a) Find the unit tangent, normal, binormal vectors T N B and the curvature and torsion at a general point on the following curves;
r = $t$i + $\frac{t^2}{2}$j + $\frac{t^3}{3}$k, ($0 \le t \le 1$).
(Note: the torsion $\tau$ is defined as $d$B$/ds$ = $\tau$N)
b) Determine the point where the curve has the minimum torsion, and write the form of the
osculating plane and osculating circle of the curve through that point.
I feel comfortable with part a, my only problem is dealing with part b. Not entirely sure how to find the minimum torsion and go about the rest of the problem. Any help and/or solution would be much appreciated. Feel free to answer part a as well, it would be nice to compare answers and see if I'm actually doing these problems properly.
Best Answer
For part (a)
Computing $\bf T$, $\bf N$ and $\bf B$ from their definition might be messy. But using
$$\kappa = \frac{\| r' \times r'' \|}{\left\| {r'} \right\|^3}$$
$$\tau = \frac{\left( {r' \times r''} \right)\cdot r'''}{\left\| {r' \times r''} \right\|^2}$$
Should be simple enough and then use the Frenet-Serret equations to back calculate $\bf N$ and $\bf B$. I think $\bf T$ is simple enough by a direct computation.
For part (b) I got
$$\tau = \frac{2}{t^4+4t^2+1}$$
(Double check I did it quickly). To find the minimum take the derivative and set it to zero and solve for t. And you get t = CP ( the critical point).
And then the osculating circle... You must have some formula or form of a circle to use here. You know that the point ${\bf r}(CP)$ is on the circle and that the radius is $$R = \frac{1}{\kappa(CP)}$$
The osculating circle is in the plane of ${\bf T}$ and ${\bf N}$. So one formula for your circle could be...
$${\bf r}(t) = R{\bf N}(CP)+ {\bf r}(CP) + {\bf N}(CP) \cdot \cos(t) + {\bf T}(CP) \cdot \sin(t) $$
Check your book or notes, there may be another equation.