Download Final - UD Math

MATH 243 Final Exam, Winter 2014
Instructor: Longfei Li
February 7, 2014
Name:
Instructions:
• Check your booklet before you start. There should be 10 problems on 10 pages. There are 100
points in total.
• No partial credit will be given if appropriate work is not shown.
• Answer questions in the space provided. If you need more space for an answer, continue your
answer on the back of the page. Do NOT take pages apart from the booklet.
• Carefully work out each problem and clearly indicate your final answer to any problem.
• You may NOT use calculators, dictionaries, notes, or any other kinds of aids.
• The duration of this exam is 120 minutes.
• DISHONESTY WILL NOT BE TOLERATED. Cheating receives a Failing grade.
Formulas You Might Need:
The line segment from r0 to r1 is: r(t) = (1 − t)r0 + tr1 , 0 ≤ t ≤ 1
The distance from a point P1 (x1 , y1 , z1 ) to the plane ax + by + cz + d = 0 is D =
Z
Arc length:
b
|r0 (t)|dt, arc length function: s(t) =
a
Z
t
|ax1 + by1 + cz1 + d|
√
a2 + b2 + c2
|r0 (t)|dt
a
T(t) =
r0 (t)
,
|r0 (t)|
N(t) =
T0 (t)
,
|T0 (t)|
B(t) = T(t) × N(t)
dT |T0 (t)|
|r0 (t) × r00 (t)|
=
κ(t) = = 0
ds
|r (t)|
|r0 (t)|3
aT = v 0 =
r0 (t) · r00 (t)
|r0 (t) × r00 (t)|
2
,
a
=
κv
=
N
|r0 (t)|
|r0 (t)|
ZZ I
Green’s Theorem:
P dx + Qdy =
C
D
∂Q ∂P
−
dA
∂x
∂y
For spherical coordinates: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ and dV = ρ2 sin φ dρdθdφ
s
ZZ
Surface Area: A(S) =
1+
D
∂z
∂x
ZZ
2
+
∂z
∂y
2
ZZ
|ru × rv |dA
dA =
D
ZZ
Surface Integrals:
ZZ
f (r(u, v))|ru × rv | dA,
f (x, y, z)dS =
S
D
ZZ
ZZ
F · dS =
Divergence Theorem:
S
S
ZZZ
F · n dS =
S
divF dV
E
ZZ
F · dS =
F · (ru × rv ) dA
D
1. (a) Find the area of the parallelogram with vertices A(−2, 1), B(0, 4), C(4, 2), and D(2, −1)
(b) Find the volume of the parallelepiped determined by the vectors a =< 1, 2, 3 >,
b =< −1, 1, 2 > and c =< 2, 1, 4 >. [10 points]
Solution:
−−→
(a) The parallelogram ABCD is the same as the parallelogram determined by vectors AB and
−−→
AD. The area of the parallelogram is
−−→ −−→
S = |AB × AD|
−−→
−−→
Since cross product is defined only for 3D, we will consider vectors AB and AD in 3D. Thus,
−−→
−−→
AB =< 2, 3, 0 >, AD =< 4, −2, 0 >
So
i j k
−−→ −−→ AB × AD = 2 3 0 = −16k
4 −2 0 So the area is S = | − 16k| = 16
(b) The triple product is
1 2 3
1 2 −1 2
−1 1
− 2
a · (b × c) = −1 1 2 = 1 2 4 + 3 2 1 = 2 + 16 − 9 = 9
1 4
2 1 4
So
V = |a · (b × c)| = |9| = 9
So the volume is 9.
1
2. Find the area of the region inside the circle (x − 1)2 + y 2 = 1 and outside the circle x2 + y 2 = 1.
[10 points]
Solution:
In polar coordinates, we have
x = r cos θ, y = r sin θ
So the equation for the circle
(x − 1)2 + y 2 = 1 ⇒ x2 + y 2 = 2x ⇒ r2 = 2r cos θ ⇒ r = 2 cos θ
And the equation for the circle x2 + y 2 = 1 is, in polar coordinates, r = 1. The intersection
happens when 2 cos θ = 1 ⇒ θ = ± π3 . So we have
−
π
π
≤ θ ≤ , 1 ≤ r ≤ 2 cos θ
3
3
So
Z
A(D) =
π
3
− π3
Z
2 cos θ
Z
r drdθ =
1
π
3
− π3
√
Z π
Z π
3
3 1
3
1 2 2 cos θ
1
π
2
r 2 cos θ− dθ =
+cos 2θ dθ = +
dθ =
π
π 2
2 1
2
3
2
−
−
3
2
3
3. Determine whether L1 : x = y = z, L2 : x + 1 =
skew, find the distance between them. [12 points]
z
y
= are parallel, skew, or intersecting. If
2
3
Solution:
v1 =< 1, 1, 1 > and v2 =< 1, 2, 3 > are the directions of L1 and L2 . They are not scalar multiple
of each other, thus not parallel.
