Exercise 2 Exercise 2. Show that T(t) and N(t) are orthogonal. (Hint: Differentiateboth sides of the expression (T(t), T(t)) = 1).So, if a is a planar curve, {T(t), N(t)} form a moving frame for R2, i.e.,any element of R2 may be written as a linear combination of T(t) and N(t)for any choice of t. In particular, we may express the derivatives of T and'Last revised: September 13, 20211N in terms of this frame. The definition of N already yields that, when a isparametrized by arclength,T'(t) = x(t)N(t).To get the corresponding formula for N', first observe thatN'(t) = aT(t) + bN(t).for some a and b. To find a note that, since (T, N) = 0, (T', N) = -(T, N').Thusa = (N'(t), T(t)) = –(T'(t), N(t)) = –x(t).

Answered: eorthogonal. (Hint: Differentiate…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Exercise 2

Expert Solution

Unit Tangent Vector

Given $α (t)$ is a planner curve. A unit tangent vector $T (t)$ to the planner curve $α (t)$ is defined as $T (t) = \frac{α' (t)}{||α' (t)||}$ and unit normal vector $N (t)$ is defined as $N (t) = \frac{T' (t)}{||T' (t)||}$ . Since $T (t) . T (t) = \frac{α' (t)}{||α' (t)||} . \frac{α' (t)}{||α' (t)||}$ therefore differentiating both side of it with respect to $t$ :

$\begin{array}{rcl} \frac{d}{d t} [T (t) . T (t)] & = & \frac{d}{d t} [\frac{α' (t)}{||α' (t)||} . \frac{α' (t)}{||α' (t)||}] \\ T' (t) . T (t) + T (t) . T' (t) & = & \frac{d}{d t} [\frac{α' (t) . α' (t)}{{||α' (t)||}^{2}}] \\ 2 T (t) . T' (t) & = & \frac{d}{d t} [\frac{{||α' (t)||}^{2}}{{||α' (t)||}^{2}}] \\ 2 T (t) . T' (t) & = & \frac{d}{d t} [1] \\ 2 T (t) . T' (t) & = & 0 \end{array}$