Exponential Wire Dynamics

$y=e^{-x}\implies \dfrac{dy}{dx}=-e^{-x},\dfrac{d^2y}{dx^2}=e^{-x}$ .

So, $\cos α=\dfrac{1}{\sqrt {1+e^{-2x}}}, \rho =\dfrac{\left (1+(\frac{dy}{dx})^2\right )^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\dfrac{(1+e^{-2x})^{\frac{3}{2}}}{e^{-x}}$ ,

where $α$ is the angle subtended by the tangent to the curve $y=e^{-x}$ at $(x, e^{-x})$ with the negative $x$ -axis and $\rho$ is the radius of curvature of this curve at that point.

Normal reaction offered by the curve on the particle is

$N(x)=mg\cos α+\dfrac{mv^2}{\rho}$ .

The velocity $v$ of the particle at that point is determined by using the energy conservation principle :

$mg(1-y)=\dfrac{1}{2}mv^2\implies mv^2=2mg(1-e^{-x})\implies$

$N(x)=\dfrac{mg(1+2e^{-x}-e^{-2x})}{(1+e^{-2x})^{\frac{3}{2}}}=$

$\dfrac{mge^{2x}\sqrt {1+e^{-2x}}(e^{2x}+2e^x-1)}{(1+e^{2x})^2}$ .

So, $a=2,b=2,c=2,d=1$ and $a+b+c+d=\boxed 7$ .

In the limit when $x$ approaches infinity, $N(x)$ tends to $mg$ , which is quite natural, since in that limit the curve flattens to a straight line parallel to the $x$ -axis, assymptotically meeting the axis, so that curvature of the curve and hence the centrifugal force becomes zero, and the component of weight force generating the normal reaction is $mg$ .