A classical mechanics problem by Steven Chase

The first derivative of the function is:

$\large{\frac{dy}{dx} = e^{-x/10}\Big[-sin(x) - \frac{1}{10} cos(x)\Big]}$

The second derivative of the function is:

$\large{\frac{d^2y}{dx^2} = e^{-x/10}\Big[-cos(x) + \frac{1}{10} sin(x) + \frac{1}{10}sin(x) + \frac{1}{100}cos(x)\Big] \\ e^{-x/10}\Big[\frac{1}{5} sin(x) - \frac{99}{100}cos(x)\Big]}$

Determine where the derivative equals zero:

$\large{-sin(x) - \frac{1}{10} cos(x) = 0 \implies x = atan(-\frac{1}{10}) \pm 2\pi n}$

The $x$ -coordinate of the peak near $x = 4 \pi$ is:

$\large{x_p = atan(-\frac{1}{10}) + 4\pi}$

In general, there may be both normal and tangential forces at this point. We can rule out the presence of a tangential force using the following argument (note that the velocity is purely horizontal at the peak):

$\large{Kinetic \, Energy = E = \frac{1}{2}m v^2 = mg(1 - y) \\ \frac{dE}{dt} = -mg \frac{dy}{dt} = 0 (for \, x = x_p) \\ \frac{dE}{dt} = \frac{dE}{dv} \frac{dv}{dt} = mv \frac{dv}{dt} = 0 \implies \frac{dv}{dt} = 0 \implies F_{tangential} = 0 }$

So the net force is equal to the normal force at the peak. We can use the familiar equation:

$\large{F_{normal} = m \frac{v^2}{R}}$

Except that here, $R$ is the radius of curvature at that point. The general formula is given below (the apostrophes denote derivatives):

$\large{R_{curvature} = \Big|\frac{(1+y'^2)^{3/2}}{y''}\Big|}$

Since the first derivative is zero, this reduces to:

$\large{R_{curvature} = \Big|\frac{1}{y''}\Big|}$

The normal force is then:

$\large{F_{normal} = m v^2 |y''| = 2mg(1-y)|y''|}$

Using the above expressions for $y$ and $y''$ and plugging in numbers gives $F_{normal} = F_{total} \approx \boxed{4.125}$ . I also confirmed this result by numerically differentiating the velocity.