I have to find the speed (as a function of $t$) of a particle whose position at time $t$ seconds is represtented by
$$c(t)=(\sin t+t, \cos t+t)$$
How would I go about finding the maximum speed? Here's what I've attempted to do so far, not sure if it's the right path though.
$$x=\sin t+t$$
$$x'=\cos t+1$$
$$y=\cos t+t$$
$$y'=-\sin t+1$$
The speed equation can be found with
$$\frac{ds}{dt}=\sqrt{(\cos t+1)^2+(-\sin t+1)^2}$$
This is where I get lost though, any ideas?
Best Answer
$(cost + 1)^2 + (1 - sint)^2 = cos^2t + 2cost + 1 + 1 - 2sint + sin^2t = 3 + 2(cost - sint)$.
To make the problem easier, you find the max value of $v^2(t) = c(t) = 3 + 2cost - 2sint$ , $t > 0$.
$c'(t) = -2cost - 2sint = 0 \iff cost + sint = 0 \to (cost + sint)^2 = 0 \to 1 + 2sintcost = 0 \to sin(2t) = -1$, so $ 2t = \dfrac{(4n-1)\pi}{2}$ , $n \in \mathbb{N}$. So:
$t = \dfrac{(4n - 1)\pi}{4}$, $n \in \mathbb{N}$.
The first value of $t$ which maximizes $c(t)$ is: $t = \dfrac{3\pi}{4}$ which corresponds to $n = 1$.
So: $v_{max} = \sqrt{c\left(\frac{3\pi}{4}\right)} = \sqrt{3 + 2cos\left(\frac{3\pi}{4}\right) - 2sin\left(\frac{3\pi}{4}\right)} = \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$