Respuesta :
Answer:
1. Parametrization: [tex](2\cos(t), 2\sin(t))[/tex] and [tex]t\in [0,\pi][/tex]
2. In case that [tex]t\in [0,4\pi][/tex], the desired parametrization is [tex](2\cos(\frac{t}{4}), 2\sin(\frac{t}{4}))[/tex]
Step-by-step explanation:
Consider the particle at the point (2,0) and the circle of equation [tex]x^2+y^2=4[/tex]. Recall that the general equation of a circle of radius r is given by [tex]x^2+y^2=r^2[/tex]. 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 [tex](r\cos(t),r\sen(t)), t\in [0, 2\pi][/tex]. Since, for all t we have that
[tex](r\cos(t))^2+(r\sin(t))^2 = r^2(\cos^2(t)+\sin^2(t)) = r^2[/tex]
If we want to draw only the upper half of the circle, we must have [tex] t\in[0,\pi][/tex].
So, with r=2 the desired parametrization is [tex](2\cos(t), 2\sin(t))[/tex] and [tex]t\in [0,\pi][/tex]. 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 [tex]\in[0,4\pi][/tex] but keeping the same particle's motion, we must do a transformation. We know that if parameter t is in the interval[tex][0,\pi][/tex] 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 [tex][0,4\pi][/tex] we get the parametrization [tex](2\cos(\frac{s}{4}), 2\sin(\frac{s}{4}))[/tex] which is obtained by replacing t in the previous parametrization.
Note that since when [tex]s=4\pi[/tex] we have that [tex]t=\pi[/tex] 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).
