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18.02 Multivariable Calculus
Fall 2007
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18.02 Practice Exam 1B Solutions
Problem 1.
−−→
−−→
a) P = (1, 0, 0), Q = (0, 2, 0) and R = (0, 0, 3). Therefore QP = ı̂ − 2ˆ and QR = −2̂ + 3k̂.
−→
−−→ −
h1, −2, 0i · h0, −2, 3i
4
QP · QR
√
=√
b) cos θ = −−→ −−→ = √
2
2
2
2
65
1 +2 2 +3
QP QP Problem 2.
−→
−−→
a) P Q = h−1, 2, 0i, P R = h−1, 0, 3i.
̂ k̂ −−→ −→ ı̂
ˆ
P Q × P R = −1 2 0 = 6ˆı + 3ˆ + 2k.
−1 0 3 1 −−→ −→ 1 p 2
1√
7
49 = .
6 + 32 + 22 =
Then area(Δ) = P Q × P R =
2
2
2
2
→
−
−−→ −→
b) A normal to the plane is given by N = P Q × P R = h6, 3, 2i. Hence the equation has the form
6x + 3y + 2z = d. Since P is on the plane d = 6 · 1 + 3 · 1 + 2 · 1 = 11. In conclusion the equation of the
plane is
6x + 3y + 2z = 11.
−
→
c) The line is parallel to h2 − 1, 2 − 2, 0 − 3i = h1, 0, −3i. Since N · h1, 0, −3i = 6 − 6 = 0, the line is
parallel to the plane.
Problem 3.
a)
�B
�
�
�
�
O
A
−→
−−→
OA = h10t, 0i and AB = hcos t, sin ti, hence
−−→ −→ −−→
OB = OA + AB = h10t + cos t, sin ti.
The rear bumper is reached at time t = π and the position of B is (10π−1, 0).
−
→
b) V = h10 − sin t, cos ti, thus
−
→
| V |2 = (10 − sin t)2 + cos2 t = 100 − 20 sin t + sin2 t + cos2 t = 101 − 20 sin t.
√
The speed is then given by 101 − 20 sin t. The speed is smallest when sin t is largest i.e. sin t = 1. It occurs
when t = π/2. At this time, the position of the bug is (5π, 1). The speed is largest when sin t is smallest;
that happens at the times t = 0 or π for which the position is then (0, 0) and (10π − 1, 0).
Problem 4.
a) |M | = −12.
b) a = −5, b = 7.



x
1
1
1
 −5
7
c)  y  =
12
z
7 −5
d~r
1 7
5
d)
.
=
, ,−
dt
12 12 12
  

4
0
t/12 + 1
−8   t  =  7t/12 − 2 
4
3
−5t/12 + 1
Problem 5.
→ →
−
−
→
a) N · −
r (t) = 6, where N = h4, −3, −2i.
−
→ →
b) We differentiate N · −
r (t) = 6:
→ d−
−
−
− −
→ →
→ d→
d−
d →
→
− →
r (t) = N · −
r (t) + N · →
r (t) = 0 · −
r (t) + N · −
r (t)
0=
N ·→
dt
dt
dt
dt
−
→
and hence N ⊥
→
d−
dt r (t).