Question

Differential Geometry ( Work Shop for Test 1)

(5) Prove that a regular curve (i.e., curve with positive curvature at all points) is a helix iff the ratio of the torsion to curvature is a constant. please use Differential Geometry Form not Calculus.

Answer 1

Let α(t)=(at,bt2,ct3), α(t)=(at,bt2,ct3) is a helix ⟺⟺ 4b2=9a2c24b2=9a2c2.
I think τ/κ must be constant that is torsion to the curvature is constant.

so lets prove that α(t)=(at,bt2,ct3) is a regular curve

Helices are defined by requiring that the unit tangent makes a constant angle with a fixed line in space. It is possible to prove that this implies that

τ(t)/k(t)=cotατ(t)/k(t)=cotα

and thus, for any helix,the ratio of curvature to torsion (the pitch) is constant. When calculating the curvature and torsion

I got:

29c2a2t2+b2a2+9b2c2t4−−−−−−−−−−−−−−−−−−−√(4b2t2+9c2t4+a2)3/229c2a2t2+b2a2+9b2c2t4(4b2t2+9c2t4+a2)3/2

And for the torsion:

3abcb2a2+9c2a2t2+9b2c2t43abcb2a2+9c2a2t2+9b2c2t4

The ratio is:

τ(t)k(t)=3abc(4b2t2+(a2+9c2t4))3/22(9a2c2t2+b2(a2+9c2t4))3/2τ(t)k(t)=3abc(4b2t2+(a2+9c2t4))3/22(9a2c2t2+b2(a2+9c2t4))3/2

In the above computation I assumed that a, b, c and t are real values. Except if I did a computational mistake (which is possible) I don't think the curve is an helix when the condition 4b2=9a2c24b2=9a2c2 is satisfied. In principle you should differentiate the ratio τ(t)k(t)τ(t)k(t) and impose that the results is zero.

As Upax already calculated

τ(t)k(t)=3abc(4b2t2+(a2+9c2t4))3/22(9a2c2t2+b2(a2+9c2t4))3/2.τ(t)k(t)=3abc(4b2t2+(a2+9c2t4))3/22(9a2c2t2+b2(a2+9c2t4))3/2.

This can be simplified to

τ(t)k(t)=3abc2b2⎛⎝4b2t2+(a2+9c2t4)9a2c2b2t2+(a2+9c2t4)⎞⎠3/2.τ(t)k(t)=3abc2b2(4b2t2+(a2+9c2t4)9a2c2b2t2+(a2+9c2t4))3/2.

Since you have quotient of two polynomials of degree 4 it is a constant iff they are proportional. Since the coefficients of the highest order are equal (9c29c2 in both cases) they have to be equal, therefore the curve is helix iff

4b2=9a2c2b24b2=9a2c2b2

which simplifies to what you need. Btw, there is a typo in the question.