### COURSE OF DIFFERENTIAL GEOMETRY

answers (partially)

100)
i) straight lines through 0.
ii) xu = (1, v, v2) and xv = (0, u, 2uv) are dependent if u=0. But this point is singular.
iii) The normal vector has the same or opposite direction as xuxv, or, equivalently, (v2, -2v, 1). Tangent plane v2x1 - 2v x2 + x3 = 0.
iv) The equation of the tangent plane is independent of u.

110) The first fundamental form is 5(du)2 + u2(dv)2, so a1 1 = 5, a1 2 = 0, a2 2 = u2.
From cos(φ) = (Σ ai j dai dbj)/(√(Σ ai j dai daj)√(Σ ai j dbi dbj)) with a1 = b2 = t2 en a2 = b1 = t3, follows, if t=1, cos(φ) = 36/(7√29).

111) xu = (1,2u,v) and xv = (1,2v,u), so a1 2 = 0 if uv = -1/5, so x = (v - 1/(5v), v2 + 1/(25v2), -1/5), so y - x2 = 2/5 and z = -1/5.

112) (ds)2 = (dθ)2 + sin2(θ) (dφ)2.
The surface element is then √ sin2(θ) dθ dφ.
So the required surface is 0 θ2θ1 sin(θ) dθ dφ = 2π (cos(θ2) - cos(θ1)).

113) Take the following parametrisation of the plane: (u√2) (cos(v/√2), sin(v/√2), 0). Then both surfaces have first fundamental form 2(du)2 + u2 (dv)2.

117) With φ(u,v) = u we find φu = 1, φv = 0 ; so the differential equation becomes:
du/dv = (a2 2φu - a1 2φv)/(a1 1φv - a1 2φu) = -a2 2/a1 2 = -(2+u2)/(uv).
We get u du/(2+u2) = -dv/v, so ln(2+u2) = -2 ln |v| + c. Hence it follows that v2(2+u2) = μ.

118) The meridians are determined by φ(θ,φ) = λ, with φ(θ,φ) = φ, so by φ = λ.
We get φθ = 0 and φφ = 1, so the differential equation becomes: cos(α) = (-a1 1dθ - a1 2dφ)/(√a1 1*√(a1 12 + 2 a1 2dθ dφ + a2 22).
With a1 1 = 1, a1 2 = 0 and a2 2 = sin2(θ) we find cos(α) = -dθ / √(dθ2 + sin2(θ) dφ2), so dθ/sin(θ) = +tg(α) dφ.
Then we get for the loxodromes: φ + c = + (1/2) cotg(α) ln(|(1+cos(θ))/(1-cos(θ))|).

127) f ' = 0, so f is constant; then we have a right helicoid surface. Or h=0, surface of revolution.

128) (ds)2 = (du)2 + (u2+h2) (dv)2 , so a1 1 = 1, a1 2 = 0, a2 2 = u2+h2.
φ(u,v) = u, hence φu=1, φv=0. We get in 116:
(1/2)√2 = (u2+h2) dv / (√(u2+h2) √((du)2 + (u2+h2)(dv)2)), dus: +du/√(u2+h2) = dv.
Hence we find: ln | u + √(u2+h2) | = +v + c.

129) The parameter lines are perpendicular to each other in the intersection points where uv = b2-a2.

130) Let x(u1(t),u2(t)) be such a bissectrix.
By equating the angle with the u1-line and the angle with the u2-line, we find via 116 (with φ(u1,u2) = u1 and φ(u1,u2) = u2 respectively):
(a1 2 du1 + a2 2 du2)/√a2 2 = +(a1 1 du1 + a1 2 du2)/√a1 1.
Hence we find (provided that a1 22 - a1 1 a2 2 ≠ 0) : a1 1 (du)2 = a2 2 (dv)2.

131) Apply 120.

137) The parameter lines form an orthogonal net because a1 2 = 0.
The equation of the indicatrix is du dv = 1 (unless b=0), because h1 1 = h2 2 = 0 and h1 2 = -b/√(b2+(a+u)2) (check this). This is an orthogonal hyperbola with asymptotes du = 0 and dv = 0.
The net of the asymptotic lines is du dv = 0, this is the net of the parameter lines.

145) Check by calculation that a1 1 = 1 + (f ' )2, a1 2 = 0, a2 2 = u2; h1 1 = f " / √(1 + (f ' )2), h1 2 = 0, h2 2 = u f ' / √(1 + (f ' )2).
With h1 1a2 2 - 2 h1 2a1 2 + h2 2a1 1 = 0 we get f ' / (1 + (f ' )2) = -uf ".
Consider this as a differential equation in f ' (let f ' = g), and find after separating the variables with partial fraction decomposition (f ')2 = c2/(u2-c2).
We find f = +c log((u + √(u2-c2)/c), so u = c cosh(f/c).

