative on all of R2 .]
Answer:
(b) If C is the loop (x − 4)2 + y 2 = 1 positively oriented. What is
Solution:
"
C
NO
F! · d!r?
The vector field is conservative inside this loop because of item (1). This implies the integral around the loop is
zero.
Answer:
(c) If C is the path from (3, 0) to (1, 0) going counterclockwise around (x − 2)2 + y 2 = 1, what is
Solution:
"
C
0
F! · d!r?
Again the vector field is conservative on the right half plane x > 0, so it is path-independent. Any line integral
from (1,0) to (3,0) that stays in this half plane will equal 4 by item (3). Going backwards along this path reverses
the sign.
Answer:
(d) If C is the oriented path from question B6, what is
"
C
Solution:
-4
F! · d!r?
This nasty loops winds around (0, 0) counterclockwise once. This means that on the region between this loop and
∂P
x2 + y 2 = 1, we have ∂Q
∂x − ∂y = 0. By Green’s theorem, the integral around the oriented boundary of this region
is zero. But this means the integral around the nasty loop and the loop from item (2) are the same.
Answer:
12
2
Name:
Problem B8: [6 pts]
(a) Find a parametrization for the algebraic curve
C :
x2/3 + y 2 = 1 .
Solution:
Rewriting the equation as
[x1/3 ]2 + y 2 = 1 ,
we guess the parametrization
[x(t)]1/3 = cos(t)
y(t) = sin(t)
0 ≤ t ≤ 2π .
(b) Find
x(t) = cos3 (t) ,
⇒
-
y dx .
C
where C is the oriented curve from part (a), positively oriented.
Solution:
Using the parametrization from part (a),
-
y dx =
"
2π
sin(t) 3 cos2 (t) (− sin(t)) dt
0
C
= −3
"
2π
[sin(t) cos(t)]2 dt
0
"
3 2π 2
= −
sin (2t) dt
4 0
3
= − π,
4
where we used two well-known facts:
sin(t) cos(t) =
and
"
b
sin2 (ωt) dt =
a
as long as (b − a)ω = 2πn for some integer n.
"
b
1
sin(2t) ,
2
cos2 (ωt) dt =
a
13
b−a
2
Name:
Problem B9: [8 pts]
A planar region is given by
D = {(x, y) : x2 + y 2 ≤ a2 and y ≥ 0} (and a > 0).
The density is constant ρ(x, y) = 1. Find the center of mass.
Solution:
Since ρ = 1, the mass just equals the area. Since D is half of the circle of radius a, we have
m = Area(D) =
1 2
πa .
2
By symmetry, we know x
¯ = 0. All that remains is to calculate y¯:
""
m y¯ =
y dA .
D
In polar coordinates,
D = {(r, θ) : 0 ≤ r ≤ a , 0 ≤ θ ≤ π} .
(Note that θ only goes up to π because we only have the upper half of the circle.)
Since y = r sin(θ) and dA = r dr dθ,
""
" a" π
y dA =
r sin(θ) · r dr dθ
D
0
0
" π
r2 dr
sin(θ) dθ
0
0
!π /
.
!
1
= a3 − cos(θ)!!
3
=
"
a
2
= a3 .
3
Therefore,
y¯ =
0
(2/3)a3
4a
.
=
2
(1/2)πa
3π
14