Finding arc length and binormal vector for a given curve Can somebody show me the arc length of a curve formula, and the binormal vector formula. The curve C with equation $r(t)=(\sqrt{3}\cos t,t,\sqrt{3}\sin t)$
How do you find the arc length from $t=0$ to $t=\dfrac{\pi}{2}$ and the binormal vector at $t=\dfrac{\pi}{2}$.
 A: *

*Length of $r$
for the length of the curve $r$ you just need to apply the definition found here, i.e.
$$L(r)=\int_0^{\frac{\pi}{2}}\sqrt{\|r'(t)\|^2}dt,$$
denoting by $r'(t)$ the fist derivative of your curve at 'time' $t$, i.e.
$$r'(t)=(-\sqrt{3}\sin t, 1, \sqrt{3}\cos t), $$
which implies $\|r'(t)\|^2=3+1=4$. Integration is now straighforward.

*

*Binormal vector $B(t)$
The binormal vector $B(t)$ to the curve $r$ at $t$ is defined as
$$B(t)= r(t)\times n(t), $$
where $n(t)$ is the normal vector of $r$ at $t$: it is given by
$$n(t)=\frac{q(t)'}{\|q'(t)\|},$$
where $q(t):=\frac{r'(t)}{\|r'(t)\|}$ is the unit tangent vector at $t$.
By definition, the binormal vector $b(t)$ is orthogonal to the plane spanned by $r(t)$ and $n(t)$. Let us compute it.
From the computations of the above section, we have
$$q(t)=\frac{1}{2}(-\sqrt{3}\sin t, 1, \sqrt{3}\cos t),$$
which implies
$$q'(t)=\frac{1}{2}(-\sqrt{3}\cos t, 0, -\sqrt{3}\sin t) $$
$$\|q'(t)\|^2=\frac{3}{4},$$
and
$$n(t)=\frac{1}{\sqrt{3}}(-\sqrt{3}\cos t, 0, -\sqrt{3}\sin t).$$
We are left with
$$B(t)=r(t)\times n(t)=\frac{1}{\sqrt{3}}(-\sqrt{3}t\sin t, 0, \sqrt{3}t \cos t),$$
which follows from using the definition of cross product, as explained here.
A: And for the binormal, use the formula for $\mathbf B(t)$ given here.
