$\begingroup$Edit: I've revised this to clear up any confusion. The $T(0), N(0)$, etc were just to help me.
Suppose we have a space curve $r(t)=<3t,\cos t,\sin t>$. I'm asked to find the curvature at $P(0,1,0)$.
So, I do the following:
$r'(t)=<3,-sin(t),cos(t)>$
$||r'(t)||=\sqrt{(3^2)+(-sin(t))^2+(cos(t))^2}=\sqrt{10}$
$T(t)=\frac{r'(t)}{||r'(t)||} = \frac{<3,-sin(t),cos(t)}{\sqrt{10}}= \frac{1}{\sqrt10}(<3,-sin(t),cos(t)>)$
$T'(t)=\frac{1}{\sqrt10}(<0,-cos(t),-sin(t)>)$
$||T'(t)||=\sqrt{(\frac{1}{\sqrt10})^2+(0)^2+(-sin(t))^2+(cos(t))^2}=\sqrt{\frac{11}{10}}$
$k(curvature)=\frac{||T'(t)||}{||r'(t)||}=\frac{<0,\frac{-cos(t)}{\sqrt{10}},\frac{-sin(t)}{\sqrt{10}}>}{\sqrt{10}} = <0,\frac{-cos(t)}{10},\frac{-sin(t)}{10}>$
My problem is that I'm not really sure how to do it at $P(0,1,0)$. Usually it's just at $t=2$ or something similar. Sorry, it may seem like a silly question, but I really can't see where to plug in all 3 points. Do you just multiply $\frac{1}{10}*P(0,1,0)?$
$\endgroup$61 Answer
$\begingroup$Your curve is $r(t)=(3t,\cos(t),\sin(t))$. It takes a number $\mathbb{R}$ (like time) and "maps" it to $\mathbb{R}^3$ (i.e. 3D space). Think of it as the curve of an object traveling in space, say a missile or something. At time $t$, it is at point in space $r(t)$. So, if you ask for the curvature of a point $P\in\mathbb{R}^3$, it may not even make sense if the curve does not pass through that point. You first need to find when the missile passes through $P$, if it does at all.
In this case, where $P=(0,1,0)$, notice something special: when $t=0$, we have $r(0)=(0,1,0)$. So, the time when $r$ passes through $P$ is right at the start, at time $t=0$. So all the question is asking is what the curvature is at $t=0$. (I.e. what is $\kappa(0)$.)
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