As Little Work As Possible

If $y\,:\, [0,\tfrac12\pi] \to \mathbb{R}$ is a smooth map such that $y(0) = 0$ and $y(\tfrac12\pi) = 1$ , then the work done moving the particle along the path defined by the function $y$ is $W[y] \; = \; \int_0^{\frac12\pi} e^{-y(x)} \,ds \; =\; \int_0^{\frac12\pi} e^{-y(x)} \sqrt{1 + \big(y'(x)\big)^2}\,dx$ The theory of Calculus of Variations tells us that the function $y$ which extremizes $W[y]$ must satisfy the Euler equation $\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{\partial F}{\partial y}\; = \; 0$ where $F(x,y,y') \; = \; e^{-y} \sqrt{1 + (y')^2}$ Thus $\begin{array}{rcl} \displaystyle \frac{d}{dx}\left(\frac{y' e^{-y}}{\sqrt{1 + (y')^2}}\right) + e^{-y}\sqrt{1 + (y')^2} & = & 0 \\ \displaystyle \frac{(y'' - (y')^2)e^{-y}}{\sqrt{1 + (y')^2}} - \frac{(y')^2y'' e^{-y}}{(1 + (y')^2)^{\frac32}} + e^{-y}\sqrt{1 + (y')^2} & = & 0 \\ \displaystyle \big(y'' - (y')^2\big)\big(1 + (y')^2\big) - (y')^2 y'' + \big(1 + (y')^2\big)^2 & = & 0 \\ \displaystyle y'' + 1 + (y')^2 & = & 0 \end{array}$ Putting $z = e^y$ , this equation becomes $z'' + z \; = \; 0$ and hence $z = a\sin x + b\cos x$ for some constants $a,b$ . Since $y(0) = 0$ and $y(\tfrac12\pi) = 1$ we deduce that $e^y\,=\, z \,=\, e\sin x + \cos x$ , and so the optimal curve is $y(x) \; = \; \ln\big(e\sin x + \cos x) \hspace{1cm} 0 \le x \le \tfrac12\pi\;.$ For this curve we calculate $\left(\frac{ds}{dx}\right)^2 \; = \; 1 + \left(\frac{dy}{dx}\right)^2 \; = \; \frac{e^2+1}{(e\sin x + \cos x)^2}$ and hence that $\frac{ds}{dx} \; = \; \frac{\sqrt{e^2+1}}{e\sin x + \cos x} \; = \; \sec(x-\alpha)$ where $\alpha = \tan^{-1}e$ .

Thus the length of the optimal curve is $\begin{array}{rcl} \displaystyle s & = & \displaystyle \int_0^{\frac12\pi} \sec(x - \alpha)\,dx \; = \; \Big[ \ln(\sec (x-\alpha) + \tan (x-\alpha)) \Big]_0^{\frac12\pi} \\ & = & \displaystyle \ln(\mathrm{cosec}\,\alpha + \cot\alpha) - \ln(\sec\alpha - \tan\alpha) \; = \; \ln\big[(\sec\alpha+\tan\alpha)(\mathrm{cosec}\,\alpha + \cot\alpha)\big] \\ & = & \displaystyle\ln \left( \frac{1 + e + e^2 + (1+e)\sqrt{e^2+1}}{e}\right) \end{array}$ If we write $s = 2\tanh^{-1}v$ then $v \; = \; \tanh\tfrac12s \; =\; \frac{e^s-1}{e^s+1} \; = \; \frac{1+e^2 + (1+e)\sqrt{e^2+1}}{(1+e)^2 + (1+e)\sqrt{e^2+1}} \; = \; \frac{\sqrt{e^2+1}}{e+1}$ so we deduce that $s \; = \; 2\tanh^{-1}\left(\frac{\sqrt{e^2+1}}{e+1}\right)$ making the answer $2 + 2 + 1 + 1 + 1 = \boxed{7}$ .