Answer:
To prove the Frenet-Serret formulas, we start with a curve in three-dimensional space parameterized by the arc length, denoted as r(s) = (x(s), y(s), z(s)), where s is the arc length parameter.
We define three vectors associated with the curve:
1. Tangent vector T(s) = r'(s) represents the instantaneous direction of the curve at any point.
2. Principal normal vector N(s) is defined as N(s) = T'(s) / ||T'(s)||, where || || denotes the magnitude or length of the vector. N(s) represents the direction of the instantaneous curvature of the curve.
3. Binormal vector B(s) = T(s) x N(s) represents a vector perpendicular to both T(s) and N(s), completing the orthogonal frame.
Now we can derive the Frenet-Serret formulas:
1. Derivative of T(s):
Taking the derivative of T(s) = r'(s) with respect to s, we have:
T'(s) = r''(s)
2. Derivative of N(s):
Taking the derivative of N(s) = T'(s) / ||T'(s)||, we have:
N'(s) = (T''(s)||T'(s)|| - T'(s)(T''(s) · T'(s))) / ||T'(s)||²
3. Orthogonality of T(s), N(s), and B(s):
We can show that T'(s) · N(s) = 0, which means that T'(s) is perpendicular to N(s).
Similarly, N'(s) · N(s) = 0, showing that N'(s) is also perpendicular to N(s).
Because T(s), N(s), and B(s) form an orthogonal frame, we have B'(s) · T(s) = 0 and B'(s) · N(s) = 0.
4. Derivative of B(s):
Using the orthogonality relationships, we can express B'(s) in terms of T(s) and N(s):
B'(s) = αT(s) + βN(s), where α and β are some scalar functions to be determined.
Taking the dot product of B'(s) with T(s), we get:
0 = B'(s) · T(s) = α(T(s) · T(s)) + β(N(s) · T(s)) = α
Similarly, taking the dot product of B'(s) with N(s), we have:
0 = B'(s) · N(s) = α(T(s) · N(s)) + β(N(s) · N(s)) = β
Therefore, α = 0 and β = 0, which implies B'(s) = 0.
In conclusion, the Frenet-Serret formulas are:
T'(s) = κ(s)N(s)
N'(s) = - κ(s)T(s) + τ(s)B(s)
B'(s) = - τ(s)N(s)
where κ(s) represents the curvature of the curve at the point parameterized by s, and τ(s) represents the torsion of the curve at that point.