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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!

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  • $\begingroup$ 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
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You should be able to get down to

$$y'(x)^2-y(x)y''(x)+1=0,$$

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:

$$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}$$

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

so that we have

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

Therefore

$$\frac{d^2}{dx^2}\ln(y)=y^{-2}...$$

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

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  • $\begingroup$ 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
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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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