Given two points $(x_1,y_1)$ and $(x_2,y_2)$, find the curve $\gamma$ connecting them such that the surface area of the volume obtained when rotating the curve along the $x$-axis is minimized.

First assume that the curve is given by $(x,y(x))$. Then the surface described has area $$ 2\pi \int_{x_1}^{x_2} y(x)\sqrt{1 + \dot{y}(x)^2}dx = 2\pi \int_{x_1}^{x_2} F(y,\dot{y})dx.$$ The Euler-Lagrange equations tell us that such a minimizing curve satisfies $$\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial \dot{y}} = 0.$$ Now I can work out these derivates but the term $\frac{d}{dx}\frac{\partial F}{\partial \dot{y}}$ becomes a complete mess (in the sense that solving the DE that arises looks impossible). Is there another way to solve? Thanks in advance!

  • $\begingroup$ The resulting equation is a bit of a mess, but there is a 'standard' trick (multiply the resulting ODE by $\dot{y}(x)$ and simplify). $\endgroup$ – copper.hat May 11 '16 at 16:22
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    $\begingroup$ Minimizing $\int_a^b y(x)\sqrt{1+y'2(x)}\>dx$ under constraints leads to a catenary, hence we shall see a catenary surface. $\endgroup$ – Christian Blatter May 11 '16 at 17:45

You should be able to get down to


and I'm sure that somebody on this site can solve that.

What you need to do once you have calculated the derivatives is to multiply by $(1+y'(x)^2)^{3/2}$ to simplify.

EDIT: Actually consider the quotient rule:


with the role of $v$ played by $y(x)$ and that of $u$ played by $y'(x)$

so that we have

$$y\cdot y''-y'\cdot y'=1=\frac{y^2}{y^2}\Rightarrow \frac{d}{dx}\left(\frac{y'}{y}\right)=\frac{1}{y^2}.$$



thought I'd be able to go somewhere with that... as I said somebody will have no problem at all from here.

  • $\begingroup$ Ok thanks, so I should soldier through the equations and solve $\endgroup$ – Slugger May 11 '16 at 16:23
  • $\begingroup$ The solution usually boils down to a first order equation... $\endgroup$ – copper.hat May 11 '16 at 16:29
  • $\begingroup$ I solved the DE, check out my answer. $\endgroup$ – Slugger May 12 '16 at 10:30
  • $\begingroup$ @Slugger Nice; well done. $\endgroup$ – JP McCarthy May 12 '16 at 11:29

I figured out how to solve the DE

We can solve the DE $$\dot{y}^2-y\ddot{y}+1=0$$ using the substitution $u(x) = \dot{y}(x)$. Then the differential equation turns into \begin{eqnarray*}\label{zoveel} -y\frac{du}{dy}u +u^2 + 1 = 0. \end{eqnarray*} This can be seen by realizing that \begin{equation} \dot{u} = \frac{du}{dx} = \frac{du}{dr} \dot{r}. \end{equation} Rearranging yields \begin{equation} \frac{dy}{y} = \frac{udu}{1 + u^2}. \end{equation} Integration yields \begin{equation} \frac12 \ln |1 +u^2| = \ln |r| +C \end{equation} So that \begin{equation} 1 + \dot{y}^2 = Cy^2. \end{equation} Now we obtain \begin{equation} y' = \sqrt{Cy^2 - 1}, \end{equation} so that \begin{equation} \frac{dy}{\sqrt{Cy^2 - 1}} = \pm dx \end{equation} The solution to this equation is given by \begin{equation} y = \frac{1}{c_1} \cosh(c_1 x + c_2). \end{equation} Now we have two initial conditions, namely that $y(0) = y_0$ and $y(x_1) = y_1$. The first condition gives \begin{equation} c_1 = \frac{\cosh(c_2)}{y_0}. \end{equation} The second initial condition then says that \begin{equation} y_1 = y_0 \frac{\cosh(c_1 x_1 + c_2)}{\cosh(c_2)} \end{equation}


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