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Concordia University February 16, 2011
MATH 380
Differential Geometry Solutions to the Midterm
(1) Consider
vp = (1, − 1 , 2) a tangent vector of R^3 at the point p = (0, 1 , −2),
ϕ = yz dx + y dy − x^2 dz a 1-form,
f = x^2 − yz a real-valued function,
V = U 1 + 2xyU 2 − z^2 U 3 a vector field,
F : R
3 → R 3 , F (u, v, w) = (u − v, u 2 , u − w) a mapping.
(a) (10 points) Compute vp[f ] and ∇vp (f V ).
Solution: vp[f ] =
∑^3
i=
∂f
∂xi
(p) (vp)i = 0 − 2 − 2 = − 4.
Recall that ∇vp (f V ) = vp[f ]V (p) + f (p)∇vp (V ).
We calculate ∇vp (V ) =
∑^3
i=
vp[Vi]Ui(p), where V =
∑^3
i=
ViUi. Since V 1 is the
constant function 1, we get vp[V 1 ] = 0. On the other hand, vp[V 2 ] = vp[2xy] = 2−0 = 2 and vp[V 3 ] = vp[−z] = −2(−2)2 =
Thus, ∇vp (f V ) = −4(U 1 (p) − 4 U 3 (p)) + 2(2U 2 (p) + 8U 3 (p)) = − 4 U 1 (p) + 4U 2 (p) + 32 U 3 (p).
(b) (10 points) Evaluate ϕ(vp) and dϕ.
Solution: ϕ(vp) = (−2) · 1 + 1 · (−1) − 0 · 2 = − 3. dϕ = z dy ∧ dx + y dz ∧ dx − 2 x dx ∧ dz = −z dx ∧ dy − (2x + y) dx ∧ dz.
(c) (8 points) Describe the image of the y-axis under the map F. Find F⋆(vp).
Solution: The y-axis is described by the equations: x = 0, z = 0, thus we consider F (0, y, 0) or F (0, v, 0) = (−v, 0 , 0) = v(− 1 , 0 , 0). Hence, the image of the y-axis under the map F is the x-axis (described from the positive to the negative direction, or we may say with the orientation reversed). By definition, F⋆(vp) = (vp[F 1 ], vp[F 2 ], vp[F 2 ]) ∈ TF (p)R^3 , where F = (F 1 , F 2 , F 3 ), hence each Fi : R 3 → R. Since F 1 (u, v, w) = u−v, we see that vp[F 1 ] = 1·1+(−1)· (−1) = 2 and, similarly we evaluate the directional derivatives of F 2 (u, v, w) = u 2
1
and F 3 (u, v, w) = u − w, concluding that F⋆(vp) = (2, 0 , −1) ∈ T(− 1 , 0 ,2)R^3. (One may also the Jacobian of F at p to calculate F⋆(vp).)
(d) Extra Credit (6 points): Calculate dϕ(vp, vp) and dϕ(vp, V(p)). Is the mapping F from part (c) regular? Explain.
Solution: dϕ(vp, vp) = 0 from the definition (which has the skew symmetry built in). See below the second calculation and you will understand.
The mapping F is not regular, as its Jacobian
2 u 0 0 1 0 − 1
(^) is singular for u = 0
(hence F⋆ is not injective).
The last one, dϕ(vp, V(p)), requires more calculations. Since dϕ = −z dx ∧ dy − (2x + y) dx ∧ dz, we need to find dx ∧ dy and dx ∧ dz on (vp, V(p)). Note that V(p) = U 1 (p) + 0 · U 2 (p) − 4 U 3 (p). As we have,
dx ∧ dy (vp, V(p)) = det
dx(vp) dy(vp) dx(V(p)) dyV(p)
= det
and
dx ∧ dz (vp, V(p)) = det
dx(vp) dz(vp) dx(V(p)) dzV(p)
= det
dϕ(vp, V(p)) = 2 − 1(−6) = 8. (Note that the above determinants will be zero if V (p) is replaced by vp which clarifies that dϕ(vp, vp) = 0.
(2) (8 points each)
(a) Show that a curve has constant speed if and only if its acceleration is everywhere orthogonal to its velocity.
Solution: (This is a homework problem from Assignment No. 3.) Let α(t), t ∈ I be a curve in R^3. Then
constant = ||α ′ (t)|| 2 =< α ′ (t), α ′ (t) > ⇔ 0 =< α ′′ (t), α ′ (t) >,
simply by differentiating the left-hand side, respectively integrating the right-hand side.
(b) Show that the curve α(t) = (cosh t, sinh t, t), t ∈ R has arc length function s(t) = √ 2 sinh t, and find the unit speed reparametrization of α.
Solution: Any unit speed reparametrization is found by s(t) =
∫ (^) t
a
∥α ′ (τ )∥ dτ
for some a in the domain of α. Calculating α ′ (t) = (sinh t, cosh t, 1), we see that
∥α ′ (t)∥ =
1 + sinh
2 t + cosh
2 t = √^1 2 (e t
2 cosh t. Choosing a = 0,
