Author Topic: Quiz-B P1  (Read 3453 times)

Victor Ivrii

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Quiz-B P1
« on: April 02, 2018, 08:25:31 PM »
If the surface is a surface of revolution $z=u(r)$ with $r^2=x^2+y^2$, then

S=2\pi\int_{D} \sqrt{1+u_r^2}\,rdr .
\tag{1}

Write Euler-Lagrange equation and solve it (find general solution).

Jingxuan Zhang

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Re: Quiz-B P1
« Reply #1 on: April 02, 2018, 09:33:04 PM »
Define
$$L(r,u_r)=r\sqrt{1+u_r^2}.$$
Then E.-L. $$(L_{u_r})_r=L_u$$ gives
$$\label{1-1}\Bigl(\frac{ru_r}{\sqrt{1+u_r^2}}\Bigr)_r=0,$$
from which we derive
$$\frac{ru_r}{\sqrt{1+u_r^2}}=A\implies u_r^2(r^2-A^2)=A^2\implies u_r=\frac{A}{\sqrt{r^2-A^2}}\implies u=A\cosh^{-1}(r/A)+B.$$
« Last Edit: April 03, 2018, 07:52:28 AM by Victor Ivrii »

Victor Ivrii

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Re: Quiz-B P1
« Reply #2 on: April 04, 2018, 06:53:02 AM »
Typical blinder: after $L_u=0$ instead of integrating $(L_{u'})'=0\implies L_{u'}=c$ started to really differentiate $L_{u'}$, getting to $u''$...

Error $u'L_{u'}-L=c$ : it would work if $L$ did not depend on the argument $r$, but here it did not depend on function $u$.

Bonus:
The same problem (surface of revolution) but now argument is  $z$ (used to be $u$) and the function is $r=R(z)$.
Then
$$S=2\pi \int_{z_0}^{z_1} \sqrt{1+R'^2} R\,dz.$$
Solve it! (the answer should be indeed the same as in the original problem)
« Last Edit: April 04, 2018, 06:56:52 AM by Victor Ivrii »

Andrew Hardy

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Re: Quiz-B P1
« Reply #3 on: April 04, 2018, 11:12:31 AM »

My Lagrangian is $$\sqrt{(1+R'^2)}R$$
so my Euler Lagrangian equations are

$$\frac{\partial L}{\partial R'} = \frac{ R R'}{\sqrt{1+R'^2}}$$
$$\frac{\partial L}{\partial R} = \sqrt{1+R'^2}$$
$$\frac{\partial L}{\partial z} = \frac{R R''} {\sqrt{1+R'^2}} + \frac{R'^2}{\sqrt{1+R'^2}} - \frac{R'^2R''R}{(1+R'^2)^{3/2}}$$

I'm not sure how to get to the solution you're looking for though. The argument isn't explicitly in the Lagrangian so I don't know how I'm going to integrate.
« Last Edit: April 04, 2018, 12:50:40 PM by Andrew Hardy »

Victor Ivrii

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Re: Quiz-B P1
« Reply #4 on: April 04, 2018, 11:15:56 AM »
Definitely not. What is E-L equation? Does $L$ depend on $z$ (you denoted it by $x$) explicitly?

Andrew Hardy

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Re: Quiz-B P1
« Reply #5 on: April 04, 2018, 11:56:26 AM »
I think I fixed my EL equations, but I still see no explicit dependence on z

Victor Ivrii

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Re: Quiz-B P1
« Reply #6 on: April 04, 2018, 12:08:50 PM »
No, you need to solve them, and you need to use a property mentioned above. And please change the colour: red belongs to admins/mods

Andrew Hardy

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Re: Quiz-B P1
« Reply #7 on: April 04, 2018, 12:58:00 PM »
Via the Energy Equation,

$$\frac{ R R'^2}{\sqrt{1+R'^2}} - \sqrt{(1+R'^2)}R = c$$
$$R(R'^2-R'^2-1) = c \sqrt{(1+R'^2)}$$

$$R = -c \sqrt{(1+R'^2)}$$
$$R^2 = c^2(1+(\frac{dR}{dz})^2)$$
$$R^2 - c^2 = c^2(\frac{dR}{dz})^2$$
$$c\frac{dR}{dz} = \sqrt{R^2 - c^2}$$
$$dz = \frac{cdR}{\sqrt{R^2 - c^2}}$$
$$z = c \text{ arcosh}(R/c) + b$$
« Last Edit: April 04, 2018, 01:19:46 PM by Andrew Hardy »