Planar Curve $(a\cos(\theta), a\sin(\theta), f(\theta))$

Question

Find $f$ such that following represents a planar Curve $(a\cos(\theta), a\sin(\theta), f(\theta))$ for parameter $\theta$.

I have a gut feeling that $f(\theta)= constant$ as otherwise it would become similar to helix for infinitesimal change in $\theta$, thus rendering the curve non planar. But I am not able to formally derive any result. Any hint is much appreciated. Thanks

Answer 1

That is not true. For instance, it is clear that$$\theta\mapsto\bigl(a\cos(\theta),a\sin(\theta),a\cos(\theta)\bigr)$$is also a plane curve. In fact, you can take $f(\theta)=\alpha+\beta\cos(\theta)+\gamma\sin(\theta)$ (and these are the only solutions).

Answer 2

Let

$$r(t)=\left(\,a\cos t\,,\,\,a\sin t\,,\,\,f(t)\,\right)\implies r'(t)=\left(\,-a\sin t,\,a\cos t,\,f'(t)\,\right),$$

$$r''(t)=\left(\,-a\cos t,\,-a\sin t,\,f''(t)\,\right)\,,\,\,r'''(t)=\left(\,a\sin t,\,-a\cos t,\,f'''(t)\,\right)\implies$$

$$ r'\times r''=\begin{vmatrix}i&j&k\\ -a\sin t&a\cos t&f'(t)\\ -a\cos t&-a\sin t&f''(t)\end{vmatrix}=\left(\,a(f''\cos t+f'\sin t)\,,\,\,a(f''\sin t-f'\cos t)\,,\,\,a^2\,\right)$$

so

$$(r'\times r'')\cdot r'''=\require{cancel}\cancel{a^2f''\sin t\cos t}+a^2f'\sin^2t-\cancel{a^2f''\cos t\sin t}+a^2f'\cos^2t+a^2f'''=$$

$$=a^2(f'+f''')$$

Since $\;r\;$ is planar iff $\;\tau=0\;$ (its torsion), we get that it is planar iff

$$f'+f'''=0\iff f''+f=C,\,\,C=\text{constant}$$

and now you solve this easy differential equation.