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Concordia University April 14, 2008
Differential Geometry MATH 380 Solutions to Final Exam Professor: Alina Stancu
(1) (15 points) Let α(t) = (cos^2 t, cos t sin t, sin t), t ∈ R.
(a) Calculate the curvature of α at the point corresponding to t =
π 2
Solution: By calculating the velocity of the curve α′(t) = (−2 cos t sin t, − sin^2 t + cos^2 t, cos t)
= (− sin(2t), cos(2t), cos t), we see immediately that the curve does not have constant (unit) speed. Hence, we will calculate its curvature at the point required with the formula:
k(α(π/2)) =
‖α′^ × α′′‖ ‖α′‖^3
t=π/ 2
Thus, we need also
α′′(t) = (−2 cos(2t), −2 sin(2t), − sin t),
and we substitute t = π/2 in both α′(t) and α′′(t) as it will simplify the calculation to do so before evaluating the cross product. We conclude that α′(π/2) = (0, − 1 , 0) and α′′(π/2) = (2, 0 , −1), and thus (α′^ × α′′)(π/2) = (1, 0 , 2). As ‖(α′^ × α′′)(π/2)‖ =
√ 5 and^ ‖α′(π/2)‖^ = 1, we have that^ k(α(π/2)) =
§ (b) Give the formula of a vector field defined along the curve which is orthogonal to α′(t) at each point α(t). 1
Solution: Since α′(t) is tangent to the curve at every point, it suffices to find the vector field N from the Frenet frame, or a multiple of it. Of course, the vector field B will also be orthogonal to T , thus orthogonal to α′(t) at each point along the curve, but it takes longer to find it. Recall that for a curve of speed v(t) = ‖α′(t)‖, we have
T =
α′(t) ‖α′(t)‖
, and T ′^ = kvN.
Therefore, it suffices to evaluate T ′(t) and the vector field so obtained will be orthogonal to α′^ at each point α(t) as T · T ′^ = T · (kvN ) = (kv) T · N = 0. From part (a),
T =
1 + cos^2 t
(− sin(2t), cos(2t), cos t),
thus
T ′^ = −
1 + cos^2 t
(2 cos(2t), 2 sin(2t), sin t)
(1 + cos^2 t)^3
(− sin(2t), cos(2t), cos t).
(No need to simplify the answer although that it is possible.) §
(2) (10 points) Let {T, N, B} be the Frenet frame along a unit- speed regular curve in R^3 with positive curvature. Addition- ally, let k and τ denote the curvature function, respectively, the torsion function of this curve. Express T · (T′^ × T′′) in terms of k and τ. Solution: This problem uses the Frenet equations for a unit speed curve.
T · (T ′^ × T ′′) = T · (kN × (kN )′) = T · (kN × (k′N + kN ′)) = T · ((kN × k′N ) + (kN × kN ′)) = T · [k^2 (N × N ′)] = k^2 T · [N × (−kT + τ B)] = k^2 T · (−kN × T + τ N × B) = k^2 T · (kB + τ T ) = k^2 τ,
(4) (10 points) Prove that an isometry F = T C carries the plane through p orthogonal to q 6 = 0 to the plane through F (p) orthogonal to C(q). Solution: The plane through p and orthogonal to q is the set of all x’s in R^3 such that (x − p) · q = 0. Hence (F (x)−F (p))·F?(q) = (C(x)+Ta −C(p)−Ta)·C(q) = C(x − p) · C(q) = (x − p) · q = 0. §
(5) (20 points) Let φ be the 1-form in R^2 defined by φ = (x + y) dx + xy dy.
(a) If {U 1 (p), U 2 (p)} is the natural orthonormal frame at p = (x, y, z), find φ(U 1 ) and φ(U 2 ).
(b) Write an expression for
α
φ, where α : [a, b] → R^2 is a regular curve defined by α(t) = α 1 (t)U 1 (α(t)) + α 2 (t)U 2 (α(t)).
