Minimum velocity

Question

There is a particle moving along the $x$-axis at any time $t \geq 0$. The velocity of this particle is given by

$$v(t) = \cos(πt) - t(6-2\pi).$$

I am supposed to find the particle's minimum velocity over the interval $[0,2]$. I found the derivative of this velocity function and set it equal to zero, but I am having trouble with getting the right answer. I used the option on my calculator to find the min, but I can’t seen to find it using calculus. Help is greatly appreciated!

Answer 1

The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $\tau$ when $v'(\tau) = 0$ so that the minimum speed is given by $v(\tau).$

$$v'(t) = -\pi \sin \pi t + 2\pi - 6 = 0.$$

So we have

\begin{align} \sin \pi t &= \frac{2\pi - 6}{\pi} \\ \pi t &= \begin{cases} \arcsin (2 - 6/\pi) + 2\pi n \\ \pi - \arcsin (2 - 6/\pi) + 2\pi n \end{cases} \end{align}

where $n \in \mathbb{Z}$ such that $t \in [0,2]$. Taking the inverse sine results in 2 solutions over $2\pi$ since there are 2 values of $y \in (0,\pi)$ that satisfy $\sin y \in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t \notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for $$ \tau = \max \begin{cases} \frac{1}{\pi} \arcsin\frac{2\pi - 6}{\pi} \\ 1 -\frac{1}{\pi} \arcsin\frac{2\pi - 6}{\pi} \end{cases}. $$

Hope that helps.

Edit-

$y = \arcsin a$ gives only 1 real $y$ for a given $a$, $|a| \leq 1$ since it is, after all, a function; however, $\sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.

$\hskip 1 in$

If we say $\sin(\alpha) = b,$ then it's also true that $\sin(\alpha + \beta) = b.$ But $2\alpha + \beta = \pi,$ so $ \sin(\pi - \alpha) = b.$ That is, $\arcsin$ is restricted from $[-\pi/2,\pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2\pi)$ that satisfy $\sin y = b.$ Those angles are $\arcsin b$ and $\pi - \arcsin b.$

Answer 2

The derivative would be $$v'(t) = -\pi \sin(\pi t)-6+2\pi$$

See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$

Answer 3

The derivative of $v(t)$ is $$v'(t) = \pi\cos\left(\pi t\right) + 2\pi - 6.$$

Setting the derivative equal to $0$ and solving for $t$, we get $t \approx 0.31831$ which is within the interval $\left[0, 2\right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get \begin{align} &v(0) = 1,\\ &v(0.31831) = 0.630,\\ &v(2) = 1.566. \end{align}

Thus, the minimum velocity over the interval $\left[0, 2\right]$ is $0.630$.