### DIFFERENTIAL GEOMETRY COURSE

16) x(t) = (a.cos(t),a.sin(t),bt), so s'(t) = ||x'(t)|| = √(a2+b2), so s = t √(a2+b2) + c (kies c=0);
x(s) = (a.cos(s/√(a2+b2)),a.sin(s/√(a2+b2)),bs/√(a2+b2)).

17) x(t) = (t,t3/2), so s'(t) = √(1+9t/4), so s(t) = (8/27)(1+9t/4)3/2 + c (kies c=-8/27), dus t = (4/9)(27s/8+1)2/3 - 4/9.

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) = λv.
The points of intersection are f(t)(cos(t),sin(t),1), and the tangent vectors t = f '(t)(cos(t),sin(t),1) + f(t)(-sin(t),cos(t),0).
The angle φ between the tangent vector and the straight line satisfies cos(φ) = (v.t)/(||v|| ||t||) = 2f '(t)/(√(2f(t)2+4f '(t)2).
When φ is constant, we have for some constant c : (f ')2 = c(f2+2(f ')2), and f '/f = d. So f(t) = α eβt.

20) Use the quotient rule and the chain rule for differentiation.
Since s'= ||x'|| = √(x'.x'), we have s'' = (x',x'')/||x'|| = ||x''|| cos(φ), where φ is the angle between x' and x''.
Also use that ||x'⊗x''|| = ||x'|| ||x''|| sin(φ).

28) Combine the answers of problems 18 and 20, and find κ(s) = (8as-s2)-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)=0 for all s;
iii) x .(s)=(a,b,c) for all s;
iv) x(s)=(as+p,bs+q,cs+r) for all s.

Now suppose c is a fixed vector with x(s)+λx .(s)=c for all s.
Since x .. is perpendicular to x ., we find (1+λ.)x . = λx .. = 0.
Unless λ = 0 for all s, we have x .. = 0 for all s, so the curve is a straight line.
Finally, suppose the fixed point is a point at infinity, say x . = c for all s. Then x .. = 0 for all s, so the curve is a straight line.

35) 3t2x - 3ty + z = t3

37) The binormal has direction +(3t2,-3t,1) (see 35));
the tangent has direction +(1,2t,3t2);
then the principal normal has direction + bt, so +(9t3-2t,1-9t4,6t3+3t).
So the normal plane has direction x + 2ty + 3t2z = t+2t3+3t5, and the rectifying plane (-9t3-2t)x + (1-9t4)y + (6t3+3t)z = ...

41) Use x ' = x . s' , x" = x ..(s')2 + x . s", x ''' = x ...(s')3 + 3x ..s's" + x . s''' , and sweep propositions for determinants. Combine with 20) en 40).

45) d2 = α2 + β2 + γ2 = h2 - κ2h4/3 + κ2h4/4 + O(h5), so
d = h√(1 - κ2h2/12 + O(h3)) = h(1 - κ2h2/24 + O(h3)).
So (d-h)/h3 → -κ2/24 if h→0.
The deviation between d and h increases with increasing curvature.

47)
i) κ = a(4(b2+a2)sin(t/2))-1, τ = b(4(b2+a2)sin(t/2))-1;
ii) κ = 3/(25t), τ = 4/(25t);
iii) κ = τ = (1+t2+t4/4)-1/6.

48) x'(t) = a*ect(-sin(t),cos(t),0) + ac*ect(cos(t),sin(t),1).
s'(t) = ||x'(t)|| = a*ect√(1+2c2).
s(t) = (a/c)(√(1+2c2))(ect-1) = B(ect-1), so t=(1/c)ln(1+s/B).
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 c such that for all s we have x(s) + λ(s)n(s) = c.
Differentiation yields t + λ .n = 0, so t(1-λκ) + λ .n + λτb = 0.
Hence τ=0, λ . = 0, and λ = 1/κ; so the curve is planar, each point is lying at a fixed distance c, and this distance is 1/κ (so κ is also constant).

55) Let α have equation ax1 + bx2 + cx3 = d (normal (a,b,c)).
Then g(t) = ax1 + bx2 + cx3 - d;
g(to)=0 means that x(to) lies on α;
if, moreover, g'(to)=0, then this means that ax1'(to) + bx2'(to) + cx3'(to) = 0, so x(t) is in x(to) tangent to α;
if, moreover, g''(to)=0, then this means that ax1''(to) + bx2''(to) + cx3''(to) = 0, so (a,b,c) is perpendicular to x'(to) and to x''(to); then α is the osculating plane in x(to).

59) Denoting R:=1/κ and T:=1/τ, the problem becomes: show that m = x + Rn + R. Tb is constant if R2 + R.2 T2 = c.
This is easy done by differentiation, using the formulas of Serret and Frenet.

