Question 9.3.2: The position of a moving particle is given by r(t) = 2 cos t......
The position of a moving particle is given by \mathrm{r}(t)=2 \cos t \mathrm{i}+2 \sin t \mathrm{j}+3 t \mathrm{k}. Find the vectors \mathrm{T}, \mathrm{N}, and \mathrm{B}. Find the curvature.
Since \mathrm{r}^{\prime}(t)=-2 \sin t \mathrm{i}+2 \cos t \mathrm{j}+3 \mathrm{k},\left\|\mathrm{r}^{\prime}(t)\right\|=\sqrt{13}, and so from (1) we see that a unit tangent is
\mathrm{T}(t)=-\frac{2}{\sqrt{13}} \sin t \mathrm{i}+\frac{2}{\sqrt{13}} \cos t \mathrm{j}+\frac{3}{\sqrt{13}} \mathrm{k}.
Next, we have
\frac{d \mathrm{T}}{d t}=-\frac{2}{\sqrt{13}} \cos t \mathrm{i}-\frac{2}{\sqrt{13}} \sin t \mathrm{j} \text { and }\left\|\frac{d \mathrm{T}}{d t}\right\|=\frac{2}{\sqrt{13}}.
Hence, (7) gives the principal normal
\mathrm{N}(t)=-\cos t \mathrm{i}-\sin t \mathrm{j}.
Now, the binormal is
\begin{aligned} \mathrm{B}(t)=\mathrm{T}(t) \times \mathrm{N}(t) & =\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ -\frac{2}{\sqrt{13}} \sin t & \frac{2}{\sqrt{13}} \cos t & \frac{3}{\sqrt{13}} \\ -\cos t & -\sin t & 0 \end{array}\right| \\ & =\frac{3}{\sqrt{13}} \sin t \mathrm{i}-\frac{3}{\sqrt{13}} \cos t \mathrm{j}+\frac{2}{\sqrt{13}} \mathrm{k} . \end{aligned}
Finally, using \|d \mathrm{T} / d t\|=2 / \sqrt{13} and \left\|\mathrm{r}^{\prime}(t)\right\|=\sqrt{13}, we obtain from (4) that the curvature at any point is the constant
\kappa=\frac{2 / \sqrt{13}}{\sqrt{13}}=\frac{2}{13}.