Problem Set 1
1. (Exercise 1.1, page 2) What is the parametrization of the line through (−1, 0, 5) and
(3, −1, −2)?
The direction vector is v = (−1, 0, 5) − (3, −1, −2) = (−4, 1, 7). A parametrization is
then
σ(t) = (3, −1, −2) + tv = (3 + 4t, −1 − t, −2 − 7t).
2. (Exercise 1.2, page 2) In R4 , what is the parametrization of the line through (−1, 6, 5, 0)
and (0, 1, −3, 9)?
This is no different from the previous exercise. The direction vector is v = (0, 1, −3, 9)−
(−1, 6, 5, 0) = (1, −5, −8, 9). A parametrization is then
σ(t) = (−1, 6, 5, 0) + tv = (−1 + t, 6 − 5t, 5 − 8t, 9t).
3. (Exercise 1.4, page 3) Newton asked the question, what is the curve which, when revolved
about an axis, gives a surface offering the least resistance to motion through a “rare”
fluid, (e.g. air)? Newton’s answer is the following curve
λ 1
λ
1
3 4
7
3
2
(x(t), y(t)) =
+ 2t + t ,
ln
+t + t − λ ,
2 t
2
t
4
8
where λ is a constant and x ≥ 2λ. Graph this curve and compute its velocity and
acceleration vectors. What angle does the curve make with the x-axis at the intersection
point (2λ, 0)?
1
First, the velocity and acceleration vectors are:
λ
1
1
0
2
3
v(t) = α (t) =
− 2 + 2 + 3t , − + 2t + 3t
2
t
t
λ 2
1
00
2
a(t) = α (t) =
+ 6t, 2 + 2 + 9t
2 t3
t
To find the angle the curve makes with the x-axis at the point (2λ, 0), we need to find
the tangent vector at that point and find the angle between the tangent vector and
the vector pointing along the x-axis. We need to know the value of t at which the
curve crosses the x-axis. To do this we need to solve the equation (x(t), y(t)) = (2λ, 0).
Setting x(t) = 2λ we get
λ 1
3
+ 2t + t
= 2λ
2 t
1
+ 2t + t3 = 4
t
t4 + 2t2 − 4t + 1 = 0, since t = 0 cannot be a solution
Solving this gives
t = 1q
√
13
1
8
1
t =
26 + 6 33 − p
√ −
3
3
3 26 + 6 33 3
q
√
13
1
4
1 1 √
t = −
26 + 6 33 + p
√ − + ı 3
3
6
3 26 + 6 33 3 2
q
√
4
1
1 1 √
13
26 + 6 33 + p
t = −
√ − − ı 3
3
6
3 26 + 6 33 3 2
!
1
8
26 + 6 33 + p
√
3 3 26 + 6 33
!
q
√
13
8
1
26 + 6 33 + p
√
3
3 3 26 + 6 33
1
3
q
3
√
or
t = 1, t ≈ 0.2955977425, t ≈ −0.6477988715 + 1.721433238 i, t ≈ −0.6477988715 − 1.721433238 i
Ignoring the complex solutions, we check the two real solutions in the second component
of the curve. This value of t must give us zero.
λ
1
3 4
7
2
y(t) =
ln
+t + t − λ
2
t
4
8
λ
3
7
ln(1) + 12 + 14 − λ
y(1) =
2
4
8
λ 7
7λ
−
=0
=
2 4
8
y(0.2955977425) = −.2190700304λ 6= 0
2
Thus, the curve passes through the point (2λ, 0) when t = 1. Now,
λ 1
λ
1
3 4
7
3
2
+ 2t + t ,
ln
+t + t − λ
α(t) =
2 t
2
t
4
8
λ
1
λ
1
α0 (t) =
− 2 + 2 + 3t2 ,
− + 2t + 3t3
2
t
2
t
0
α (1) = (2λ, 2λ)
(1)
(2)
(3)
Now, choose a vector in the positive x-direction, namely ı = (1, 0). The angle that the
curve makes with the x-axis can be found from the dot product.
α0 (1) · (1, 0) = |α0 (1)| |(1, 0)| cos(θ).
This gives us
√
2λ = 2λ 2 cos(θ)
√
cos(θ) = 1/ 2
θ = π/4.
4. (Exercise 1.5, page 9) suppose a circle of radius a sits on the x-axis making contact at
(0, 0). Let the circle roll along the positive x-axis. Show that the path α followed by the
point originally in contact with the x-axis is given by
α(t) = (a(t − sin(t)), a(1 − cos(t)))
where t is the angle formed by the (new) point of contact with the axis, the center and
the original point of contact. This curve is called a cycloid.