The parametric equations of L1 and L2 are, respectively,
L1 : x = t, y = t, z = t,
L2 : x = −1 + s, y = 2s, z = 3s
If they are intersecting, then for some t and s we have
t = −1 + s
(1)
t = 2s
(2)
t = 3s
(3)
Solve (2) and (3) to get t = s = 0, however, this pair of values do not satisfy (1). Therefore, L1
and L2 do not intersect. Hence, they must be skew.
We know two skew lines determine two parallel planes. One common normal vector of the parallel
planes is given by
n = v1 × v2 =< 1, −2, 1 >
So the plane P1 which contains L1 is x − 2y + z = 0, and we know the point (−1, 0, 0) is on L2 .
So the distance is
√
| − 1 − 2(0) + 0|
1
6
d= √
=√ =
6
1+4+1
6
3
4. Find the velocity and position vectors of a particle that has the given acceleration and the given
initial velocity and position a(t) = 2i + 6tj + 12t2 k, v(0) = i, r(0) = j − k. [10 points]
Solution:
Velocity:
Z
v(t) =
a(t)dt =< 2t, 3t2 , 4t3 > +c
Since v(0) =< 1, 0, 0 >, we have
< 0, 0, 0 > +c =< 1, 0, 0 >⇒ c =< 1, 0, 0 >
Therefore,
v(t) =< 2t + 1, 3t2 , 4t3 >
Position:
Z
r(t) =
v(t)dt =< t2 + t, t3 , t4 > +c
Initial position gives
r(0) =< 0, 1, −1 >= c
Therefore, the position function is
r(t) =< t2 + t, t3 + 1, t4 − 1 >
4
Z
F · dr, where F =< x + 2y, x2 > and C consists of the line segments
5. Evaluate the line integral
C
from (0, 0) to (2, 1) and from (2, 1) to (3, 0). [12 points]
Solution:
The equations for C1 and C2 are
1
C1 : y = x, 0 ≤ x ≤ 2
2
C2 : y = −x + 3, 2 ≤ x ≤ 3
So the line integral
Z
Z
F · dr =
C
(x + 2y) dx + x2 dy
ZC
=
Z
2
(x + 2y) dx + x dy +
C1
2
Z
=
0
(x + 2y) dx + x2 dy
C2
1
2x + x2 dx +
2
5
Z
2
3
−x + 6 − x2 dx =
5
2
6. A solid E lies within the cylinder x2 + y 2 = 1, below
p the plane z = 4 and above the paraboloid
2
2
z = 1 − x − y . The density function is ρ(x, y) = x2 + y 2 . Find the mass of E. [10 points]
Solution:
The mass of E is given by the triple integral:
ZZZ p
M=
x2 + y 2 dV
E
We use cylindrical coordinates:
x = r cos θ, y = r sin θ, z = z
Then,
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, 1 − r2 ≤ z ≤ 4
Therefore,
Z
1 Z 2π
Z
4
M=
0
Z
0
1
= 2π
0
So the mass of E is
√
r2 r dzdθdr =
1−r2
Z
0
1 Z 2π
r2 (4 − 1 + r2 ) dθdr
0
1 1 12π
3r2 + r4 dr = 2π(r3 + r5 ) =
5
5
0
12π
.
5
6
7. Find the absolute maximum and minimum values of f = x2 + y 2 − 2x on the set D, where D is
the closed triangular region with vertices (2, 0), (0, 2), and (0, −2). [12 points]
Solution:
Since f is continuous and D is closed and bounded, f has an absolute max and absolute min
either at the critical points or on the boundaries.
fx = 2x − 2 = 0 ⇒ x = 1, fy = 2y = 0 ⇒ y = 0
So (1, 0) is the critical point and f (1, 0) = −1.
On L1 , x = 0, −2 ≤ y ≤ 2: f (x, y) = f (0, y) = y 2 . The max of y 2 is obtained when y = ±2 and
min of y 2 is obtained when y = 0. So we have on L1 : f (0, 0) = 0, f (0, 2) = f (0, −2) = 4.
