Acceleration problem (Second Derivative Application) The position of an oscillating particle is given by $x(t) = 7\sin(3t)$. What is the acceleration of the particle at its maximum position?
I found the second derivative, which gives me the function $a(t)= -63\sin(3t)$. I don't know how I'm supposed to find the max position.
 A: Maximum occurs when derivative vanishes. Second derivative should be negative. 
$$x^{'}(t) = 21\cos(3 t)=0,\, t = (4 k+1) \pi/6$$.
A: Since the equation $x'(t)=21\cos(3t) = 0$ has solutions of the form $t=\frac{\pi\cdot(4k+1)}{6}, ~k\in\mathbb{Z}$, we may conclude that $x(t$) has extremum points at $t=\frac{\pi\cdot(4k+1)}{6}$. Extremum point is the local maximum (minimum) if $x''(t) < 0$ ($x''(t) > 0$). So $x''(t)=-63\sin(3t) < 0$ if $0<3\cdot\frac{\pi(4k+1)}{6}<\pi~(\mbox{mod}~2\pi)$. This inequality is true for even $k$. Take, for example, $k=2$ then $t=\frac{\pi}{3}$ and $x''(\frac{\pi}{3})=-63\sin\pi=0$.
A: In this example, you can also find the maximum position of the particle without using the first derivative of the position function. We know that the greatest possible value of the sine function is $1$, so the maximum of $x(t)=7\sin(3t)$ is equal to $7\cdot1=7$, the amplitude of the oscillation, attained whenever $\sin(3t)=1$ $\implies$ $3t=\frac{\pi}{2}+2\pi n$ $\implies$ $t=\frac{\pi}{6}+\frac{2\pi n}{3}$.
UPDATE. Moreover, we don't even need to solve for $t$. Once we figure out that the maximal position occurs whenever $\sin(3t)=1$, we can immediately plug that into the acceleration function: $a(t)=-63\sin(3t)=-63\cdot1=\ldots$.
A: $x(t) = 7\sin(3t),$ $t \in (0,\infty)$.
$a(t) = -63 \sin(3t)$.
Maximum extension from $x =0$ : $+7,-7$ units.
In the positive direction:
$3t = π/2 +n 2π,$ $n = 0,1,2,3,...$
In the negative direction: 
$3t = 3π/2 + n 2π,$ $n=0,1,2,3,...$
The first maximum extension, turning point of the particle, occurs at 
$3t = π/2$, $t = π/6$.
Acceleration at $t =π/6$:
$a(t=π/6) = -63 $.
