Solutions to Math 227 Final Exam, April, 2009
1. A curve in IR3 is given by the parametric equation x(t) = et , e−t ,
(a) Find the length of the curve between t = 0 and t = 1.
(b) Find the curvature at a general point x(t).
√ 2t .
Solution. (a) The velocity vector for this curve is
p
p
√ v(t) = et , −e−t , 2
=⇒ |v(t)| = e2t + e−2t + 2 = (et + e−t )2 = et + e−t
So the length of the curve is
Z
1
0
|v(t)| dt =
Z
0
1
1
t
e + e−t dt = et − e−t 0 = e1 − e−1 = 2 sinh 1
(b) The unit tangent vector at time t is
T̂ =
1
et +e−t
et , −e−t ,
√ 2
=⇒
√ t
t
−t
−e−t
et , e−t , 0 − [eet +e
, 2
−t ]2 e , −e
√ t
t
t
−t
t
−t
+e−t
−e−t
= [eet +e
, 0 − [eet +e
, 2
−t ]2 e , e
−t ]2 e , −e
√
= [et +e1−t ]2 2 , 2 , − 2[et − e−t ]
T̂′ =
1
et +e−t
The curvature is |T̂′ (t)|/|v(t)| which is
1
[et +e−t ]3
p
4 + 4 + 2[et − e−t ]2 =
1
[et +e−t ]3
p
2[et + e−t ]2 =
√
2
[et +e−t ]2
2. Let f (x, y) = xey − y 2 ex .
(a) Find the first– and second–order Taylor polynomials P1 (x, y) and P2 (x, y) at (0, 1). (It is not
necessary to simplify your answers.)
(b) Prove that f (x, y) − P1 (x, y), where P1 is the first–order Taylor polynomial from (a), has a local
maximum at (0, 1).
(Hints: (1) e ≈ 2.72, (2) you have done most of the necessary calculations in (a).)
Solution. (a) Solution 1. Using ez = 1 + z + 21 z 2 + O(z 3 ), we have
f (x, y) = xey − y 2 ex = exey−1 − ([y − 1] + 1)2 ex
= ex 1 + [y − 1] + O([y − 1]2 ) − ([y − 1] + 1)2 1 + x + 21 x2 + O(x3 )
= ex + ex[y − 1] − 1 − x − 12 x2 − 2[y − 1] − 2x[y − 1] − [y − 1]2 + O(x[y − 1]2 + x3 + x2 [y − 1])
so that
P1 = −1 + (e − 1)x − 2[y − 1]
P2 = −1 + (e − 1)x − 2[y − 1] − 21 x2 + (e − 2)x[y − 1] − [y − 1]2
(a) Solution 2.
f (x, y) = xey − y 2 ex
2 x
y
fx (x, y) = e − y e
y
fy (x, y) = xe − 2ye
2 x
fxx (x, y) = −y e
fxy (x, y) = ey − 2yex
y
fyy (x, y) = xe − 2e
1
x
x
f (0, 1) = −1
fx (0, 1) = e − 1
fx (0, 1) = −2
fxx (0, 1) = −1
fxy (0, 1) = e − 2
fyy (0, 1) = −2
so that
P1 = −1 + (e − 1)x − 2[y − 1]
P2 = −1 + (e − 1)x − 2[y − 1] − 21 x2 + (e − 2)x[y − 1] − [y − 1]2
(b) Set g(x, y) = f (x, y) − P1 (x, y). Then
gx (0, 1) = gy (0, 1) = 0
so that (0, 1) is a critical point for g(x, y). Using the results of part (a),
gxx (0, 1) = −1
gxy (0, 1) = e − 2
gyy (0, 1) = −2
As gxx (0, 1) gyy (0, 1) − gxy (0, 1)2 = 2 − (e − 2)2 > 2 − 1 > 0 we have that (0, 1) is either a local maximum
or a local minimum. As gxx (0, 1) < 0, it is a local maximum.