On L2 , y = x−2, 0 ≤ x ≤ 2: f (x, y) = f (x, x−2) = x2 +(x−2)2 −2x = 2x2 −6x+4 = 2(x− 23 )2 − 12 .
This is a quadratic which attains min when x = 32 , where f ( 32 , − 12 ) = − 21 and attains its max
when x = 0, where f (0, −2) = 4.
On L3 , y = −x + 2, 0 ≤ x ≤ 2: f (x, y) = f (x, −x + 2) = 2x2 − 6x + 4 = 2(x − 32 )2 − 12 . It attains
the min when x = 23 , where f ( 32 , 21 ) = − 12 and attains the max when x = 0, where f (0, 2) = 4
Compare all the values. The absolute max of f on D is f (0, ±2) = 4 and absolute min is
f (1, 0) = −1.
7
8. Evaluate the integral
Z
1Z
√
1−x2
Z √2−x2 −y2
√
0
0
xydzdydx
x2 +y 2
[12 points]
Solution:
p
The region E for this integral is the region above the cone z = x2 + y 2 and below the sphere
x2 + y 2 + z 2 = 2 in the first octant. So in spherical coordinates, we have
0≤ρ≤
√
And the integral becomes
Z Z √
Z √
1−x2
1
0
√
2−x2 −y 2
√
0
π
π
, 0≤φ≤
2
4
2, 0 ≤ θ ≤
Z
π
2
2Z
xydzdydx =
x2 +y 2
0
√
Z
0
Z
0
2
√
4 2−5
=
15
8
ρ sin φ cos θρ sin φ sin θρ2 sin φ dφdθdρ
0
ρ4 dρ
=
π
4
! Z
0
π
2
! Z
cos θ sin θ dθ
0
π
4
!
sin3 φ dφ
ZZ
F · dS, where F =< 3xy 2 , xez , z 3 > and S is the surface of the
9. Evaluate the surface integral
S
solid bounded by the cylinder y 2 + z 2 = 1 and the planes x = −1 and x = 2. [12 points].
Solution:
The surface S is a closed surface, so we are able to use divergence theorem to evaluate the surface
integral:
ZZZ
ZZ
F · dS =
divF dV
E
S
where E is the region enclosed by S. Since E is a circular cylinder with axis being the x−axis.
So we use cylindrical coordinates to evaluate the triple integral.
y = r cos θ, z = r sin θ, x = x
with
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π,
−1≤x≤2
In addition,
divF = 3y 2 + 0 + 3z 2 = 3r2
Therefore the integral equals:
ZZ
ZZZ
Z
F · dS =
divF dV =
S
E
0
9
1 Z 2π
0
Z
2
−1
3r2 · r dxdθdr =
9π
2
x2 y 2 z 2
10. (a) Show that the equation of the tangent plane to the ellipsoid 2 + 2 + 2 = R2 at P (x0 , y0 , z0 )
a
b
c
can be written as
xx0 yy0 zz0
+ 2 + 2 = R2
a2
b
c
√
x2 y 2
2
2
(b) Use the result of part (a) to find the tangent plane to
+ +z = 1 at the point (−1, 1,
).
4
4
2
[Extra 10 points]
Solution:
Let
F (x, y, z) =
x2 y 2 z 2
+ 2 + 2
a2
b
c
The ellipsoid can be regarded as the level surface of F , i.e. F (x, y, z) = R2 . In addition,
∇F (x, y, z) =<
2x 2y 2z
, ,
>
a2 b2 c2
At P (x0 , y0 , z0 )
∇F (x0 , y0 , z0 ) =<
2x0 2y0 2z0
,
,
>
a2 b2 c2
So the tangent plane to the ellipsoid at P is
2x0
2y0
2z0
(x − x0 ) + 2 (y − y0 ) + 2 (z − z0 ) = 0
a2
b
b
xx0 yy0 zz0
x20 y02 z02
+
+
=
+ 2 + 2
a2
b2
c2
a2
b
c
Since P (x0 , y0 , z0 ) is on the ellipsoid, we have
⇒
x20 y02 z02
+ 2 + 2 = R2
a2
b
c
Therefore, the tangent plane can be written as
xx0 yy0 zz0
+ 2 + 2 = R2
a2
b
c
√
2
). Apply the formula from (a), we know the equation of
(b) In the case, the point is (−1, 1,
2
the tangent plane is
√
x y
2
− + +
z=1
4 4
2
10