Find all functions f(t) such that x = (cost, sint, f(t)) is a plane curve

Question

Okay so I have a question

How do we find all function f(t) such that x = (cost, sint, f(t)) is a plane curve

I know this means the torsion is 0. So I know that we can find the pieces of the TNB needed from x and then find an equation that involves torsion (example t = N' • B).

However, my issue is that this curve is not unit speed.

So T is going to be

X'(t)/|x'(t)|

N is going to be

T'(t)/(|T'(t)|)•(|x'(t)|)

And then We need to find B but that cross product is going to look really ugly isn't it?

For reference T will be

(-sint, cost, f'(t))/sqrt(1+f'^2(t))

Then N will be

(-cost, -sint, f''(t))/(sqrt(1+f''^2(t))•(sqrt(1+f'^2(t))

And then I'll have to cross T and N to get B.

Thanks in advance

Answer 1

$r'(t)=(-\sin t,\cos t,f'(t))$
$r''(t)=(-\cos t,-\sin t,f''(t))$
$r'(t)\times r''(t)=(\sin(t) f'(t)+\cos(t)f''(t),-\cos tf'(t)+\sin t f''(t)),1)$
$r'''(t)=(\sin t,-\cos t,f'''(t))$

The formula for the torsion for any parameter is: $$\tau=\frac{r'(t)\times r''(t)\cdot r'''(t)}{|r'(t)\times r''(t)|^2}$$ So $\tau=0$ when $0=r'(t)\times r''(t)\cdot r'''(t)=f'(t)+f'''(t)$, i.e. $$f'''(t)=-f'(t)$$

Now solve for $f(t)$. The answer is an ellipse, which is the intersection of a plane with a cylinder.

Answer 2

Different approach:

Points $P$ are on a plane with normal $n$ when $n\cdot P=d$ .

Hence, for fixed constants $n_x,n_y,n_z,d$ we should have $n_z f(t)=d-n_x \cos(t) -n_y \sin(t)$