3
In the above figure, P is the original point of contact. We need to find the x- and
y-coordinates for P . The distance from the original point of contact to the new point
of contact is the distance that the circle has rolled. This is the arclength of the arc
from the old contact point to the new contact point. The length of an arc of a circle is
the radius times the angle subtended, in radians. Thus, s = at. So in the figure above
x + w = at or x = at − w. Consider the small triangle formed by cutting a right angle
out of the angle t. w is the leg adjacent to this angle, t − π/2, and let z denote the leg
opposite this angle. Then,
x = at − w
= at − a cos(t − π/2)
= a(t − sin(t))
y = a+z
= a + a sin(t − π/2)
= a(1 − cos(t)).
5. (Exercise 1.6, page 10) Consider a cycloid of the form
(x(t), y(t)) = (A + a(t − sin(t)), B − a(1 − cos(t))),
Graph this cycloid to see that it is an inverted form of the one found in the prevous
exercise. Suppose a unit mass particle starts at rest at a point (x̄, ȳ) on the cycloid
corresponding to an angle θ in the parametrization above. Under the influence of gravity
(and assuming no friction), show that, no matter what initial θ is chosen, it always
takes a time of
r
a
T =
π where g is the gravitational constant
g
for the particle to slide down to the bottom of the cycloid (i.e., t = π).
Potential energy is turned into kinetic energy by, v 2 /2 = g(ȳ − y). Time is distance
divided by speed, so
Z xbot p
1 + y02
1
√
dx
T =√
ȳ − y
2g x̄
Now
y0 =
dy
dy/dt
sin(t)
=
=−
dx
dx/dt
1 − cos(t)
and ȳ = B − a(1 − cos θ), y = B − a(1 − cos(t)), so
r Z πr
a
1 − cos t
T =
dt
g θ
cos θ − cos t
4
The claim is that this integral is
T =
r
a
g
" √
2 cos 2t
−2 arcsin √
1 + cos t
#π
θ
We need to check this antidifferentiation. We do so by differentiating the last function:
d
dt
r
a
g
d
dt
r
a
g
" √
#!
√
r
2 cos 2t
a
2 sin(t/2)
q
−2 arcsin √
=
p
g 1 + cos(θ) 1+cos(θ)−2 cos2 (t/2)
1 + cos θ
1+cos(θ)
√
r
a
2 sin(t/2)
p
=
g 1 + cos(θ) − 2 cos2 (t/2)
r
cos(t) + 1
cos(t/2) =
2
r
1 − cos(t)
sin(t/2) =
2
" √
#!
√
r
t
2 cos 2
a
1 − cos t
√
−2 arcsin √
=
g 1 + cos θ − 1 − cos t
1 + cos θ
r r
a
1 − cos t
=
g cos θ − cos t
Thus, we have a correct antiderivative. Now, we need to evaluate it:
#π
" √
r
2 cos 2t a
T =
−2 arcsin √
g
1 + cos t #!
"θ √
r
2 cos 2θ
a
=
−2 arcsin(0) + 2 arcsin √
g
1 + cos θ
√
r a
1 + cos θ
0 + 2 arcsin √
=
g
1 + cos θ
r
a
=
π
g
6. (Exercise 1.11, page 15) Identify the mystery curve
1
1
α(t) = √ cos t, sin t, √ cos t ,
2
2
Find α0 (t), α00 (t), and L(α).
5
0 ≤ t < 2π.
This curve is a circle, but not in the xy-plane. Note that |α(t)| = 1 for all t. Further,
note that for any value of t, the vector α(t) is perpendicular to the vector (−1, 0, 1),
since
1
1
α(t) · (−1, 0, 1) = − √ cos t + 0 + √ cos t = 0.
2
2
Thus, α(t) lies in the plane through the origin with normal vector (−1, 0, 1). The
equation of this plane is x = z. The curve α(t) is the circle of radius 1 in the plane
x = z centered at the point (0, 0, 0).
1
1
α (t) =
− √ sin t, cos t, − √ sin t
2
2
1
1
00
α (t) =
− √ cos t, − sin t, − √ cos t
2
2
Z 2π
Z 2π
L(α) =
|α0 (t)| dt =
dt = 2π
0
0
(4)
(5)
(6)
0
7. (Exercise 1.12, page 15) Parametrize a circle which is centered at the point (a, b).
The circle centered at the origin is parametrized by α(t) = (r cos t, r sin t). To center
it at the point (a, b), just move it over:
β(t) = (a + r cos t, b + r sin t).
8. (Exercise 1.13, page 15) Parametrize the ellipse
x2
a2
+
y2
b2
= 1.
The easiest way to do this is to note that if we let x = a cos t and y = b sin t, then
2
2
clearly xa2 + yb2 = 1.
6
In the above figure, the outer circle has radius a, the inner circle has radius b, and the
ray makes an angle of t with the x-axis. If we take the point whose x-coordinate is the
x-coordinate of the intersection of the ray with the outer circle and whose y-coordinate
is the y-coordinate of the intersection of the ray and the inner circle, we trace out an
ellipse. The parametrization of this ellipse will come from the parametrizations of the
two circles and will be:
β(t) = (a cos t, b sin t),
just as before.
7