Math 114 Quiz 4
Oct. 19, Wednesday
NAME:
Recitation: 8-9am
9-10am
−r (t) = (cos t cos α, sin t cos α, sin α), in which 0 ≤ t ≤ 2π is the
1. Consider the space curve given by →
parameter, 0 ≤ α < π is a fixed constant.
(a) Find a quadratic surface and a plane such that the curve is their intersection.
(b) Calculate the curve’s total length.
→
−
→
−
−r (t)?
(c) Calculate the curvature κ and the normal vector N. What’s the angle between N and →
Proof. (a) Because
(cos t cos α)2 + (sin t cos α)2 + (sin α)2 = cos2 α(cos2 t + sin2 t) + sin2 α = cos2 α + sin2 α = 1
so the curve lies on the sphere x2 + y2 + z2 = 1. The plane can simply be taken as z = sin α(since α is a
constant now). The curve is the intersection of the sphere x2 + y2 + z2 = 1 and plane z = sin α.
2
2
Note: The answer is not unique. The curve lies on cylinders, too, for example: cosX2 α + cosY 2 α = 1. The key
point is to understand β is just a constant in this problem, so feel free to use it in any coefficients in any way
as you like.
Actually, this is just a circle on a sphere; if you view the sphere to be our earth, then it is just a latitude –
there is one such curve right under our feet! α is just the latitude angle if we put the equator as the xy-plane
in our coordinate system. To see this, think about the geometric meaning of z-coordinate. What is the vertical
distance from a latitude with latitude angle α to the equator?
(b)
→
−v = (− sin t cos α, cos t cos α, 0)
−v | = |cosα|
|→
So the curve’s total length is
L=
Z
2π
−v | d t = 2π| cos α|
|→
0
Note: You can also solve geometrically: you know it is a latitude on a standard sphere with latitude angle
equal to α, so it is a circle of radius R = 1 · | cos α|, so the length is 2πR = 2π| cos α|.
(c) unit tangent vector
→
−v
→
−
= (− sin t, cos t, 0)
T = →
|−v |
→
−
dT
= (− cos t, − sin t, 0)
dt
→
−
dT
|
|=1
dt
→
−
1 dT
1
κ= →
|
|=
−
| cos α|
|v| d t
→
−
N=
→
−
dT
= (− cos t, − sin t, 0)
→
−
| dd Tt | d t
1
→
−
−r (t) is θ. Then
Assume the angle between N and →
→
− →
N · −r (t)
cos θ = →
= − cos α
− −
| N||→
r (t)|
Therefore θ = π − α
1
2
Note: You can also solve geometrically: it is a circle of radius | cos α| so the curvature is | cos1 α| . The normal
vector is always pointing towards the center Q of the circle, whose coordinate is (0, 0, sin α). So it is
−−→ →
OQ − −r (t)
(0, 0, sin α) − (cos t cos α, sin t cos α, sin α)
→
−
N = −−→
=
= (− cos t, − sin t, 0)
→
−
|(0,
0, sin α) − (cos t cos α, sin t cos α, sin α)|
|OQ − r (t)|
−r (t) is the complement of the latitude
Since the angle between the normal vector and the position vector →
angle, so it is π − α.
Comment This is a very good example to show the power of geometric intuition. With that in mind, the
solution will be very succint and need no real calculations at all.
Three common mistakes are:
1. One did not notice α is just a constant. When you make derivatives, you don’t make derivatives with
respect to α, but with respect to t. So treat α as if any your favorite number, 2, 3, or 2012. To tell which is
treated as a constant and which is as a variable is going to be very essential in the future as well. So make
sure you have got that!
→
−
−
−r (t). Notice that the frame is TNB frame, so →
2. One took for grant that N is perpendicular to →
N is always
→
−
−r (t). You should distinguish the position vector →
−r (t) and the
perpendicular to T , but not necessarily to →
−v (t).
tangent vector→
3. Although I have pointed out for times, there are still very few people putting an absolute value on what
you take out of a square root. In this example, α is between 0 and π. So it indeed can be negative, so the
absolute value cannot be omitted. If you get 2π cos α in your answer of total length, then when α is between
π
2 and π, you will get a curve with negative length, which is absolutely impossible.
I hope this example has helped you learn something!