Solution for Advanced Calculus 2 Mid-Term
• Here we only give brief outlines.
1. The cycloid is a curve
with parametric equation x = t − sin t, y = 1 − cos t. Show that the curvature
√
of the cycloid equals 1/ 8y, for y 6= 0.
Sol. First note that
r0 (t) = (1 − cos t, sin t, 0),
r00 (t) = = (sin t, cos t, 0).
Therefore, the curvature K is
| cos t − 1|
kr0 × r00 k
1
√ .
=p
3 =
kr0 k3
8y
2(1 − cos t)
2. Find the area of the region common to the regions bounded by r = −6 cos θ, and r = 2 − 2 cos θ.
Sol.
1
A=2×
2
Z
2π/3
π/2
1
(−6 cos θ) dθ + 2 ×
2
2
Z
π
(2 − 2 cos θ)2 dθ = 5π
2π/3
3. The thrust of an airplane’s engine produces a speed of 600 mph in still air. The plane is aimed in the
direction of (2,2,1) and the wind velocity is (10,20,0) mph. Find the velocity vector of the plane with respect
to the ground and find the speed.
Sol. The velocity vector of the plane is
2 1 1
600
, ,
+ (10, 20, 0) = (410, 420, 200),
3 3 3
and therefore the speed is
p
4102 + 4202 + 2002 .
4. Consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls,
it generates the cycloid
r(t) = b(ωt − sin ωt)i + b(1 − cos ωt)j
where ω is the constant angular velocity of the circle and b is the radius. Using this, find the maximum speed
of a point on the circumference of an automobile tire of radius 1 foot when the automobile is traveling at 55
miles per hour.
Sol. For a general cycloid, the velocity vector is
r0 (t) = (bω − bω cos ωt, bω sin t).
Therefore, the speed is
√
kr0 (t)k = bω 2 − 2 cos ωt.
From this, we see that the maximum speed is 2bω when ωt = π, 3π, . . .. The angular velocity w of the tire is
55 miles
(rads per hour).
1 ft
In sum, the max speed is 110 miles per hour.
5. Find the distance between the point (1,-2,4) and the line defined by the parametric equation: x = 2t,
y = t − 3, and z = 2t + 2.
Sol. Let P be a point (0, −3, 2), which is on the line. Then, the distant between Q = (1, −2, 4) and the line
is given by
−−→
kP Q × uk
,
kuk
where u is the direction vector for the line. Therefore, we obtain
√
5
k(1, 1, 2) × (2, 1, 2)k
D=
=
.
k(2, 1, 2)k
3
6. Sketch the graph of r = θ.
Sol. Omitted.
7. Find
x2 + y 2
(if it exists).
x2 y
(x,y)→(0,0)
lim
Sol. Since
lim(x,y)→(0,0)
2
x + y 2 x2 1
≥
x2 y x2 y ≥ |y| ,
x2 +y 2
x2 y
does not exist.
8. Find the length of the space curve
r(t) = (a cos t, a sin t, bt)
over the interval [0, 2π].
Sol. Since
r0 (t) = (−a sin t, a cos t, b),
we obtain
Z
L=
0
2π
p
a2 + b2 dt = 2π
p
a2 + b2 .