Consider a particle following the parametric equations below. What is the speed of the particle, and is it moving clockwise or counter-clockwise?

\(\begin{align*} x &= 3 + 2 \sin (2t),\\ y &= 4 + 2 \cos (2t). \end{align*}\)

Guest May 7, 2021

#1**0 **

I think

\(\frac{dx}{dt}=2cos(2t)\\ \frac{dy}{dt}=-2sin(2t)\\~\\ \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=-2sin(2x)*\frac{1}{2cos(2x)}=-tan(2x)\)

I admit it. I don't know.

Melody May 10, 2021

#3**+1 **

Here's a non-calculus answer.

\(\displaystyle x-3=2\sin(2t)\\y-4=2\cos(2t).\)

Squaring and adding,

\(\displaystyle (x-3)^{2}+(y-4)^{2}=2^{2},\)

so the particle is moving on a circle centre(3, 4) radius 2.

At t = 0, the particle is at the point (3, 6), (the top of the circle), and as t increases sin(2t) increases so x increases,

implying that the particle is moving clockwise around the circle.

The particle will be at x = 3 when sin(2t) = 0, that is, when 2t = 0, pi, 2pi, ..., t = 0. pi/2, pi, ... .

At t = 0, x = 3, y = 6.

At t = pi/2, x = 3, y = 2.

At t = pi, x = 3, y = 6, etc.

So, it takes pi seconds(?) to go once round, and since the circumference is 4pi it follows that its speed is 4.

Guest May 11, 2021