60) The curve m(s) = x(s) + n(s)/(2κ(s)) is a straight line if m' and m'' have equal or opposite directions everywhere (s is not arc length for m(s)).
Let R = 1/κ. Since 2m' = t + R.n en 2m'' = t(-κR.) + n(κ + R..), the requirement becomes: R.. + κR.2 + κ = 0.
This becomes d/ds (RR.) + 1 = 0, so RR. = c-s, so R2 = 2cs-s2. This yields the cycloid (see 28).

62)
a) The osculating plane is b(s).(x-x(s)) = 0; so for the fixed point c we have b(s).(c-x(s)) = 0.
Differentiation yields b.t + b..(x-c) =0, so (-κtb).(x-c) =0, so κt.(x-c) =0.
So if κ≠0 and τ≠0 then x-c is perpendicular to t, n and b, and then x=c (for all s).
b) Let v be the direction vector of the given straight line. Then, for all s, b.v=0, so b..v = τ n.v =0.
If τ≠0 then n.v=0, so n..v=0. Then v is perpendicular to b, n and -κtb. Since v is not 0, we get κ =0.

63) Let y(s) = x(s) + R(s)n(s). Then y' has direction equal or opposite to y'', so R.n + R τ b has direction equal or opposite to -R. κt + (R..-Rτ2)n + (2R.τ + R τ .)b.
We get (with κ≠0) R.=0 (for all s) and hence τ=0.

64) Let a = x(s) + α(s)n(s) + β(s)b(s), so 0 = t(1-ακ) + n.-τβ) + b(ατ+β.), so α=1/κ, β=-κ./(κ2τ).
Hence ||x(s)-a|| = √(α22) = √(κ-2 + κ.2κ-4τ-2).
For all points on the curve x(s), a is the center of the osculating sphere.
From ατ+β.=0 we find that the radius of the osculating sphere is constant.

65) Let a = x(s) + α(s)n(s) + β(s)b(s).
Like in 64) we get by differentiation κα=1, α.=τβ, and √(α22 is constant.
From cos(φ) = ((αnb).n)/√(α22) and the fact that φ is constant, we then find: α is constant, so κ is constant, and τ=0.

66) The equation of V(s) is (x-x(s)).n(s) =0, so according to Hesse we have a(s) = |(m-x(s)).n(s)|.
Let f(s) = r2-s2 = (m-x).(m-x) - s2. Then f . = 2t.(x-m) - 2s, and f .. = 2κn.(x-m) + 2t.t - 2 = 2κn.(x-m).So |f ..| = 2κa.

70) y(s) = x(s) + λn(s) (λ constant).
cos2(φ) = (x ..y ' )2/|| y ' ||2 = (1-λκ)2/((1-λκ)22τ2) is constant if τ=0, if λ=1/κ (then φ=90 degrees), if κ and τ are constant, and also if τ=d(λκ-1) ≠0.

Alternative solution: y ' = t(1-λκ) + λτb, and x . = t, so tan(φ) = λτ/(1-λκ). If the conditions are met, then tan(φ) constant.

71) y(s) = x(s) + Rn(s) (R is constant).
So y ' = x . + Rn . = τRb and y '' = τ . Rb + τRb . = -τ2Rn + τ . Rb and y ''' = τ2t -3ττ . Rn + (τ .. R - τ3R)b.
new| = ||y ' ⊗ y '' ||/||y ' ||3 = τ3R2/(τ3R3) = κ, κnew = -κ
(x(s) and y(s) are corresponding curves of Bertrand, they have the same principal normals but opposite curvature, and exist of each other's curvature centers).
Calculate τnew via τnew = det(y ' ,y '' ,y ''' )/|| y ' ⊗ y '' ||2.
We get ττnew = κ2.

72) Suppose y(s) = x(s) + λ(s)n(s).
Then y ' = t(1-λκ) + λ . n + λτb, and y '' = t(-2λ . κ - λκ . ) + (λ .. - λτ2 + κ - λκ2)n + (2λ . τ + λτ . )b.
Now y ' and y '' must lie in the same plane as n and b, so 1-λκ = 2λ . κ + λκ . = 0. Using λκ=1 we find λ . κ + λκ . = 0.
Then it follows that λ . κ = 0, and because κ ≠0 we get λ . =0. So λ is constant and κ=1/λ, so κ is also constant.

73) Suppose y(s) = x(s) + λ(s)b(s).
Require that b be perpendicular to y ' and that det(b, y ', y '' ) =0.
Since y ' = t - λτn + λ . b we have λ . =0, so λ is constant.
Then y '' = t . - λτ . n - λτn . = λκτt + (κ - λτ . )n - λτ2b, so det(b, y ', y '' ) = κ - λτ . + λ2κτ2 =0.