3. Find the minimum distance from the origin to the surface z(3x + 4y) = 20.
Solution. We are to minimize x2 + y 2 + z 2 subject to the constraint that z(3x + 4y) − 20 = 0. So define
L(x, y, z, λ) = x2 + y 2 + z 2 + λ(3xz + 4yz − 20). Then
0 = Lx = 2x + 3λz
=⇒
0 = Ly = 2y + 4λz
=⇒
0 = Lz = 2z + λ(3x + 4y)
=⇒
x = − 32 λz
y = −2λz
0 = 2z + λ − 29 λz − 8λz = z 2 −
0 = Lλ = 3xz + 4yz − 20
25 2
2 λ
As z cannot be 0, because then z(3x + 4y) would be 0 rather than 20, we must have λ = ± 52 and hence
x = ∓ 35 z and y = ∓ 45 z. Substituting into the last equation gives
∓ 95 z 2 ∓ 56 z 2 − 20 = 0
∓5z 2 − 20 = 0
=⇒
So we must take the lower sign and z = ±2. So the nearest points are ±
q
√
64
distance is 36
25 + 25 + 4 = 2 2.
6 8
5, 5, 2
and the minimum
4. Evaluate the following integrals:
RR
(a) D cos(x2 ) dA, where D is the triangle in the xy–plane with vertices (0, 0), (2, 0), (2, 2)
RRR
xzdV , where W is the bounded solid enclosed by the planes z = 0, z = 2, y = 0, y = x, and
(b)
W
the cylinder x2 + y 2 = 1.
Solution. The following figures show the domain of integration, D, for part (a) and the top view of the
domain of integration W , for part (b).
2
(a)
(b)
2
(a)
ZZ
cos(x2 ) dA =
2
dx
0
D
(b)
Z
ZZZ
W
xzdV =
Z
0
π
4
dθ
Z
x
dy cos(x2 ) =
0
Z 1
dr r
0
1
Z
2
dx x cos(x2 ) =
0
Z
2
dz (r cos θ)z =
0
Z
0
2
π
4
dθ
Z
0
1
1
2
2
sin(x2 ) =
0
1
2
sin 4
π 3 1
dr 2r2 cos θ = 2 sin θ 04 r3 0 =
√
2
3
5. Evaluate the line integrals below. (Use any methods you like.)
R
(a) X F · ds, where F = xi + 2yj + 4zk and X is the parametrized curve (cos t + sin t, cos t − sin t, t)
with 0 ≤ t ≤ 1.
(b) The (outward) flux of F(x, y) = (x3 + sin y)i + ex+y j across the boundary of the rectangle 0 ≤ x ≤ 1,
0 ≤ y ≤ 2 in the xy–plane.
x2
2
Solution. (a) Since F = ∇ ϕ with ϕ =
(cos 1 + sin 1 , cos 1 − sin 1 , 1),
Z
X
+ y 2 + 2z 2 , and X is a smooth curve from (1, 1, 0) to
(cos 1+sin 1 , cos 1−sin 1 , 1)
F · ds = ϕ
= 12 (cos 1 + sin 1)2 + (cos 1 − sin 1)2 + 2 −
(1,1,0)
=
3
2
cos2 1 − sin 1 cos 1 +
3
2
sin2 1 +
1
2
1
2
−1
= 2 − sin 1 cos 1
(b), by direct integration: Call the rectangle R. Then
Z
∂R
F · n̂ds =
1
Z
0
=−
Z
F(x, 0) · (−j) dx +
Z
1
=
Z
1
=
Z
0
0
2
1
Z
2
0
2
F(1, y) · (i) dy +
0
F(x, 0) · j dx +
− F(x, 0) + F(x, 2) · j dx +
(e2 − 1)ex dx +
Z
0
F(1, y) · i dy +
Z
0
2
Z
0
1
F(1 − t, 2) · (j) dt +
1
Z
0
F(x, 2) · j dx −
Z
0
Z
0
2
F(0, 2 − t) · (−i) dt
2
F(0, y) · i dy
2
− F(0, y) + F(1, y) · i dy
dy
0
= (e − 1)(e − 1) + 2
(b), by Green’s Theorem: In a problem set, you used Green’s Theorem to show that
ZZ
R
∇ · F dA =
I
C
F · n̂ ds
where C is the oriented boundary of the plane domain R, F = F1 ı̂ + F2 ̂, n̂ is the outward normal to C
and s is the arclength along C. For the current problem, this gives
ZZ
I
1 1 2
2
3x + ex+y dxdy = 2 x3 0 + ex 0 ey 0 = 2 + (e − 1)(e2 − 1)
F · n̂ ds =
[0,1]×[0,2]
C
6. (a) Find a function f (x, y) such that F = ∇f , where F(x, y) = (x2 + y 2 )i + 2xyj.
R
(b) Evaluate C F · ds, where C is any oriented piecewise C 1 curve from (1, 2) to (3, 4) and F is the
vector field in (a).
