Question

Show that the curvature \(\kappa \) is related to the tangent and normal vectors by the equation

\(\frac{{dT}}{{ds}} = \kappa N\)

Short answer

The relation among curvature \(\kappa \) , tangent vector \(T\) , and normal vector \(N\) is \(\frac{{dT}}{{ds}} = \kappa N\) .

Step-by-step solution

Expression

curvature\(\left( \kappa \right)\).

\(\kappa = \left| {\frac{{dT}}{{ds}}} \right|\)

Calculation

Multiply and divide by\(dt\)on right hand side of equation.

\(\begin{aligned}{l}\kappa = \left| {\left( {\frac{{dT}}{{ds}}} \right)\left( {\frac{{dt}}{{dt}}} \right)} \right|\\ = \left| {\frac{{\left( {\frac{{dT}}{{dt}}} \right)}}{{\left( {\frac{{ds}}{{dt}}} \right)}}} \right|\\ = \frac{{\left| {\frac{{dT}}{{dt}}} \right|}}{{\left( {\frac{{ds}}{{dt}}} \right)}}\end{aligned}\)

Re-arrange the equation.

\(\begin{aligned}{l}\kappa \left( {\frac{{ds}}{{dt}}} \right) = \left| {\frac{{dT}}{{dt}}} \right|\\\left| {\frac{{dT}}{{dt}}} \right| = \kappa \left( {\frac{{ds}}{{dt}}} \right)\end{aligned}\)

Write the expression for normal vector\(\left( N \right)\).

\(N = \frac{{\frac{{dT}}{{dt}}}}{{\left| {\frac{{dT}}{{dt}}} \right|}}\)

Substitute\(\kappa \left( {\frac{{ds}}{{dt}}} \right)\)for\(\left| {\frac{{dT}}{{dt}}} \right|\),

\(\begin{aligned}{l}N = \frac{{\frac{{dT}}{{dt}}}}{{\kappa \left( {\frac{{ds}}{{dt}}} \right)}}\\\kappa N = \left( {\frac{{dT}}{{dt}}} \right)\left( {\frac{{dt}}{{ds}}} \right)\\\kappa N = \frac{{dT}}{{ds}}\\\frac{{dT}}{{ds}} = \kappa N\end{aligned}\)

Thus, the relation among curvature \(\kappa \) , tangent vector \(T\) , and normal vector \(N\) is \(\frac{{dT}}{{ds}} = \kappa N\) .