Question

Find parametric equations and a parameter interval for the motion of a particle starting at the point ​( 2 2​,0) and tracing the top half of the circle x squared plus y squared equals 4 x2+y2=4 four four times. Find parametric equations for the​ particle's motion. Let the parameter interval for the motion of the particle be 0 less than or equals ≤t less than or equals ≤ 4 4 pi π.

Answer 1

Parametric equations for the particle's motion are x(t) = 2cos(t) and y(t) = 2sin(t), with a parameter interval from 0 to 4π radians to trace the top half of the circle x2 + y2 = 4 four times.

Step-by-step explanation:

To find the parametric equations that describe the motion of a particle tracing the top half of the circle x2 + y2 = 4 four times, we start from the standard parametric equations for a circle with radius 2 centered at the origin: x(t) = 2cos(t) and y(t) = 2sin(t). Since the problem specifies the top half of the circle and repeats this motion four times, we adjust our parameter t to cover the desired motion from 0 ≤ t ≤ 4π.

The parametric equations for the particle's motion are then: x(t) = 2cos(t) and y(t) = 2sin(t) with the parameter interval of 0 ≤ t ≤ 4π. Here, the parameter t represents the angle of rotation in radians, where the values of t from 0 to π cover the upper semicircle and the values of t from π to 2π would cover the lower semicircle, which we're ignoring in this case. Our chosen interval ensures that the top half is traced four times.

Answer 2

1. Parametrization:
`(2\cos(t), 2\sin(t))` and
`t\in [0,\pi]`

2. In case that
`t\in [0,4\pi]`, the desired parametrization is
`(2\cos((t)/(4)), 2\sin((t)/(4)))`

Step-by-step explanation:

Consider the particle at the point (2,0) and the circle of equation
`x^2+y^2=4`. Recall that the general equation of a circle of radius r is given by
`x^2+y^2=r^2`. Then, in our case, we know that the circle has radius 2.

One classic way to parametrize the movement of a particle that starts at point (r,0) and moves in a counterclockwise manner over a circular path of radius r is given by the following parametrization
`(r\cos(t),r\sen(t)), t\in [0, 2\pi]`. Since, for all t we have that

`(r\cos(t))^2+(r\sin(t))^2 = r^2(\cos^2(t)+\sin^2(t)) = r^2`

If we want to draw only the upper half of the circle, we must have
`t\in[0,\pi]`.

So, with r=2 the desired parametrization is
`(2\cos(t), 2\sin(t))` and
`t\in [0,\pi]`. Recall that in this parametrization when t=0 the particle is at (2,0) and when t=pi the particle is at (-2,0).

In the case that we want the parameter s
`\in[0,4\pi]` but keeping the same particle's motion, we must do a transformation. We know that if parameter t is in the interval
`[0,\pi]` we get the desired motion. Note that in this case we are multiplying this interval by 4. So, we have that s = 4t. If we solve for the parameter t, we get that t=s/4. Then, with the parameter s in the interval
`[0,4\pi]` we get the parametrization
`(2\cos((s)/(4)), 2\sin((s)/(4)))` which is obtained by replacing t in the previous parametrization.

Note that since when
`s=4\pi` we have that
`t=\pi` and that when s=0, we have t=0, then the motion of the particle is the same (it changes only the velocity in which the particle moves a cross the path).