Mathematical Immaturity

14.7 Exercises

• A particle moves along the parabola $x^2 + c(y - x) = 0$ in such a way that the horizontal and vertical components of the acceleration vector are equal. If it takes $T$ units of time to go from the point $(c, 0)$ to the point $(0, 0),$ how much time will it require to go from $(c, 0)$ to the halfway point $(c/2, c/4)?$
• Note that $(x')^2 + xx'' = (xx')'.$ What does it mean if $(xx')' = 0?$
• We start by differentiating the equation of the parabola with respect to the parameter $t,$ recalling that $x'' = y''.$ $$\frac{d}{dt}\left[x^2 + c\left(y - x\right)\right] = 2x\frac{dx}{dt} + c\frac{dy}{dt} - c\frac{dx}{dt} = 0$$ Differentiating once more with respect to $t,$ recalling that $x'' = y'':$ $$ \begin{align*} \frac{d}{dt}\left[2x\frac{dx}{dt} + c\frac{dy}{dt} - c\frac{dx}{dt}\right] &= \frac{d}{dt}\left[2x\frac{dx}{dt}\right] + \frac{d}{dt}\left[c\frac{dy}{dt} - c\frac{dx}{dt}\right]\\ &= 2\left(\frac{dx}{dt}\right)^2 + 2x\frac{d^2x}{dt^2}\\ &= 0 \end{align*} $$ But the above result implies that $$ \begin{align*} \frac{d}{dt}\left[x\frac{dx}{dt}\right] &= 0 \end{align*} $$ which would imply that $x\frac{dx}{dt} = k$ for some real $k.$ If we separate variables and integrate both sides, using our initial condition that it takes $T$ time units for $x$ to move from $c$ to $0$ then: $$ \begin{align*} \int_c^0x\,dx = \int_0^T k\,dt \end{align*} $$ Or in other words, $$ k = \frac{-c^2}{2T} $$ Using this, we now calculate $$ \begin{align*} \int_c^{c/2}x\,dx = \int_0^{T_0} k\,dt \end{align*} $$ Which gives us: $$ \begin{align*} \frac{c^2}{2} - \frac{c^2}{8} = \frac{c^2}{2T}T_0 \end{align*} $$ or in other words, that $T_0 = \frac{3}{4}T.\quad\blacksquare$