The catenary is the curve formed by a homogeneous cord under the influence of gravity. It looks like a parabola, but is essentially different.
The form of the catenoid is visible when we span a soap fleece between two circular thread figures whose centres are lying straight above each other.

Extra problem : a) (see picture above)
u-lines: parabolas in planes z=constant, x≥0.
v-lines: parabolas in planes x=constant, z≥0.

b) First fundamental form (4u2+v2)(du)2 + 2uv du dv + (u2+4v2)(du)2.
The parameter lines are perpendicular where uv=0, so on the x- and z-axis.

c) Second fundamental form (2/√((u2+v2)2+2u2v2) (v2(du)2 - 2uv du dv + u2(dv)2).
Indicatrix (2/√..) (v du - u dv)2 = 1, two parallel lines with direction vector xu + (v/u)xv.
Asymptotic lines du/dv = u/v, u=cv, x = v2(c2,c,1) (halfs of straight lines).

d) Orthogonal trajectories of the u-lines from xu . (xu+ xv dv/du) = 0.
We get du/dv = - (4u2+v2)/(uv) and at the end v2u2 + 4u4 = constant.

e,f) From c) we see the surface exists of half straight lines v2(c2,c,1) = λ(t2,t,1) (λ positive). The half lines start from (0,0,0) (top) and go through (t2,t,1) (direction curve, a parabola). So it is a half parabolic cone.
Tangent planes (equation v2x - 2uv y + u2z = 0) are constant on rules u = cv (equation x - 2cy + c2z = 0).

f) Parametrisation λ(t2,t,1): see e) above.
The rules are the λ-lines, so the orthogonal trajectories of the rules satisfy xλ . (xλ+ xt dt/dλ) = 0. We get dt/dλ = -(t4+t2+1)/((2t3+t)λ).
At the end we find λ||(t2,t,1)|| = constant, so an orthogonal trajectory exists of points at the same distance from (0,0,0).

153) If the meridian curves of a helicoid surface are curvature lines, then dv = 0 must be a solution of 148, so a1 1 h1 2 - a2 1 h1 1 = 0.
We get h (1 + (f ' )2 + u f ' f " ) = 0. So h = 0 of f ' f " /(1 + (f ' )2) = -1/u. This gives the alleged solutions.

154) Check by calculation that a1 2 = h1 2 = 0 and use proposition 152.

155) Check by calculation that a1 2 = h1 2 = 0 and use proposition 152.

165) x ' = 0 gives a cone; e ' = 0 gives a cylinder. Suppose x ' , e ' ≠ 0.
If y(u) = x(u) + v(u) e(u) is the turning curve, then y '(u) = x '(u) + v(u) e '(u) + v '(u) e(u) has the same or opposite direction as e(u) (since this holds for x ' + v e ' because that's how we chose v).
So the ruled surface is a surface of tangents.

EP) First prove that the singular points are the points with u=v, and determine the candidate turning curve x(t) with t:=u=v and the corresponding surface of tangents x(t) + λ x '(t).
Use a change of coordinates to show that this surface of tangents coincides with the original surface.

180) x = q(cos(t),sin(t),t/p) (circular helix) ; x = (t,t2,t3) (cubic parabola).

181) x = (t cos(t) - sin(t), t sin(t) + cos(t)) (evolvent of a circle).

189) Check that the Christoffel equations 186 are in this case u.. = v.. = 0

190) Check what the differential equations 186 are in this case. We get:
u(u .)2 - u(v .)2 + (1+u2) u .. = 2 u . v . + u v .. = 0.
Furthermore we have (ds)2 = (1+u2)(du)2 + u2(dv)2.
From the second Christoffel equation we deduce v . = c/u2, so
u4(dv)2 = c2(ds)2 = c2((1+u2)(du)2 + u2(dv)2).
Hence we get v = + ∫ (c/u) √((1+u2)/(u2-c2)) du.
For c=0 we find the meridians.

191) Start from x = (u cos(v), u sin(v), a*u). We find:
(ds)2 = (1+a2)(du)2 + u2(dv)2, -u (v .)2 + (1+a2) u .. = 2 u . v . + u v .. = 0.
Like in 190, u4(dv)2 = c2(ds)2. Here this yields:
u4(dv)2 = c2(1+a2)(du)2 + u2 c2 (dv)2, from which we get:
v = +√(1+a2) ∫ c/(u √(u2-c2)) du.
With t = √(u2-c2) we find v = +√(1+a2) arctan(√(u2-c2)/c) + d.

