#unit tangent vector T and curvature
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do the same thing I said for the last time you asked this
t not x, and assuming sec(t) > 0
so $T=\left(\frac{1}{\abs{\sec(t)}},\frac{-\tan(t)}{\abs{\sec(t)}}\right)$
Omegabet_
Oh it's - sin t right?
as your calc1 class would tell you, yes
So must put modulus?
well you havent specified the domain of the curve
actually sorry, you require cos(t) > 0, so yes, it's just sec(t)
but anyway you get $T=(\cos(t),-\sin(t))$
Omegabet_
How do you get -sin t though? From -tan^2 (t) / sec (t) + sec (t)
Oh
so it's easy to see T'=(-sin(t),-cos(t)), then the formula for curvature computes curvature
You have $\kappa=\frac{\norm{r'\times r''}}{\norm{r'}^3}$
Oh
Omegabet_
But since we already have unit tangent vector T, so we can just use |T'(t) | / | r'(t) |?
From my understanding
yeah you can use that formula too
I just know for arbitrary regular curves kappa has that formula 
but yes, by Frenet, $T'=\kappa v N\to \norm{T'}=\kappa v\norm{N}=\kappa v$, so $\kappa=\frac{\norm{T'}}{\norm{T}}$
Omegabet_
since N is unit
Just curious if I use the other formula
What does it mean
I kinda know it, but I don't know how to use it if I got only two component, usually it has 3 ( x, y, z) right?
Oh okayh
The answer for curvature, is it correct?
looks correct
Oh alright then thanks for being patient with me btw, I struggle with calculus actually, so I appreciate your kindness 😊