# Differentiable Function - Differential Geometry - Solved Exam, Exams for Computational Geometry. Aligarh Muslim University

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This is the Solved Exam of Differential Geometry which includes Normal Vector, Normal Vector, Binormal Vector, Curvature, Torsion, Binormal Vector, Speed Space Curve etc. Key important points are: Differentiable Functio...
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SOLUTIONS TO FINAL EXAM

QUESTIONS FROM PREVIOUS YEARS.

Curves.

1. Let s be the arclength parameter.

γ′(t) = (1, f ′). ds

dt = |γ′(t)| =

√ 1 + f ′2.

ds =

1√ 1 + f ′2

(1, f ′)

d

dt

ds =

1√ 1 + f ′2

(0, f ′′) − f ′f ′′

(1 + f ′2)3/2 (1, f ′)

= 1

(1 + f ′2)3/2 (−f ′f ′′, f ′′)

d2γ

ds2 =

f ′′

(1 + f ′2)2 (−f ′, 1)

curvature = ∣∣∣∣ d2γ

ds2

∣∣∣∣ = f ′′

(1 + f ′2)3/2

Curvature of the parabola y = kx2 is

2k (1 + 4k2x2)3/2

.

2. Let t be the arclength parameter for α and let s be the arclength parameter for β(t) =

Typeset by AMS-TEX 1

2 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

α′(t). Let t,n,b be the Frenet frame for α.

dt =

d2α

dt2 = kn.

ds

dt =

∣∣∣∣

dt

∣∣∣∣ = k. dβ

ds = n.

d

dt

ds =

dn dt

= −kt− τb. d2β

ds2 = −kt− τb

k = t− τ

k b.

curvature of β = ∣∣∣∣ d2β

ds2

∣∣∣∣ = √

1 + τ2

k2 .

Tangent space and derivatives of maps.

1. There are misprints in this question. Let us assume that the cylinder is x2 + y2 = 1 and that the tangent vector in T(1,0,0) is (0, 1, 1). In the given coordinates (a coordinate map is just the inverse of a parameterization), dϕn has matrix

( ∂nθ ∂θ

∂nθ ∂z

∂z ∂θ

∂z ∂z

) =

( n 0 0 1

) .

Now (1, 0, 0) corresponds to (θ, z) = (0, 0). We see that ϕn(1, 0, 0) = (1, 0, 0). The basis for the tangent space associated to the coordinates is

(x, y, z) ∂θ

= (sin θ, cos θ, 0), ∂(x, y, z) ∂z

= (0, 0, 1).

At (θ, z) = (0, 0) this basis is (0, 1, 0), (0, 0, 1).

Then at (1, 0, 0),

dϕn(0, 1, 0) = (0, n, 0), dϕ(0, 0, 1) = (0, 0, 1).

Hence dϕn(0, 1, 1) = (0, n.1).

General second fundamental form questions.

SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS. 3

1. Following Do Carmo, for the graph z = f(x, y), the matrix of the second fundamental form with respect to the basis

(1, 0, fx), (0, 1, fy),

is ( fxx fxy fxy fyy

) .

In our case, the basis at (0, 0, 0) is

(1, 0, 0), (0, 1, 0)

and the matrix of the second fundamental form is (

e f f g

) =

( 2A B B 2C

) .

The matrix of the derivative of the Gauss map is

(

E F F G

)1 ( e f f g

) =

( 1 + f2x fxfy fxfy 1 + f2y

)1 ( fxx fxy fxy fyy

) ,

which gives

K = fxxfyy − f2xy 1 + f2x + f2y

, H = (1 + f2y )fxx + (1 + f

2 x)fyy − 2fxfyfxy

2(1 + f2x + f2y ) .

In our case, K = 4AC −B2, H = 2A + 2C.

The unit vector in direction (x, y, 0) is (cos θ, sin θ, 0) = cos θ(1, 0, 0) + sin θ(0, 1, 0). The normal curvature in this direction is

(cos θ, sin θ) (

2A B B 2C

)( cos θ sin θ

) = 2A cos2 θ + 2B cos θ sin θ + 2C sin2 θ.

