Mathematical Immaturity

14.9 Exercises

• This exercise describes how the unit tangent and the principal normal are affected by a change of parameter. Suppose a curve $C$ is described by two equivalent functions $X$ and $Y,$ where $Y(t) = X[u(t)].$ Denote the unit tangent for $X$ by $T_X$ and that for $Y$ by $T_Y.$ (a) Prove that at each point of $C$ we have $T_Y(t) = T_X[u(t)]$ if $u$ is strictly increasing, but that $T_Y(t) = -T_X[u(t)]$ if $u$ is strictly decreasing. In the first case, $u$ is said to preserve orientation; in the second case, $u$ is said to reverse orientation. (b) Prove that the corresponding principal normal vectors $N_X$ and $N_Y$ satisfy $N_Y(t) = N_X[u(t)]$ at each point of $C.$ Deduce that the osculating plane is invariant under a change of parameter.
• (a) We have $$ T_X(u) = \frac{X'(u)}{\|X'(u)\|} \quad \text{and} \quad T_Y(t) = \frac{u'(t)}{|u'(t)|}\frac{X'[u(t)]}{\|X'[u(t)]\|} $$ so it follows that if $u$ is strictly increasing, then $u'(t)/|u'(t)| = 1$ and $T_Y(t) = T_X[u(t)]$ and if $u$ is strictly decreasing, then $u'(t)/|u'(t)| = -1$ and $T_Y(t) = -T_X[u(t)]$ (b) Taking the derivative of $T_X$ with respect to $u,$ we get: $$ \frac{d}{du}T_X(u) = T'_X(u) $$ and the principal normal $N_X$ is $$ N_X = \frac{T'_X(u)}{\|T'_X(u)\|} $$ Taking the derivative of $T_Y$ with respect to $t,$ we get: $$ \frac{d}{dt} T_Y = u'(t)\frac{u'(t)}{|u'(t)|}T'_X[u(t)] $$ And thus the principal normal $N_Y$ is $$ \begin{align*} N_Y &= \frac{u'(t)^2}{u'(t)^2} \frac{T'_X[u(t)]}{\|T'_X[u(t)]\|} \\ &= \frac{T'_X[u(t)]}{\|T'_X[u(t)]\|}\\ &= N_X \end{align*} $$ This demonstrates that the osculating plane is invariant under a change of parameter because the linear span of $T_X$ and $N_X$ is the same as that of $T_Y$ and $N_Y.$ Moreover, the cross products $T_X \times N_X$ and $T_Y \times N_Y$ only differ (potentially) by sign.