(c) Compute
α
φ for α the unit circle centered at (0, 0).
Solution: (a)
φ(U 1 ) = (x + y)dx(U 1 ) + xydy(U 1 ) = (x + y) · 1 + xy · 0 = x + y
and φ(U 2 ) = (x+y)dx(U 2 )+xydy(U 2 ) = (x+y)·0+xy ·1 = xy. (b) We have ∫
α
φ =
∫ (^) b
a
φ(α′(t)) dt =
∫ (^) b
a
φ(α′ 1 (t)U 1 (α(t)) + α′ 2 (t)U 2 (α(t))) dt
∫ (^) b
a
[(α 1 (t) + α 2 (t))α′ 1 (t) + (α 1 (t)α 2 (t))α 2 ′(t)] dt.
(c) Directly from part (b), using α(t) = (cos t, sin t), t ∈ [0, 2 π], thus α′(t) = (− sin t, cos t), we have
∫
α
φ =
∫ (^2) π
0
[(cos t + sin t)(− sin t) + cos t sin t cos t] dt
∫ (^2) π
0
[
cos(2t) − 1 2
− sin t cos t + cos^2 t sin t
]
dt
[
sin(2t) 4
t 2
sin^2 t 2
cos^3 t 3
] 2 π
0
= −π.
(6) (15 points) Let x : R^2 → R^3 be the mapping x(u, v) = (u, (uv)^2 , v + v^3 ).
(a) Show that x is a proper patch and denote its image by M. (b) Find the equation of the tangent plane to the surface M at the point (2, 4 , 2). (c) Is M an orientable surface? Explain. Solution: (a) First we’ll show that x is injective. If x(u, v) = x(¯u, ¯v), we have immediately u = ¯u. Note that the function f : R → R, f (x) = x + x^3 has f ′(x) = 1 + 3x^2 > 0, hence f is strictly increasing, and thus injective. Therefore f (v) = f (¯v) ⇒ v = ¯v. Next, we’ll show that x is regular by calculating ‖xu × xv‖ = EG − F 2 > 0 and showing that is strictly positive. Note that xu = (1, 2(uv)v, 0), xv = (0, 2(uv)u, 1+3v^2 ). Thus
‖xu×xv‖^2 = ‖xu‖^2 ‖xv‖^2 −(xu·xv)^2 = [1+4(uv)^2 v^2 ]·[(1+3v^2 )^2 +4(uv)^2 u^2 ]−4(uv)^3
(1 + 3v^2 )^2 + 4(uv)^2 u^2 + 4(uv)^2 v^2 (1 + 3v^2 )^2 + 16(uv)^4 − 4(uv)^3
= (1+3v^2 )^2 +2(uv)^2 (u^2 +v^2 )+2(uv)^2 (u^2 +v^2 − 2 uv)+16(uv)^4 +4(uv)^2 v^2 (6v^2 +9v^4 )
= (1+3v^2 )^2 +2(uv)^2 (u^2 +v^2 )+2(uv)^2 (u−v)^2 +16(uv)^4 +4(uv)^2 v^2 (6v^2 +9v^4 ) > 0.
(b) Note that (2, 4 , 2) = x(2, 1). Hence the normal to M at (2, 4 , 2) is U (2, 4 , 2) = xu(2, 1) × xv(2, 1) = (1, 4 , 0) × (0, 8 , 4) = (16, − 4 , 8) = 4(4, − 1 , 2). Hence the equation of the tangent plane to M at (2, 4 , 2) is (x − 2 , y − 4 , z − 2) · (4, − 1 , 2) = 0 ⇔ 4 x − y + 2z − 8 = 0.
(c) The surface is orientable as it is the image is a single reg- ular patch and therefore there exists a continuous, globally de- fined, unit vector field, U (p) = U (x(u, v)) = xu(u, v)×xv(u, v). §
