DIFFERENTIAL GEOMETRY COURSE

*answers (partial)*

*16)* * x*(t) = (a.cos(t),a.sin(t),bt), so s'(t) = ||

*17)* * x*(t) = (t,t

*18)* * x*'(t) = a(1-cos(t),sin(t)), s'(t) = 2a.|sin(t/2)|, s(t) = 4a - 4a.cos(t/2) (t ∈ [0,2π]).

*19)* The straight lines are λ(cos(t),sin(t),1) = λ* v*.

The points of intersection are f(t)(cos(t),sin(t),1), and the tangent vectors

The angle φ between the tangent vector and the straight line satisfies cos(φ) = (

When φ is constant, we have for some constant c : (f ')

*20)* Use the quotient rule and the chain rule for differentiation.

Since s'= ||* x*'|| = √(

Also use that ||

*28)* Combine the answers of problems 18 and 20, and find κ(s) = (8as-s^{2})^{-1/2}.

*29)* The following four assertions are equivalent (a,b,c,p,q,r are constant):

i) κ(s)=0 for all s;

ii) __x__** ^{..}**(s)=

iii)

iv)

Now suppose * c* is a fixed vector with

Since

Unless λ = 0 for all s, we have

Finally, suppose the fixed point is a point at infinity, say

*35)* 3t^{2}x - 3ty + z = t^{3}

*37)* The binormal has direction __+__(3t^{2},-3t,1) (see 35));

the tangent has direction __+__(1,2t,3t^{2});

then the principal normal has direction __+__ * b*⊗

So the normal plane has direction x + 2ty + 3t

*41)* Use * x* ' =

*45)* d^{2} = α^{2} + β^{2} + γ^{2} = h^{2} - κ^{2}h^{4}/3 + κ^{2}h^{4}/4 + O(h^{5}),
so

d = h√(1 - κ^{2}h^{2}/12 + O(h^{3})) = h(1 - κ^{2}h^{2}/24 + O(h^{3})).

So (d-h)/h^{3} → -κ^{2}/24 if h→0.

The deviation between d and h increases with increasing curvature.

*47)*

i) κ = a(4(b^{2}+a^{2})sin(t/2))^{-1}, τ = b(4(b^{2}+a^{2})sin(t/2))^{-1};

ii) κ = 3/(25t), τ = 4/(25t);

iii) κ = τ = (1+t^{2}+t^{4}/4)^{-1}/6.

*48)* * x*'(t) = a*e

s'(t) = ||

s(t) = (a/c)(√(1+2c

Now calculate κ(t) using 20 and τ(t) using 41; finally substitute t=(ln(1+s/B))/c.

*51)* Suppose * x*(s) is such a curve. Then there exist a function λ(s) (≠0) and a point

Differentiation yields

Hence τ=0, λ

*55)* Let α have equation ax_{1} + bx_{2} + cx_{3} = d (normal (a,b,c)).

Then g(t) = ax_{1} + bx_{2} + cx_{3} - d;

g(t_{o})=0 means that * x*(t

if, moreover, g'(t

if, moreover, g''(t

*59)* Denoting R:=1/κ and T:=1/τ, the problem becomes: show that * m* =

This is easy done by differentiation, using the formulas of Serret and Frenet.

*60)* The curve * m*(s) =

Let R = 1/κ. Since 2

This becomes d/ds (RR

*62)*

a) The osculating plane is * b*(s).(

Differentiation yields

So if κ≠0 and τ≠0 then

b) Let

If τ≠0 then

*63)* Let * y*(s) =

We get (with κ≠0) R

*64)* Let * a* =

Hence ||

For all points on the curve

From ατ+β

*65)* Let * a* =

Like in 64) we get by differentiation κα=1, α

From cos(φ) = ((α

*66)* The equation of V(s) is (* x*-

Let f(s) = r

*70)* * y*(s) =

cos

Alternative solution: * y* ' =

*71)* * y*(s) =

So

|κ

(

Calculate τ

We get ττ

*72)* Suppose * y*(s) =

Then

Now

Then it follows that λ

*73)* Suppose * y*(s) =

Require that

Since

Then

*78)* φ_{2}(s) = φ_{1}(s)+c (c constant);

using λ = (κ cos(φ))^{-1} we get * y* =

using φ = 0 we get λ = κ

using τ=0 and φ=φ

*79)* The intersection point of the plane z=0 and the tangent in (a cos(t), a sin(t), bt) to the circular helix is (a cos(t) + at sin(t), a sin(t) + at cos(t), 0). This point lies on the
tangent in (a cos(t), a sin(t), 0) to the circle, and, at the same time, this tangent is perpendicular to the intersection figure (check all of this by calculation).

*80)* Let * y*(t) be the evolute. Then we have

Since

So

*85)* Using Serret and Frenet we find 0 = det(__x__** ^{..}**,

From κ

*86)* Using τ = a κ en κ^{-2} + κ ^{.}^{2}τ^{-2}κ^{-4} = b^{2} we find
R^{2} + c^{2}R^{2}R ^{.}^{2} = b^{2}.

So RR ** ^{.}**/√(b

So κ = 1/√(b

*87)* Using 47 we find κ = τ. The direction of the axis is * t* cos(α) +

The direction of

From the equation of the osculating plane it follows that

*88)*

a) s' = ||* x* ' || = √(((1/2)√3 cos(t) - 1/2)

κ = ||

b) Matrix ((0, 0, 1), ((1/2)√3, 0, 1/2), (-1/2, 0, (1/2)√3)).

*89)* Arbitrary sphere (x-a)^{2} + (y-b)^{2} + (z-c)^{2} - r^{2} = 0, so g(t) = (t-a)^{2} + (t^{4}-b)^{2} + (t^{2}-c)^{2}
- r^{2}.

From g(1) = g ' (1) = g '' (1) = g ''' (1) = 0 we get a=32, b=15/2, c=-55/2, r=42.6087.

*90)*

a) K_{1} : * x*(s), K

Require det(

b) Require det(

We get λτ { λτ(λτ

If λ is constant we get λ

*91)*

a) s' = ||* x* ' || = 5, so K

b)

c)

d) K

e) cos(α) =

*92)* K_{1} : * x*(s), K

Requirement 1:

Requirement 2:

With requirement 1) we get cos(α) + λ

With requirement 2) we get (κ cos(α) - τ sin(α)

(The lines with direction vector cos(α)