Answer to Question #138850 in Differential Geometry | Topology for Mubina

Question

Prove them

K=|r'×r"|/|r'|^3

T=[r', r",r"]/K^2(r')^6

Answer 1

We express r'(t) and r"(t) in terms of T and T'.Then compute their cross product.

Computation of r'. Since T=r'/||r'|| and ds/dt=||r'|| we get that: r'=(ds/dt)*T

Computation of r'' Taking the derivative with respect to t of the previous formula gives us: r''=(d²s/dt²)*T+ (ds/dt)*T'

Computation of r'(t)×r"(t). From 2 previous formulas and using the properties of cross products, we see that: r'×r"=(ds/dt)*(d²s/dt²)(T×T)+(ds/dt)²(T×T')

Since the cross product of a vector by itself is always the zero vector, we see that: |r'×r"|=(ds/dt)²|T×T'|=(ds/dt)²|T||T'|sinf

Where f is the angle between T and T'. Since |T|=1, we know that T is perpendicular to T'

thus: |r'×r"|=(ds/dt)²|T'|=|r'|²|T'|

Therefore: |T'|=(|r'×r"|)/|r'|²

so

K=|T'|/|r'|

K=|r'×r"|/|r'|³