2. elliptic points = {q : K(q) (0,∞)}.

4 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

Since K is a continuous (indeed a smooth) function on S and (0,∞) is open, the set of elliptic points of S is open, and is hence an open neighborhood of each of its points. The same is true for the hyperbolic points, but not for parabolic or planar points. For example, the graph

z = x4 + y2

is parabolic at (0, 0, 0) and elliptic elsewhere, while the graph

z = x4 + y4

is planar at (0, 0, 0) and elliptic elsewhere.

Gauss curvature questions.

1. (a).

yu = xu + cNu ⇒ 〈yu,N= xu,N+ c〈Nu,N= Nu,N= 12

∂u 〈N,N= 0.

Similarly, N is perpendicular to yv, and hence is normal to the surface parameterized by y. (b). Since xu and xv are lines of curvature, we have

Nu = µ1xu, Nv = µ2xv

for smooth scalar functions µ1, µ1 (we don’t specify which is bigger.)

yu × yv = (xu + cNu)× (xv + cNv) = xu × xv + cxu ×Nv + cNu × xv + c2Nu ×Nv = xu × xv − c(µ1 + µ2)xu × xv + c2µ1µ2xu × xv = (12cH + c2K)xu + xv.

(c). We think of N as being a function of (u, v). Since x and y have the same normal, N gives the Gauss map for either surface. Consider (u0, v0) and set x(u0, v0) = x0 and y(u0, v0) = y0. Consider a small disc R around (u0, v0). Then

curvature of x at x0 = lim diam R→0

areaN(R) areax(R)

,

curvature of y at y0 = lim diam R→0

areaN(R) areay(R)

.

Hence

curvature of y curvature of x

= lim diam R→0

areax(R) areay(R)

= lim diam R→0

R |xu × xv| dudv

R |yu × yv| dudv =

1 12cH + c2K .

SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS. 5

2. K = 1: the sphere x2 + y2 + z2 = 1. K = 0: the plane z = 0. K = 1: the surface of revolution obtained by revolving (e−s, 0,

s 0

1− e−2s ds), for s > 0, about the z-axis.

We derived the formula for the curvature surface obtained by revolving the curve (φ(s), 0, ψ(s)) around the z-axis in the case when s is the arclength - that is φ′2 + ψ′2 = 1. The Gauss curvature is

−φ ′′

φ .

In particular it is zero precisely when φ′′ = 0. This implies that

φ(s) = as + b, ψ(s) = √

1− a2s + c.

This means that the curve is a straight line and the surface of revolution is a cone or cylinder.

3. xs = α′(s) + ub(s) = t(s) + n(s), xu = b(s),

so

E = 1 + u2τ2

F = 0 G = 1

Furthermore

xs × xu = (t + n)× b = n + t, N = n + t1 + u2τ2

.

g = xuu,N= 0,N= 0. f = xuv,N= 〈τn, −n + t

1 + u2τ2 = −τ√

1 + u2τ2 .

K = eg − f2

EG− F 2 = − τ2

(1 + u2τ2)2 .

Isometric surfaces.

1. Done in class.

2. Consider the parameterizations x(u, v) and

x̄(u, v) = r(u, v + d) = (f(v + d) cos u, f(v + d) sin u, g(v + d)).

6 SOLUTIONS TO FINAL EXAM QUESTIONS FROM PREVIOUS YEARS.

Then r̄ r1

is the map (u, v) (u, v + d). This is an isometry if and only if E = Ē, F = , G = Ḡ. But

( E F F G

) =

( f2(v) 0

0 (f ′(v))2 + (g′(v))2

) ,

( Ē F̄ F̄ Ḡ

) =

( f2(v + d) 0

0 (f ′(v + d))2 + (g′(v + d))2

) .

Equating E and we get f(v + d) = ±f(v) for all d. Assuming f is not identically zero, this gives that f is constant and the surface is a cylinder.