mark
Compute the TNB frame for the helix
\vec{p}(t) = \langle 2\cos(t), 3t, 2\sin(t) \rangle.
Archived May, 2026.
TNB frame for a helix
User 001
\begin{aligned}
\mathbf{p}(t) &= (2\cos t,\, 3t,\, 2\sin t) \\[2mm]
\mathbf{T}(t) &= \frac{\mathbf{p}'(t)}{\|\mathbf{p}'(t)\|} \\[1mm]
\mathbf{T}(t) &= \frac{1}{\sqrt{13}}\,(-2\sin t,\, 3,\, 2\cos t) \\[2mm]
\mathbf{N}(t) &= \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \\[1mm]
\mathbf{N}(t) &= \frac{\frac{1}{\sqrt{13}}(-2\cos t,\, 0,\, -2\sin t)}{2/\sqrt{13}} \\[1mm]
\mathbf{N}(t) &= (-\cos t,\, 0,\, -\sin t) \\[2mm]
\mathbf{B}(t) = \mathbf{T}(t)\times \mathbf{N}(t) &=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2\sin t & 3 & 2\cos t \\
-\cos t & 0 & -\sin t
\end{vmatrix} \\[1mm]
&= (-3\sin t,\, -2\sin^2 t + 2\cos^2 t,\, 3\cos t) \\[1mm]
\mathbf{B}(t) &= (-3\sin t,\, -2,\, 3\cos t)
\end{aligned}
... i think
❤️ 1
User 001
Forgot something, here to fix the mistake.
Earlier I wrote vector \langle-3\sin(t),\,-2,\,3\cos(t)\rangle, but I forgot the normalization from T(t), which is 1/\sqrt{13}.
Therefore I believe the correct answer should have been
\displaystyle \left\langle-3\sin t,\, -2,\, 3\cos t\right\rangle \cdot \frac{1}{\sqrt{13}}