COURSE OF DIFFERENTIAL GEOMETRY

*answers (partially)*

*100)*

i) straight lines through * 0*.

ii)

iii) The normal vector has the same or opposite direction as

iv) The equation of the tangent plane is independent of u.

*110)* The first fundamental form is 5(du)^{2} + u^{2}(dv)^{2}, so a_{1 1} = 5, a_{1 2} = 0, a_{2 2} = u^{2}.

From cos(φ) = (Σ a_{i j} da_{i} db_{j})/(√(Σ a_{i j} da_{i} da_{j})√(Σ a_{i j} db_{i} db_{j}))
with a_{1} = b_{2} = t^{2} en a_{2} = b_{1} = t^{3}, follows, if t=1, cos(φ) = 36/(7√29).

*111)* __x___{u} = (1,2u,v) and __x___{v} = (1,2v,u), so a_{1 2} = 0 if uv = -1/5, so * x* =
(v - 1/(5v), v

*112)* (ds)^{2} = (dθ)^{2} + sin^{2}(θ) (dφ)^{2}.

The surface element is then √ sin^{2}(θ) dθ dφ.

So the required surface is _{0} ∫ ^{2π} _{θ2} ∫ ^{θ1} 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} +
u^{2} (dv)^{2}.

*117)* With φ(u,v) = u we find φ_{u} = 1, φ_{v} = 0 ; so the differential equation becomes:

du/dv = (a_{2 2}φ_{u} - a_{1 2}φ_{v})/(a_{1 1}φ_{v} - a_{1 2}φ_{u}) = -a_{2 2}/a_{1 2} =
-(2+u^{2})/(uv).

We get u du/(2+u^{2}) = -dv/v, so ln(2+u^{2}) = -2 ln |v| + c. Hence it follows that v^{2}(2+u^{2}) = μ.

*118)* The meridians are determined by __φ__(θ,φ) = λ, with __φ__(θ,φ) = φ, so by φ = λ.

We get __φ___{θ} = 0 and __φ___{φ} = 1, so the differential equation becomes:
cos(α) = (-a_{1 1}dθ - a_{1 2}dφ)/(√a_{1 1}*√(a_{1 1}dθ^{2} + 2 a_{1 2}dθ dφ + a_{2 2}dφ^{2}).

With a_{1 1} = 1, a_{1 2} = 0 and a_{2 2} = sin^{2}(θ) we find cos(α) = -dθ / √(dθ^{2} + sin^{2}(θ) dφ^{2}),
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} + (u^{2}+h^{2}) (dv)^{2} , so a_{1 1} = 1, a_{1 2} = 0, a_{2 2} = u^{2}+h^{2}.

φ(u,v) = u, hence φ_{u}=1, φ_{v}=0. We get in 116:

(1/2)√2 = (u^{2}+h^{2}) dv / (√(u^{2}+h^{2}) √((du)^{2} + (u^{2}+h^{2})(dv)^{2})), dus:
__+__du/√(u^{2}+h^{2}) = dv.

Hence we find: ln | u + √(u^{2}+h^{2}) | = __+__v + c.

*129)* The parameter lines are perpendicular to each other in the intersection points where uv = b^{2}-a^{2}.

*130)* Let * x*(u

By equating the angle with the u

(a

Hence we find (provided that a

*131)* Apply 120.

*137)* The parameter lines form an orthogonal net because a_{1 2} = 0.

The equation of the indicatrix is du dv = 1 (unless b=0), because h_{1 1} = h_{2 2} = 0 and h_{1 2} = -b/√(b^{2}+(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 a_{1 1} = 1 + (f ' )^{2}, a_{1 2} = 0, a_{2 2} = u^{2}; h_{1 1} = f " / √(1 + (f ' )^{2}), h_{1 2} = 0,
h_{2 2} = u f ' / √(1 + (f ' )^{2}).

With h_{1 1}a_{2 2} - 2 h_{1 2}a_{1 2} + h_{2 2}a_{1 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} = c^{2}/(u^{2}-c^{2}).

We find f = __+__c log((u + √(u^{2}-c^{2})/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 (4u^{2}+v^{2})(du)^{2} + 2uv du dv + (u^{2}+4v^{2})(du)^{2}.

