- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.15 Exercises
- If a point moves so that the velocity and acceleration vectors always have constant lengths, prove that the curvature is constant at all points of the path. Express this constant in terms of $\|\mathbf{a}\|$ and $\|\mathbf{v}\|.$
- Use the definition of acceleration from Equation 14.9. $$ \mathbf{a}(t) = v'(t)T(t) + v(t)T'(t) \tag{14.9} $$ If velocity has a constant length, what does this imply about the accleration vector?
- Recall the component-wise definition of acceleration from Equation 14.9 $$ \begin{align*} \\ \mathbf{a}(t) = v'(t)T(t) + v(t)T'(t) \tag{14.9} \end{align*} $$ Then, since velocity has a constant length, ie $v'(t) = 0,$ we can see that acceleration is a constant scalar multiple of $T'(t):$ $$ \begin{align*} \\ \mathbf{a}(t) &= v(t)T'(t) \\ \end{align*} $$ If we take the length of both sides of the equation, noting that acceleration also has a constant length, we get $$ \begin{align*} \\ \|\mathbf{a}\| &= \|\mathbf{v}\|\|T'(t)\| \\ \end{align*} $$ But, we know by definition that $\kappa(t) = \|T'(t)\|/v(t),$ giving us $$ \begin{align*} \\ \|\mathbf{a}\| &= \|\mathbf{v}\|^2\kappa(t) \\ \end{align*} $$ Rearranging terms, we can express curvature for all $t$ in terms of the constants $\|\mathbf{a}\|$ and $\|\mathbf{v}\|$ $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}\|}{\|\mathbf{v}\|^2} \quad \blacksquare \end{align*} $$