Calculus problem #2189

We have $\frac{dx}{dt}=-4 \sin t$ , $\frac{dy}{dt} = 2 \cos t$ , $\frac{d^2x}{dt^2}=-4 \cos t$ , and $\frac{d^2y}{dt^2} = -2 \sin t$ . Thus, the velocity and acceleration at $t=\frac{\pi}{4}$ are $\overrightarrow{v} = \left(\frac{dx}{dt}, \frac{dy}{dt} \right) = (-4 \sin t, 2 \cos t) = (-2\sqrt{2}, \sqrt{2})$ and $\overrightarrow{\alpha} = \left(\frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right) = (-4 \cos t, -2 \sin t) = (-2\sqrt{2}, -\sqrt{2})$ , respectively.

Thus, the magnitudes of velocity and acceleration at $t=\frac{\pi}{4}$ are $\left| \overrightarrow{v} \right| = \sqrt{(-2\sqrt{2})^2+(\sqrt{2})^2} = \sqrt{10}$ and $\left| \overrightarrow{\alpha} \right| = \sqrt{(-2\sqrt{2})^2+(-\sqrt{2})^2} = \sqrt{10}$ , respectively.

Therefore, $m^2=10$ and $n^2=10$ , hence $m^2+n^2=20$ .