Notice that the solutions in 190 en 191 satisfy the fact that in each point a geodesic departs in each direction.

199)
a) Check by calculation that the indicatrix becomes (2/√...) (du - dv)2 = 1.
Each point is parabolic, but there are no singular points (because N is nowhere 0). So the surface must be a cylinder.
We find asymptotic curves with du = dv, u = v + c, so their parametrisations are (1+c, c, c2) + v(0, 1, 1). So we have the direction curve (1+c, c, c2) and the axis has direction (0,1,1).

b) A simpler parametrisation is (1+w, w, w2) + v(0, 1, 1).
The differential equations become:
(2+4w2)(w .)2 + 2(1+2w)(w .)(v .) + 2(v .)2 = 1
((4w-2)/(4w2-4w+3))(w .)2 + (w ..) = 0
((4-4w)/(4w2-4w+3))(w .)2 + (v ..) = 0
Solutions are, among others: w(s) = c1, v(s) = +(1/2)√2 s + c2 (the rules).

200) Both surfaces have first fundamental form r2 (du)2 + 2 (dr)2. The first surface is right circular cone, the second a plane.

201)
a) We find the asymptotic lines from (1/u)(du)2 - (2/v)(du)(dv) + (u/v2)(dv)2 = 0 = (du/u - dv/v)2, so each point is parabolic.
Solving du/dv = u/v yields u = cv. Parameter representation of the asymptotic lines v(c2, c, 1/c).
b) From a) we see the surface has parameter representation v(c2, c, 1/c), so it exists of straight lines through 0. Hence it is a cone, so it's developable.

203)
a) (xu . xv)/(||xu||.||xv||) = cos(α), so a1 2 = cos(α) √(a1 1a2 2).
b) Let x(u,v(u)) be such a trajectory. Require ((xu) . (xu + xv dv/du))/(||xu||.||xu + xv dv/du||) = (1/2)√3, so (a1 1 + a1 2 dv/du)/(√a1 1 . √(a1 1 + 2 a1 2 dv/du + a2 2(dv/du)2)) = (1/2)√3.

204)
a) The u-lines are congruent to the graph of g; the v-lines are straight lines; U is a 'wave-like plate' (cylinder with sinusoid as direction curve).
b) We get a simpler parametrisation of U by taking (w,u,g(u)) with first fundamental form (1+g'(u)2) (du)2 + (dv)2.
The plane z=0 has parametrisation (w,f(u),0) with first fundamental form f '(u)2 (du)2 + (dv)2.
Now take f so that f '(u) = √(1+g'(u)2).

205)
a) Consider two surfaces x(u,v) and y(u,v).
If two curves on the first surface have a common tangent in a certain point, they are image curves of two curves in the u,v-plane that in the original of this contact point have a common tangent with direction coefficient dv/du. So then the common tangent to these image curves has in that contact point direction xu + xv dv/du.
Under the diffeomorfism yox-1 they are mapped to two curves on the second surface that have a common tangent with direction yu + yv dv/du.
So having a common tangent is invariant under all diffeomorfisms, not only under isometries.
b) If we bend a right circular cylinder to a plane, the circle becomes a straight line, so curvature is not invariant.

207)
a) Using u dv + v du = 0, we find a family of asymptotic lines on the surface. We get uv = c, and hence we find the alternative parametrisation (c, c2, c3) + u(c, c, c2). With this parametrisation, the u-lines are straight lines.
b) xuxc = (-c3 - uc2, 3c3 + uc2 - c2, c - 2c2) is dependent on u, so the tangent plane is not constant along the rule, so the answer is no.
Alternative : Using xuxc = 0 we find two singular points, namely (0,0,0) and (1/4,0,0). A cylinder doesn't have any singular points, a cone has one singular point, a surface of tangents has a whole curve of singular points.
c) Take v=1 and u=t (or c=u=t).
d) The normal vector on the surface has the same or opposite direction as (-2t3, 4t3-t2, t-2t2), the principal normal is perpendicular to x ' ⊗ x " = (24t2, -12t-12t2, 4).
If k is a geodesic, N and n must have the same or opposite directions in all points of k, so (-2t3, 4t3-t2, t-2t2) and (24t2, -12t-12t2, 4) must be perpendicular for all t. This doesn't even hold for t=1. So k is no geodesic.