How do you find the maximum velocity of a particle using its first derivative?

[Swerting]

The problem states that a particle's position at time t is given by the equation:
$$x(t)=2\pi t+cos(2\pi t)$$
Therefore, the velocity of the particle at time t would be the first derivative of the above equation:
$$x'(t)=v(t)=2\pi-2\pi(sin(2\pi t))$$
I was asked what the maximum velocity of the particle was, and was able to determine that it is 2, but that is only because I used a graphing aid. I am not quite sure where to start on how to find the maximum, other than know what the graph looks like in one's mind's eye, but that seems a little too extreme. I have also calculated the second derivative (acceleration of the particle) if it is needed:
$$x''(t)=a(t)=4\pi^2cos(2\pi t)$$
I would show other attempt at work, but we really weren't shown how to do this.
But yes, we still have to do it.
Thank you for your help.

 

[Dick]

The max of v isn't 2. Working this in your minds eye is a perfectly good way to do it. sin(2*pi*t) starts at 0 goes up to 1, goes down to -1, goes back to zero and then just keeps repeating. For which of those values is v a maximum?

 

[Swerting]

The max of v isn't 2. Working this in your minds eye is a perfectly good way to do it. sin(2*pi*t) starts at 0 goes up to 1, goes down to -1, goes back to zero and then just keeps repeating. For which of those values is v a maximum?

hmmm...that does make sense. Thank you very much for your reply, i think the $$2\pi$$ outside the sine section of the function was confusing me.

Now that I have double checked my graphing aid again, i find that it gives me a completely different answer, something to the tune of 12...
now i am quite sure that you are correct! thank you again for the help, i appreciate it a lot.

 

[HallsofIvy]

You have $x'= 2\pi - 2\pi sin(2\pi t)$ and want to find its maximum value. Okay, forget about the fact that it is a first derivative and just think of maximizing $y= 2\pi - 2\pi sin(2\pi t)$. $y'= 4\pi^2 cos(2\pi t)= 0$ when [itex]2\pi t= \pi/2[itex]. or when $2\pi t= 3\pi/2$. When $2\pi t= \pi/2$ [itex]x'= y= 2\pi- 2\pi sin(\pi/2)= 0. When [itex]2\pi t= 3\pi/2[/itexs] $x'= y= 2\pi- \2pi sin(3\pi/2)= 2\pi+ 2\pi= 4\pi$. The maximum value is $4\pi$.