How would you solve this equation by graphing? Or any other method?

Question

1. A particle is moving so that its position at time $t$ is given by the parametric equations \begin{align*} x &= 5\sin(-2t) \\ y &= 5\cos(2t). \end{align*}What is the speed of the particle?

2. A particle's position in the plane $t$ seconds after it starts moving is\begin{align*} x &= 6\cos(kt),\\ y &= 6\sin(kt), \end{align*}for some constant $k$. If the particle moves a distance of $120\pi$ every second, then what is $k$?

Answer 1

HINT

1. The position of the particle is given by $s(t) = (5\sin(-2t),5\cos(2t))$. Consequently, the speed of the particle is given by $\|s'(t)\|$. Can you take it from here?
2. At this case, the distance is given by \begin{align*} d = \int_{t_{0}}^{t_{0} +1}\|s'(t)\|\mathrm{d}t \end{align*}

where $d = 120\pi$ and $s(t) = (6\cos(kt),6\sin(kt))$. Can you take it from here?

EDIT

In the first problem, notice that $s'(t) = (-10\cos(-2t),-10\sin(2t))$. Thus we have that \begin{align*} \|s'(t)\| = \sqrt{10^{2}\cos^{2}(-2t) + 10^{2}\sin^{2}(2t)} = 10 \end{align*} In the second problem, notice that $s'(t) = (-6k\sin(kt),6k\cos(kt))$. Hence we get that \begin{align*} 120\pi = \int_{t_{0}}^{t_{0}+1}\|s'(t)\|\mathrm{d}t = \int_{t_{0}}^{t_{0}+1}\sqrt{36k^{2}\sin^{2}(kt) + 36k^{2}\cos^{2}(kt)}\mathrm{d}t = 6|k| \end{align*}

Therefore $k = \pm 20\pi$.

Hopefully this helps.