18.02 Exam 4 Solutions
Problem 1.
/2 1 1
0
0
r2 r dz dr d.
0
Problem 2.
a) sphere: = 2a cos .
b) plane: = a sec .
2 /4 2a cos
c)
2 sin d d d.
0
0
a sec
Problem 3.
a) y (2xy + z 3 ) = 2x =
2
z (x
+ 2yz ) = 2y =
2
3
2
2
x (x + 2yz );
z (2xy + z ) = 3z
18.02 Exam 1 Solutions
Problem 1.
a) P = (1, 0, 0), Q = (0, 2, 0) and R = (0, 0, 3). Therefore QP = 2 and QR = 2 + 3k.
j
j
QP QR
1, 2, 0 0, 2, 3
4
b) cos = =
=
2 + 22 22 + 32
65
1
QP QP
Problem 2.
a) P Q = 1, 2, 0, P R = 1, 0, 3.
j
k
P Q P R = 1 2 0
18.02
Final Exam Solutions
Problem 1.
a) Line L has direction vector v = 1, 2, 3 which lies in P .
To get a point P0 on L take t = 0 P0 = (1, 1, 2).
P0 Q = 1, 1, 2 1, 1, 2 = 2, 0, 0 also lies in P .
A normal to P is
i j k
n = v
P0 Q = 1 2 3 = i(0)
18.02 Practice Exam 4 A Solutions
Problem 1.
a)
My = ex z = Nx
M = ex A - S
18.02 PracticezExamy4= Px
olutions
NZ = ex + 2y = Py
b) We begin with
fx = ex yz
fy = ex z + 2yz
fz = ex y + y 2 + 1
l b) We begin with
Integrating fx we get f = ex yz + g (y, z
18.02 Practice Exam 1 Solutions
Problem 1.
1
1
a) OQ = + + k; OR = + + k.
j
j
2
2
(1, 1, 1) ( 1 , 1, 1 )
22
OQ OR
2
2
b) cos =
=
=
.
3
|OQ | |OR |
3
3
2
Problem 2.
V
dR
V
Velocity: V
=
= (3 sin t, 3 cos t, 1).
dt
Problem 3.
V
Speed: |V | =
9 sin2 t
18.02 Practice nal-Solutions
Problem 1.
P : (1, 1, 1),
Q : (1, 2, 0),
R : (2, 2, 2)
i
PQ PR = 0
3
Plane: 2x 3y + 3z = 4 (substitute any of the pts. into
P Q =< 0, 1, 1 >, P R =< 3, 1, 3 >
Problem
1 0
2 c
1 1
10
2 2
1 1
j k
1 1
1 3
2x-3y+3z=d)
2.
= (2c
18.02 Exam 3 Solutions
(1,2)
y = 2x
x=1
1. a)
(1,1)
y = x
1 y
2 1
b)
dxdy +
0
dxdy.
y /2
1
y /2
(the rst integral corresponds to the bottom half 0 y 1, the second
integral to the top half 1 y 2.)
r sin
rdrd = sin drd.
r2
3
M=
dA =
sin drd =
2. a) dA
18.02 Practice Exam 3 A Solutions
1. a) The area of the triangle is 2, so y =
1
1
2
22y
y dxdy.
0
2y 2
b) By symmetry x = 0.
r2 rdrd =
2. = |x| = |r cos |. I0 =
D
2
1
/2
1
2
r
|r cos |rdrd = 4
0
0
/2
r4 cos drd = 4
0
0
0
1
4
cos d =
5
5
3. a) Nx = 6x2
18.02 Practice Exam 2 A Solutions
Problem 1.
a) f = (y 4x3 ) + x ; at P , f = 3, 1.
j
b) w 3 x + y .
Problem 2.
dh
h
.2.
a) By measuring, h = 100 for s 500, so
ds u
s
b) Q is the northernmost point on the curve h = 2200; the vertical distance between con