Solutions for Exam 1 (Green Version)
Qinfeng Li
September 29
ANSWER KEY: BC EA AC AE DC BD
Solution for Problem 1. Since P~Q = Q − P = (3, 2, 0) − (1, 0, 0) = (2, 2, 0)
and P~R = R − P = (0, 2, 1) − (1, 0, 0) = (−1, 2, 1), we obtain one normal
candidate n as
i j
n = P~Q × P~R = 2 2
−1 2
k 0 = (2, −2, 6)
1
k (1, −1, 3)
Hence the equation must be x − y + 3z = d where d is a constant specified later.
By pluging (x, y, z) = P = (1, 0, 0) into the equation, we get d = 1. Therefore,
the equation for the plane containing P, Q and R is x − y + 3z = 1.
Solution for Problem 2. Clearly, n1 = (3, 0, −3), and n2 = (1, −1, 0), hence
cosθ =
This implies θ =
3
1
n1 · n2
= √ √ = .
|n1 ||n2 |
2
(3 2)( 2)
π
3.
Solution for Problem 3. The answer is −x2 + y 2 − z 2 = 1, since this is a
hyperboloid of 2 sheet, and no picture looks like a hyperboloid of 2 sheet. You
may check the table in your textbook Section 12.6.
Solution for Problem 4. Let’s first find the intersection of the curve and
surface by solving
√
(a, b, c)
= ( 2t, t2 + 1, 1 − 4t)
a2 + 2b − c = 0
Therefore,
2t2 + 2(t2 + 1) − (1 − 4t) = 0.
Doing a little algebra, we obtain
4t2 + 4t + 1 = 0,
i.e.
1
(2t + 1)2 = 0.
Hence t = − 12 . Therefore,
a=
√
√
2t = −
2
5
1
, b = t2 + 1 = + 1 = .
2
4
4
Therefore,
a2 + 2b =
1 5
+ = 3.
2 2
Solution for Problem 5. γ(t) = (t − sint, 1 − cost) ⇒ γ 0 (t) = (1 − cost, sint).
Therefore the length of the curve between 0 ≤ t ≤ π is
Z πp
Z π
√
(1 − cost)2 + sin2 t dt =
2 − 2cost dt.
0
0
Using the double angle formula 1 − cost = 2sin2 (t/2), we have
Z π
Z πp
π
2 · 2sin2 (t/2)dt =
length =
2sin(t/2)dt = −4cos(t/2) = 0−(−4) = 4.
0
0
0
Remark. In the above calculation,
we use the fact that when 0 ≤ t ≤ π,
p
sin(t/2) ≥ 0, and that’s why sin2 (t/2) = sin(t/2).
Solution for Problem 6.
γ(t) = (9cost, 9sint) ⇒ γ 0 (t) = (−9sint, 9cost), |γ 0 (t)| = 9
γ 0 (t)
= (−sint, cost)
⇒ T (t) = 0
|γ (t)|
⇒ T 0 (t) = (−cost, −sint), |T 0 (t)| = 1
Therefore, the curvature
κ(t) =
1
|T 0 (t)|
= .
0
|γ (t)|
9
In particular at t = π, the cuvature is also 91 .
Remark. Actually, the curvature of a circle is always 1r where r is the radius
of the circle. In this problem, the radius of the circle is clearly 9, hence the
curvature is 19 , no matter at what point on the circle.
3
Solution for Problem 7. Since a(t) = (2, −cost, 4√
), by finding the ant
√
3
tiderivative we obtain v(t) = (2t, −sint, 2 t) + C1 . Since v(0) = (1, 0, 0),
C1 = (1, 0, 0), and thus
v(t) = (2t, −sint,
3√
3√
t) + (1, 0, 0) = (2t + 1, −sint,
t).
2
2
Consequently,
γ(t) = (t2 + t, cost, t3/2 ) + C2 ,
2
thus γ(0) = (0, 1, 0) + C2 . Since we are given that γ(0) = (0, 1, 1), we obtain
C2 = γ(0) − (0, 1, 0) = (0, 1, 1) − (0, 1, 0) = (0, 0, 1). So,
γ(t) = (t2 + t, cost, t3/2 ) + (0, 0, 1) = (t2 + t, cost, t3/2 + 1),
thus
γ(π) = (π 2 + π, cosπ, π 3/2 ) = (π 2 + π, −1, π 3/2 + 1).
Solution for Problem 8. One can easily see that function f (x, y) =
is a continuous function near (0, 0), hence
lim
2+xy 3
1+x2 −y 2
f (x, y) = f (0, 0) = 2.
(x,y)→(0,0)
Solution for Problem 9. By the chain rule,
∂u ∂v ∂w ,
)
Px (x, y) = ∇P ·( ,
.
∂x ∂y ∂x (x,y)
(u,v,w)
√
∂v
. Since v = yex , ∂x
= yex .
Since P = u2 + v 2 + w2 , ∇P = √u(u,v,w)
2 +v 2 +w 2
0
Hence at (x, y) = (0, 1), by hypothesis we have (u, v, w) = (0, 1 · e , 2) = (0, 1, 2),
∂v ∂w
and ( ∂u
∂x , ∂x , ∂x ) = (2, 1, 0). Therefore,
∂u ∂v ∂w (0, 1, 2)
1
Px (0, 1) = ∇P ·( ,
,
)
=√
·(2, 1, 0) = √ = 5−1/2 .
2
2
2
(0,1,2) ∂x ∂x ∂x (0,1)
5
0 +1 +2
√
2
2
2
Solution for Problem 10. Since A = πr r + h + πr ,
p
r
πrh
Ar = π r2 + h2 + πr √
+ 2πr and Ah = √
2
2
r +h
r2 + h2
√
When r = 3, h = 4, r2 + h2 = 5, hence Ar = 5π + 9π
5 + 6π = 12.8π and
Ah = 12π
=
2.4π.
Since
the
error
in
measurement
is
at most 1, we have
5
dr = dh = 1, and therefore, the error in total surface area can be approximated
by
dA = Ar dr + Ah dh = 12.8π + 2.4π = 15.2π.
Solution for Problem 11. Since R(p, q) = tan−1 (pq 2 ),
1
q2
∂R
2
=
·
q
=
,
∂p
1 + (pq 2 )2
1 + p2 q 4
and thus
∂2R
∂
q2
2q(1 + p2 q 4 ) − q 2 (4p2 q 3 )
2q − 2p2 q 5
=
(
)
=
=
.
∂q∂p
∂q 1 + p2 q 4
(1 + p2 q 4 )2
(1 + p2 q 4 )2
Solution for Problem 12. Since the the direction vector is from (3, −2) toward
v
√
= (−3,2)
. By direct calculation,
the origin, v = (−3, 2), and thus u = |v|
13
Tx = −
y−1
,
(x − 2)2
Ty =
1
,
x−2
hence at (3, −2), Tx = 3 and Ty = 1. So the directional derivative at (3, −2) is
√
∇T · u = (3, 1) · (−3,2)
= √−7
.
13
13
3