Question
Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_{0}\left( \theta_{0} \ll 1 \right)$ with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is:
(take $\theta(x) = sin\theta(x) = tan\theta(x) = \frac{dy}{dx},g$ is the acceleration due to gravity)
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Solution
ROC $=$ Radius of curvature at point $A$
$$\begin{matrix} & \text{~}\text{Curvature}\text{~} = \frac{1}{ROC} = \frac{\left| \frac{d^{2}y}{dx^{2}} \right|}{\left( 1 + \left( \frac{dy}{dx} \right)^{2} \right)^{\frac{3}{2}}} = \frac{\left| \frac{d^{2}y}{dx^{2}} \right|}{(1 + 0)^{\frac{3}{2}}} = \frac{d^{2}y}{dx^{2}}\ \left\lbrack \because\frac{dy}{dx} = tan\theta \simeq 0 \right\rbrack \\ & \Delta P = S \times \text{~}\text{curvature}\text{~} \\ & \ \Rightarrow \rho gy = S\frac{d^{2}y}{dx^{2}} \\ & \ \therefore\frac{d^{2}y}{dx^{2}} = \frac{\rho gy}{S} \end{matrix}$$
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