The parameter lines are perpendicular where uv=0, so on the x- and z-axis.

c) Second fundamental form (2/√((u^{2}+v^{2})^{2}+2u^{2}v^{2}) (v^{2}(du)^{2} - 2uv du dv + u^{2}(dv)^{2}).

Indicatrix (2/√..) (v du - u dv)^{2} = 1, two parallel lines with direction vector __x___{u} + (v/u)__x___{v}.

Asymptotic lines du/dv = u/v, u=cv, * x* = v

d) Orthogonal trajectories of the u-lines from __x___{u} **.** (__x___{u}+ __x___{v} dv/du) = 0.

We get du/dv = - (4u^{2}+v^{2})/(uv) and at the end v^{2}u^{2} + 4u^{4} = constant.

e,f) From c) we see the surface exists of half straight lines v^{2}(c^{2},c,1) = λ(t^{2},t,1) (λ positive). The half lines start from (0,0,0) (top) and go through
(t^{2},t,1) (direction curve, a parabola). So it is a half parabolic cone.

Tangent planes (equation v^{2}x - 2uv y + u^{2}z = 0) are constant on rules u = cv (equation x - 2cy + c^{2}z = 0).

f) Parametrisation λ(t^{2},t,1): see e) above.

The rules are the λ-lines, so the orthogonal trajectories of the rules satisfy __x___{λ} **.**
(__x___{λ}+ __x___{t} dt/dλ) = 0. We get dt/dλ = -(t^{4}+t^{2}+1)/((2t^{3}+t)λ).

At the end we find λ||(t^{2},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
a_{1 1} h_{1 2} - a_{2 1} h_{1 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 a_{1 2} = h_{1 2} = 0 and use proposition 152.

*155)* Check by calculation that a_{1 2} = h_{1 2} = 0 and use proposition 152.

*165)* * x* ' =

If

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

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) ;

*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+u^{2}) u ^{..} = 2 u ^{.} v ^{.} + u
v ^{..} = 0.

Furthermore we have (ds)^{2} = (1+u^{2})(du)^{2} + u^{2}(dv)^{2}.

From the second Christoffel equation we deduce v ^{.} = c/u^{2}, so

u^{4}(dv)^{2} = c^{2}(ds)^{2} = c^{2}((1+u^{2})(du)^{2} + u^{2}(dv)^{2}).

Hence we get v = __+__ ∫ (c/u) √((1+u^{2})/(u^{2}-c^{2})) du.

For c=0 we find the meridians.

*191)* Start from * x* = (u cos(v), u sin(v), a*u). We find:

(ds)

Like in 190, u

u

v =

With t = √(u

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

We find asymptotic curves with du = dv, u = v + c, so their parametrisations are (1+c, c, c

b) A simpler parametrisation is (1+w, w, w^{2}) + v(0, 1, 1).

The differential equations become:

(2+4w^{2})(w ** ^{.}**)

((4w-2)/(4w

((4-4w)/(4w

Solutions are, among others: w(s) = c

*200)* Both surfaces have first fundamental form r^{2} (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/v^{2})(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(c^{2}, c, 1/c).

b) From a) we see the surface has parameter representation v(c^{2}, c, 1/c), so it exists of straight lines through * 0*. Hence it is a cone, so it's developable.

*203)*

a) (__x___{u} **.** __x___{v})/(||__x___{u}||.||__x___{v}||) = cos(α),
so a_{1 2} = cos(α) √(a_{1 1}a_{2 2}).

b) Let * x*(u,v(u)) be such a trajectory. Require ((

*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

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

Under the diffeomorfism

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, c^{2}, c^{3}) + u(c, c, c^{2}). With this parametrisation, the u-lines are straight lines.

b) __x___{u} ⊗ __x___{c} = (-c^{3} - uc^{2}, 3c^{3} + uc^{2} - c^{2}, c - 2c^{2}) is dependent on u,
so the tangent plane is not constant along the rule, so the answer is no.
__Alternative__ : Using __x___{u} ⊗ __x___{c} = * 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 (-2t

If k is a geodesic,