(c) Let F = ∇f be a conservative vector field (not necessarily the same as in (a)–(b)), and let x(t) be
d
f x(t) ≥ 0.
a flow line of F. Prove that dt
Solution. (a) We need
∂f
∂x (x, y)
∂f
∂y (x, y)
= x2 + y 2
=⇒
= 2xy + g ′ (y) = 2xy
=⇒
So, for any constant C, f (x, y) =
x3
3
+ y 2 x + C works.
3
f (x, y) =
x3
3
+ y 2 x + g(y)
g(y) = C, a constant
(b) Since F = ∇f ,
Z
C
F · ds = f (3, 4) − f (1, 2) = [9 + 48] − [ 31 + 4] = 52 32
(c) Since x(t) is a flow line
x′ (t) = F x(t)
d
dt f
=⇒
x(t) = (∇
∇f ) x(t) · x′ (t) = F x(t) · F x(t) ≥ 0
7. Let F = (x + z)i + (y + 2z)j + (2x + 3y)k. What are the possible values of
radius r contained in a plane x + 3y − z = a?
R
C
F · ds, if C is a circle of
Solution. Observe that a unit normal vector to the plane x + 3y − z = a is n̂ =
ı̂
̂
k̂
∂
∂
∂
= ı̂ − ̂
∇ × F = det dx
dy
dz
x + z y + 2z 2x + 3y
√1 (1, 3, −1)
11
and that
Denote by S the part of the plane inside C. There are two possible orientations for C. So, by Stokes’
theorem,
ZZ
ZZ
ZZ
Z
dS = ∓ √211 πr2
(1, −1, 0) · √111 1, 3, −1 dS = ∓ √211
∇ × F · n̂ dS = ±
F · ds = ±
C
S
S
S
8. Let X be the parametrized surface X(s, t) = (t cos s, t sin s, 2t), 0 ≤ s ≤
following integrals:
RR
(a) X z 2 dS,
RR
(b) X F · dS, if F = y 2 i.
π
2,
1 ≤ t ≤ 2. Evaluate the
Solution. As a preliminary computation, we find
∂X
∂s (s, t)
(a)
ZZ
2
z dS =
×
∂X
∂s (s, t)
∂X
∂t (s, t)
∂X
∂t (s, t)
ZZ
[0, π
2 ]×[1,2]
X
=
ZZ
[0, π
2 ]×[1,2]
(b)
π
2
= − t sin s, t cos s, 0
=
cos s,
sin s, 2
= 2t cos s, 2t sin s, −t
(2t)2 ∂X
∂s (s, t) ×
(2t)2 ×
√
∂X
∂t (s, t)
ds dt
5t ds dt
2
√
ds
dt 4 5t3
0
1
√ t4 2
√
√
π
15
= 2 × 4 5 4 1 = 2π 5 15
4 = 2 π 5
ZZ
ZZ
∂X
t2 sin2 s i · ∂X
F · dS =
∂s (s, t) × ∂t (s, t) dsdt
=
Z
=
ZZ
Z
[0, π
2 ]×[1,2]
X
2t3 sin2 s cos s dsdt
[0, π
2 ]×[1,2]
π
2
=2
Z
=2
=
1
3
sin2 s cos s ds
0
sin3 s
Z
1
π2 0
1 4
4t
5
2
4
2
1
2
t3 dt
9. Let ω = (x + z)dx ∧ dy + (y − x)dy ∧ dz.
(a) Compute dω. Simplify your answer.
RR
(b) Find X ω if X(s, t) = (t + s, t, s2 ), −1 ≤ s ≤ 1, 0 ≤ t ≤ 1.
Solution. (a) dω = dz ∧ dx ∧ dy − dx ∧ dy ∧ dz = 0
(b) Write
x(s, t) = t + s dx = ds + dt
y(s, t) = t
z(s, t) = s
dy = dt
2
dz = 2s ds
So
ZZ
ω=
ZZ
=
ZZ
=
Z
=
Z
=
Z
[−1,1]×[0,1]
X
[−1,1]×[0,1]
Z 1
1
ds
x(s, t) + z(s, t) dx(s, t) ∧ dy(s, t) + y(s, t) − x(s, t) dy(s, t) ∧ dz(s, t)
x(s, t) + z(s, t) − 2s y(s, t) − x(s, t) ds ∧ dt
dt
0
−1
1
ds
Z
1
0
−1
1
ds
−1
1
2
t + s + s2 − 2s − s
dt t + s + 3s2
+ s + 3s2
=1+0+2=3
5