78) φ2(s) = φ1(s)+c (c constant);
using λ = (κ cos(φ))-1 we get y = x + Rn + Rb tan(φ);
using φ = 0 we get λ = κ-1 and τ=0;
using τ=0 and φ=φo we get y ' = λ . (n cos(φ) + b sin(φ)) where b is fixed, so the angle between y ' and b is fixed.

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 y(t) = (t-sin(t),1-cos(t)) + λ(t)(-sin(t),1-cos(t)).
Since y '(t) is parallel to (-sin(t),1-cos(t)), we find det( (1-cos(t)-λcos(t), sin(t)+λsin(t)), (-sin(t), 1-cos(t)) ) = 0, so (2+λ)(1-cos(t)) = 0, so λ=-2.
So y(t) = (t+sin(t),-1+cos(t)). We get this curve if we translate x(t) over (-π,-2).

85) Using Serret and Frenet we find 0 = det(x .., x ..., x ....) = det(t ., t .., t ...) = κ3det(n , n ., n ..) = -κ3.τ - κτ .).
From κ .τ - κτ . = 0 we get κ/τ = c (c constant).

86) Using τ = a κ en κ-2 + κ .2τ-2κ-4 = b2 we find R2 + c2R2R .2 = b2.
So RR ./√(b2-R2) = a, and √(b2-R2) = -as.
So κ = 1/√(b2-a2s2), τ = a/√(b2-a2s2).

87) Using 47 we find κ = τ. The direction of the axis is t cos(α) + b sin(α) = (1/2)√2 (t+b).
The direction of t is (6,6t,3t2), so t = (2,2t,t2)/√(4+4t2+t4).
From the equation of the osculating plane it follows that b = (t2,-2t,2)/√(4+4t2+t4). So the direction of the axis is (1,0,1).

88)
a) s' = ||x ' || = √(((1/2)√3 cos(t) - 1/2)2 + (-sin(t))2 + ((1/2)cos(t) + (1/2)√3)2) = √2, dus s=t√2.
κ = ||x ' ⊗ x '' || / ||x ' ||3 = 1/2.
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 - r2 = 0, so g(t) = (t-a)2 + (t4-b)2 + (t2-c)2 - r2.
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) K1 : x(s), K2 : y(s) = x(s) + λ(s)b(s) (s no arc length for K2).
Require det(b, y ' , y " ) = 0. We get λ2κτ2 = λτ . + 2λ . τ - κ.
b) Require det(b, y ' , y ' ⊗ y ") = 0.
We get λτ { λτ(λτ2 - λ .. ) - λ . (κ - 2 λ . τ - λτ .) } = λ .. - λτ2 - λλ . κτ.
If λ is constant we get λ3τ4 = - λτ2 and subsequently τ=0 (planar curve).

91)
a) s' = ||x ' || = 5, so K1 : (3 sin(s/5), 4s/5, 3 cos(s/5)).
b) t : ((3/5)cos(s/5), 4/5, (-3/5)sin(s/5)), n : (-sin(s/5), 0, -cos(s/5)), b : ((-4/5)cos(s/5), 3/5, (4/5)sin(s/5)).
c) x.b(t) = x(t).b(t), so -4 cos(t) x1 + 3 x2 + 4 sin(t) x3 = 12t.
d) K2 : (-3 sin(s/5), 4s/5, -3 cos(s/5)).
e) cos(α) = t1.t2 = 7/25, so α is about 74 degrees.

92) K1 : x(s), K2 : y(s) = x(s) + λ(s)(cos(α) t(s) + sin(α) b(s)) (s no arc length for K2).
Requirement 1: y ' ⊥ cos(α) t + sin(α) b.
Requirement 2: y " ⊥ cos(α) t + sin(α) b.
y ' = (1 + λ . cos(α)) t + (λκ cos(α) - λτ sin(α)) n + (λ . sin(α)) b.
With requirement 1) we get cos(α) + λ . = 0, dus λ = -s cos(α) + c1, and
y ' = sin2(α)) t + (-s cos(α) + c1)(κ cos(α) - τ sin(α)) n - sin(α)cos(α) b,
y " = (-κ t + τ b)(-s cos(α) + c1)(κ cos(α) - τ sin(α)) + (...)n.
With requirement 2) we get (κ cos(α) - τ sin(α)2(s cos(α) - c1) = 0, so τ/κ = cotg(α); so K1 is a helix.
(The lines with direction vector cos(α) t + sin(α) b are parallel to the axis; if K1 is a circular helix, then K2 is a circle (perpendicular to the axis) on the cylinder where K1 lies on.)