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I am trying to find the equation of the curve between two points in spherical co-ordinate whose length is shortest, i.e. find the equation of line in spherical co-ordinate system. Here is my work so far, enter image description here Did I do some mistake somewhere? Because I am stuck and don't know how to solve the differential equation, $$\frac{r}{\sqrt{r'^2+r^2}}-\frac{d}{d\theta}\frac{r'}{\sqrt{r'^2+r^2}} = 0$$ Please tell me how to proceed.

P.S. I made a mistake in the final equation, forgot the square-root and r prime. Please ignore.

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    Have you learned any spherical geometry? It might be a lot easier with that. I presume you mean the arc on the surface of a sphere, right? Yeah, it's not a differential equations problem. Nice differential equation though. You did a nice job setting up an alternate way to do it. – 2017-01-24
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    You ask for an equation in spherical coordinates but you use only two coordinates. Did you actually want the equation in _polar_ coordinates? – 2017-01-24
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    @TheGreatDuck This is a problem from calculus of variation exercise and the question asked to solve using Euler-Lagrange equation. No, the arc is not any sphere, it is like any other function in polar co-ordinate. – 2017-01-24
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    @DavidK Yes. I confuse a lot between spherical and polar co-ordinates. In my mind, they are essentially same with polar being a 2D version for spherical co-ordinates. – 2017-01-24
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    @Ayatana polar coordinates has a normal 3D extension which is just by adding z. Spherical is not a 3D version of polar coordinates. – 2017-01-24

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Why not simply $$ ax+by+c=0 \implies ar\cos\theta+br\sin\theta+c=0\implies r=-\frac{c}{a\cos\theta+b\sin\theta} $$

Anyway, you should perform the derivative, obtaining $$ \frac{r}{\sqrt{r'^2+r^2}}-\frac{r''}{\sqrt{r'^2+r^2}}+\frac{r'(2r'r''+2rr')}{2\sqrt{(r'^2+r^2)^3}} = 0 $$ and further simplify to $$ -\frac{r(rr''-2r'^2-r^2)}{\sqrt{(r'^2+r^2)^3}} = 0 $$

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    I understand that there are other ways to find the equation of line, but this is a problem from the exercise of Euler-Lagrange equation. The question is to solve using Variational Calculus. – 2017-01-23
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    @Ayatana: see edit – 2017-01-23
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    I solved $-\frac{r(rr''-2r'^2-r^2)}{\sqrt{(r'^2+r^2)^3}} = 0$ using Mathematica and I got $\frac{c_2}{\sec(\theta+c_1)}$, which is indeed a straight line in polar co-ordinates. But you have any idea how to solve that differential equation by hand? – 2017-01-24
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    @Ayatana: with the substitution $u=r'/r$ you get $u'=u^2+1$ – 2017-01-24