Download Parametrized Curve - Differential Geometry - Solved Exam and more Exams Computational Geometry in PDF only on Docsity!
Math 5378, Differential Geometry Solutions to practice questions for Test 1
- Find a parametrized curve whose trace is the set of points (x, y) in R^2
with xy = 1, x > 0.
Solution: One possible solution is α(t) = (t, 1 /t) for t ∈ (0, ∞). Re- member to specify the interval that your curve is parametrized on.
- Find the arc length of the curve
α(t) =
t sin t, t cos t,
t 3 / 2
between t = 0 and t = 1.
Solution: The velocity is
α ′ (t) = (sin t + t cos t, cos t − t sin t,
2 t 1 / 2 ),
and the speed simplifies to
‖ α ′ (t) ‖=
1 + t^2 + 2t = (1 + t).
Therefore, the length between t = 0 and t = 1 is ∫ (^1)
0
(1 + t)dt = 3/ 2.
- Show that the curve
α(t) = (sin t, t, − cos t)
has constant speed. Then find a reparametrization of this curve by arc length.
Solution: The velocity is
α(t) = (cos t, 1 , sin t)
and the speed simplifies to
‖ α ′ (t) ‖=
which is constant.
Both β 1 (s) = α(s/
- and β 2 (s) = α(−s/
- are reparametrizations by arc length.
- Give the Frenet formulas for the derivatives of the tangent, normal,
and binormal of a curve parametrized by arc length.
Solution: If t, n, and b are the tangent, normal, and binormal vectors, then the Frenet formulas say the following.
t ′ (s) = k(s)n(s)
n′(s) = −k(s)t(s) − τ (s)b(s)
b ′ (s) = τ (s)n(s)
Here k(s) is the curvature and τ (s) is the torsion. (Or you could write it in matrix form.)
- If α(s) is a curve parametrized by arc length, prove that α ′′ (s) is per- pendicular to α′(s).
Solution: If α is parametrized by arc length, then
1 =‖ α ′ (s) ‖ 2 = α ′ (s) · α ′ (s).
Differentiating this by using the Leibniz rule, we find
0 = α ′′ (s) · α ′ (s) + α ′ (s) · α ′′ (s),
or equivalently 0 = 2α ′′ (s) · α ′ (s).
Since these vectors have dot product equal to zero, they are perpendic- ular.
- Find all real numbers c so that the set
{(x, y, z) | x 2 − y 2
is a smooth surface in R^3.
Solution: This set is the set of zeros of the function
f (x, y, z) = x 2 − y 2
The gradient of this function is
∇f = [2x, − 2 y, 3 z 2 − 1],
Therefore, the length of the image is ∫ (^0)
− 1
e−^1 dt =
e−^1.
(Note a correction; I forgot the square root in the original version of this solution set.)
- With the same coordinate chart as the previous problem, find the area of the image of the entire region under x.
Solution: The area of the image is the integral ∫ (^) π
−π
−∞
EG − F 2 du dv =
∫ (^) π
−π
−∞
e u du dv
∫ (^) π
−π
dv
= 2 π.
- Consider the coordinate chart
x(u, v) = (u, u 2
Find a field N of unit normal vectors for this coordinate chart.
Solution: The tangent vectors to the surface are xu = (1, 2 u, 0) and xv = (0, 2 v, −1), and so a unit normal vector field is given by
xu × xv
‖ xu × xv ‖
(− 2 u, 1 , 2 v)
‖ (− 2 u, 1 , 2 v) ‖
− 2 u √ 4 u^2 + 4v^2 + 1
4 u^2 + 4v^2 + 1
2 v √ 4 u^2 + 4v^2 + 1
You could equally well have the negative of this vector field instead.
- Given a surface S with unit normal vector field N, give the mathemat-
ical definition of the second fundamental form IIp(v) for a vector v in the tangent space Tp(S).
Solution: The second fundamental form is
IIp(v) = −〈dNpv, v〉,
where dNp is the differential of the function N at the point p.
- Prove that a point p of a smooth surface is umbilical if and only if the
Gaussian curvature K and the mean curvature H satisfy H^2 = K.
Solution: Recall that a point is umbilical if and and only if the two principal curvatures k 1 and k 2 are equal.
If k 1 = k 2 , then K = k 1 k 2 = k^21 and H = k^1 +k^2 2 = k 1 , so H^2 = K.
Conversely, if H^2 = K, then
k 1 k 2 =
k 1 + k 2
2
k 2 1 +
k 1 k 2 +
k 2
Subtracting k 1 k 2 from both sides, we get
k 2 1 −^
k 1 k 2 +
k 2 2
k 1 − k 2
2
If this number squares to zero, then it must be equal to zero, so k 1 = k 2.
Note, it is important to show both directions:
- if the principal curvatures are equal, then K = H^2 , and
- if K = H 2 , then the principal curvatures are equal.
