Find the rate of change of a particle moving along a parabola

Consider a particle moving along the parabola $y = x^2$ .

Find the point on the curve at which the rate of change of the abscissa and ordinate are changing at the same rate.
Find the rate in part (a) if at time $t$ we have $x = \sin t$ and $y = \sin^2 t$ .

Since $y = x^2$ we have
$\frac{dy}{dt} = 2x \frac{dx}{dt}.$

Thus, $\frac{dy}{dt} = \frac{dx}{dt}$ when $x = \frac{1}{2}$ . Then $y = x^2$ so $x = \frac{1}{2}$ implies $y = \frac{1}{4}$ .
If we have $x = \sin t$ and $y = \sin^2 t$ at time $t$ and we have $\frac{dx}{dt} =\frac{dy}{dt}$ then
$\begin{align*} x &= \frac{1}{2} &\implies &t = \sin^{-1} \left( \frac{1}{2} \right) \\ \frac{dx}{dt} &= \cos t &\implies &\frac{dx}{dt} = \cos \left( \sin^{-1} \left( \frac{1}{2} \right) \right) = \frac{\sqrt{3}}{2}. \end{align*}$