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?
$\endgroup$ 13 Answers
$\begingroup$$(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: $v_{max} = \sqrt{c\left(\frac{3\pi}{4}\right)} = \sqrt{3 + 2cos\left(\frac{3\pi}{4}\right) - 2sin\left(\frac{3\pi}{4}\right)} = \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$
$\endgroup$ $\begingroup$You are definitely on the right track, and very close to the finish line.
The speed, as you have shown, is given by $v(t) = \sqrt{ (\cos t + 1)^2 + (- \sin t + 1)^2}$. You want to find a value of $t$ that makes this a maximum. One way to make this easier is to realize that $v(t)$ is maximized for the same value of $t$ for which $v(t)^2$ is maximized. So go right ahead and drop the square root sign: You want to find the value of $t$ for which the function $f(t) = (\cos t + 1)^2 + (- \sin t + 1)^2$ is a maximum.
To finish the problem:
- Expand that formula and use trigonometric identities where possible to write the function as simply as possible;
- Take the derivative with respect to $t$, set it equal to zero, and solve for $t$
- If there is more than one solution for $t$ consider the critical points one at a time and determine which are maxima, minima, etc.;
- When you have found the value of $t$ that maximizes $f(t)$, plug that value of $t$ into the formula for $v(t)$ and compute the maximum speed.
Let $v(t)$ be the last expression you wrote. The maximum speed is in $t_0$ such that $v '(t_0) = 0 $, like in single variable calculus.
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