Going Down the Exponential Slide

Relevant wiki: Conservation of Mechanical Energy

By conservation of energy, for a bead of mass $m$ starting at height $h = 1$ , we have that

$mg \times 1 = mgy + \dfrac{1}{2}mv^{2} \Longrightarrow v = \sqrt{2g(1 - y)}$ .

Now the incremental distance the bead travels along the wire is

$ds = \sqrt{dx^{2} + dy^{2}} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}} dx \Longrightarrow \dfrac{ds}{dx} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}}$ .

So the speed of the bead at any time will be $v = \dfrac{ds}{dt} = \dfrac{ds}{dx} \dfrac{dx}{dt} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}} \dfrac{dx}{dt}$ .

Equating this to $v = \sqrt{2g(1 - y)}$ we then have that $dt = \dfrac{1}{\sqrt{2g}} \sqrt{\dfrac{1 + (dy/dx)^{2}}{1 - y}} dx$ .

Then with $y = e^{-x} \Longrightarrow \dfrac{dy}{dx} = -e^{-x}$ , upon integration we have that

$\displaystyle\large t_{a,b} = \dfrac{1}{\sqrt{2g}} \int_{a}^{b} \sqrt{\dfrac{1 + e^{-2x}}{1 - e^{-x}}} dx$ ,

and so $P + Q + R + S = 1 + (-2) + (-1) + (-1) = \boxed{-3}$ .

In the resulting integral, the variable of integration is $x$ , so we will express everything in terms of $x$ . We already know $y(x) = e^{-x}.$ Since the only forces on the bead are the gravitational force and a (perpendicular) normal force, we apply conservation of mechanical energy. The initial mechanical energy is $E_0 = mg\cdot 1 = mg$ ; thus we have $mgy + \tfrac12mv^2 = mg\ \ \ \ \therefore\ \ \ \ v^2 = 2g(1-y),$ so that $v(x) = \sqrt{2g}\sqrt{1 - e^{-x}}.$ Now consider what happens as the bead slides over an infinitesimal horizontal distance $dx$ . It also moves vertically; using the derivative of $y(x)$ we have $dy = \frac{dy}{dx}dx = -e^{-x}dx.$ Thus the overall distance is $ds = \sqrt{dx^2 + dy^2} = dx\sqrt{1^2 + (-e^{-x})^2} = dx\sqrt{1 + e^{-2x}}.$ Now we can write an expression for the time it takes to slide this distance: $dt = \frac{ds}v = \frac{\sqrt{1+e^{-2x}}dx}{\sqrt{2g}\sqrt{1-e^{-x}}} = \frac{1}{\sqrt{2g}}\sqrt{\frac{1+e^{-2x}}{1-e^{-x}}}\ dx.$ The total time to slide from $x = a$ to $x = b$ is simply $t_{a,b} = \int_a^b dt,$ so when we substitute the expression we found for $dt$ we get the desired result: $t_{a,b} = \frac{1}{\sqrt{2g}}\int_a^b \sqrt{\frac{1+e^{-2x}}{1-e^{-x}}}\ dx.$ Comparing coefficients and exponents show immediately that $\boxed{P = 1,\ Q = -2,\ R = -1,\ S = -1}$