Respuesta :
Answer:
- The particle moves at [tex] \sqrt{\sin^2(t) + \frac{\cos^2(\frac{t}{2})}{4}[/tex] feets per second after t seconds.
- The first stop is at π seconds
- The particle stops at nπ seconds, with n an odd integer
Step-by-step explanation:
We can write the position with a function [tex] f: R \rightarrow R^2 [/tex] , [tex] f(t) = (cos(t), sin(t/2)) . [/tex]
In order to find the speed, we have to find the norm of the derivate of f. The derivate of cos(x) is -sin(x), and the derivate of sin(x) is cos(x). Accoring to the chain rule, in order to derivate sin(t/2), we have to multiply cos(t/2) by the derivate of t/2, 1/2. Thus the derivate of sin(t/2) is cos(t/2) * 1/2. Therefore, we have
[tex] f'(t) = (-\sin(t), \cos(t/2) * 1/2 ) [/tex]
So, the speed of the particle after t seconds is || f'(t) || = [tex]\sqrt{(-\sin(t))^2 + (\cos(\frac{t}{2})*\frac{1}{2})^2} = \sqrt{\sin^2(t) + \frac{\cos^2(\frac{t}{2})}{4}[/tex] feats per second.
The particle stops when f'(t) = (-sin(t),cos(t/2) * 1/2) = (0,0). That can only happpen if sin(t) = cos(t/2) = 0. The values where sin(t) is equal to 0 follow the expression kπ, where k in an integer. The values where cos(t) = 0 have the form π/2 + jπ, with j another integer. Thus, cos(t/2) = 0 only when t = 2 * (π/2 + jπ) = π + 2jπ. In other words, t = nπ, with n an odd integer. In order to conclude the excercise we have to note a few things:
- if t = nπ, with n odd, not only cos(t/2) = 0, but also sin(t) = 0
- The first positive value for which both sin(t) and cos(t/2) are 0 is nπ, with n equal to 1, thus, t = π.
With this information, we can conclude that the first positive time the particle stops is at π seconds, and the particle stops at nπ seconds, with n being any odd integer.
