The components of a position vector of a particle moving in the plane are . What is the distance traveled by the particle from t = 0 to t = π?

Question

The components of a position vector of a particle moving in the plane are <t^2, sin(t)> . 

What is the distance traveled by the particle from t = 0 to t = π?

anonymous · Answer

Do you know the arclength formula?

raffle_snaffle · Answer

https://www.google.com/search?q=arc+length+formula&espv=2&biw=1920&bih=969&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwiD79TmhrDLAhVHxGMKHQp6D6YQsAQIGw

czesc · Answer

@wio @raffle_snaffle now that I know the formula, how do I apply it here? thanks for the responses guys!

anonymous · Answer

Given the position vector $\mathbf p(t)$, the arc lenth between $t_1$ and $t_2$ is: $$
L=\int\limits_{t_1}^{t_2} \|\mathbf p'(t)\|~dt
$$Given that the velocity vector $\mathbf v(t)$ is equal to $\mathbf  p'(t)$ we can write: $$

L=\int\limits_{t_1}^{t_2} \|\mathbf v(t)\|~dt
$$

anonymous · Answer

`from t = 0 to t = π?`
This means $t_1=0$ and $t_2=\pi$.

anonymous · Answer

So plugging it all in, we have: $$
L = \int_0^\pi \|\langle t^2, \sin(t)\rangle\|~dt
$$

anonymous · Answer

Do you think you can do the rest?

czesc · Answer

@wio , Thanks for all of that, I understand a lot more now, but how do I integrate the vector components? (sorry if I sound arrogant)

anonymous · Answer

There is no vector, because we are taking the magnitude of the vector.

czesc · Answer

@wio So how do I integrate <t^2, sin(t)>dt from 0 to pi

anonymous · Answer

Do you know how to find the magnitude of a vector?

anonymous · Answer

$$
\|\langle t^2,\sin(t)\rangle\| = \sqrt{\langle t^2,\sin(t)\rangle\cdot \langle t^2,\sin(t)\rangle} = \sqrt{t^4+\sin^2(t)}
$$

czesc · Answer

Oh so it would be sqrt(v(x)^2+v(y)^2) then it would just be: 

$$\int\limits_{0}^{\pi} \sqrt{t^4+\sin^2(t)}=10.8265$$

correct?

czesc · Answer

@wio  Thank you very much for being so detailed and patient with me I highly appreciate it.

anonymous · Answer

Yes