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A curve $\alpha(t)=(r(t),z(t))$ in the $(r,z)$-plane, where $r(t)>0$, is rotated around the $z$-axis. We can parameterise it with $x(t,\phi)=(r(t) \cos\phi,r(t) \sin\phi,z(t))$ for $t \in (a,b)$ and $\phi \in (0,2\pi)$.

Now given the second fundamental form $(L_{ij})$

$$\frac{1}{\sqrt{\dot r^2+ \dot z^2}} \begin{pmatrix} \dot r \ddot z- \dot z \ddot r & 0\\ 0& r \dot z \end{pmatrix}$$

I've got to prove that $\det(L_{ij})=0 $ only then if every meridian is a straight line. Well, calculating the determinant, I get the equation $\dot z \ddot r= \dot r \ddot z$. Now I don't know how to go on. I'd be grateful for any help.

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  • $\begingroup$ Did not understand question. For a straight line meridian of a cone $ r = m z + c $ we have $ \dfrac{\dot r}{\dot z} = \dfrac{\ddot r}{\ddot z} = m $ by differentiation and quotient rule. $\endgroup$ – Narasimham Jan 10 '17 at 16:38
  • $\begingroup$ There is no real difficulty solving this separable equation. Where are you stuck ? $\endgroup$ – Yves Daoust Jan 10 '17 at 17:12
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Assume that : $$\dot r^2 + \dot z^2=1 \Rightarrow \dot r \ddot r + \dot z \ddot z =0$$ $$det(L_{ij})=0\Rightarrow (\dot r \ddot z-\dot z \ddot z)r\dot z = 0 $$ If $\dot z = 0$ we have : $$\dot z=0 \Rightarrow \dot r^2=1 \Rightarrow \dot r=\pm 1 $$ If $\dot z\neq 0$ we have : $$\dot r \ddot z - \dot z \ddot r=0 \Rightarrow \dot r =0 \ \vee \ \dot r \neq 0 $$ $$\dot r \neq 0 \Rightarrow \ddot z = \dot z(\frac{\ddot r}{\dot r ^2} ) \Rightarrow \ddot z =\dot z(\frac{- \dot z \ddot z}{\dot r ^2})\Rightarrow \ddot z= 0$$ sinc if $\ddot z \neq 0 \Rightarrow \dot r^2 + \dot z ^2 =0$ Contradiction

So if $\dot r \neq 0 \Rightarrow \ddot z =0 \Rightarrow \dot z =0 \Rightarrow \dot r = c$ $$ $$ Note : You can assume that : $\dot r^2 + \dot z ^2 = c$

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  • $\begingroup$ Not so. The quotient is constant $\dot r^2 / \dot z ^2 = c$ $\endgroup$ – Narasimham Aug 3 '17 at 5:27
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Hints: Using primes instead of dots to denote derivatives with respect to $t$, you have $$ r' z'' - r'' z' = 0\quad\text{for all $t$.} \tag{1} $$

  • If $r'z'$ is non-vanishing in some interval, (1) is equivalent to $$ \left[\log\left(\frac{z'}{r'}\right)\right]' = \frac{z''}{z'} - \frac{r''}{r'} = 0. $$

  • If $r'z' = 0$, either $r' = 0$ or $z' = 0$.

    Continuity and the first bullet point show that $r'z'$ cannot vanish on a set with empty interior if the profile curve $\alpha$ is regular.

    On the other hand, if $r'z'$ vanishes identically on some interval and $\alpha$ is regular, then either $r'$ or $z'$ vanishes identically.

(These claims leave non-trivial details to check, so I trust this outline doesn't spoil all your fun.)

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Assuming $\dot z\ne 0$, $$\frac{\ddot r\dot z-\dot r\ddot z}{\dot z^2}=\dot{\left(\frac{\dot r}{\dot z}\right)}=0,$$ then $$\dot z=a\dot r$$ and $$z=ar+b.$$

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Given that

$$ {\dot r}{\ddot z}={ {\ddot r}{\dot z}}, \quad \dfrac{\dot r}{\dot z} = \dfrac{\ddot r}{\ddot z} \tag1;$$

If the above is to hold good, by the Quotient Rule $ \dfrac{\dot r}{\dot z} $ has to be invariably constant, say $=m$, but not so for further integration.

EDIT1:

That is

$$ \dfrac{\dot z \ddot r -\dot r \ddot z} {\dot z^2}= 0,\quad \frac{d}{dt}(\dfrac{\dot r}{\dot z})=0,\quad \frac{\dot r}{\dot z} = \,const.= say\, m,\quad {\dot r}=m{\dot z}\tag2$$

it would result in the same.

We can rule out higher order quotient constancy, ie if $ \dfrac{\ddot r}{\ddot z} $ is constant then $\dfrac{\dot r}{\dot z} $ cannot be automatically constant due to appearance of arbitrary constants which in general need not vanish.

Integrating the last equation (first order relation of (2)) we obtain a straight line meridian for a frustum of a cone or full cone depending on what we choose for constant of integration $c$.

$$ r = m z + c. \tag3 $$

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