Related Rates: Angular Velocity The figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute.



*

*Find the angular velocity of the connecting rod, $\frac{d\alpha}{dt}$, in radians per second, when $\theta = \frac{\pi }{3}$.

*Express the distance $x = \left|OP \right|$ in terms of $\theta$.

*Find an expression for the velocity of the pin P in terms of $\theta$.


I have found a solution for all three questions, but my answer to QUESTION 1 is different from the given solution from the textbook and I am struggling to figure out why.
QUESTION 1
I use the law of sines to solve this.
$$\frac{\sin\alpha }{OA} = \frac{\sin\theta }{AP}$$
from which I find that:
$$\sin\alpha =\frac{OA}{AP}\sin\theta =\frac{40}{120}\sin\theta =\frac{1}{3}\sin\theta$$
$$\alpha = \arcsin\left(\frac{1}{3}\sin\theta  \right)$$
$$\frac{d\alpha }{dt} = \frac{1}{\sqrt{1 - \frac{{\sin\theta }^{2}}{9}}} \cdot  \frac{1}{3}\cos\theta \cdot \frac{d\theta }{dt}$$
and this is equal to $$\frac{d\alpha }{dt} = \frac{3\pi }{\sqrt{33}}$$ when $$\theta = \frac{\pi }{3}$$
The problem here is that according to the textbook the right answer is $\frac{4\pi \sqrt{3}}{\sqrt{11}}$ which I quite do not understand. Furthermore I am not supposed, at this stage, to know the derivative of inverse functions including arcsin. Is there any other way to solve this without using the arcsin funcion?
 A: We have
$$
\sin(\theta) = \frac{s}{r} \\
\sin(\alpha) = \frac{s}{l}
$$
such that equating on $s$ we get
$$
r \sin(\theta) = l \sin(\alpha) \iff \\
\alpha = \arcsin\left( \frac{r}{l} \sin(\theta) \right)
$$
Then the time derivative is
$$
\dot{\alpha} 
= \frac{1}{\sqrt{1-\left( \frac{r}{l} \sin(\theta)\right)^2}}
\frac{r}{l}\cos(\theta) \dot{\theta}
$$
with $r=0.4\,\text{m}$, $l=1.2\,\text{m}$, $\dot{\theta}=6\cdot 2\pi\,\text {rad}/\text{s}$. Then for $\theta=\pi/3$ we get:
$$
\sin(\pi/3) = \sqrt{3}/2 \\
\cos(\pi/3) = 1/2 \\
\dot{\alpha} = 
\frac{1}{\sqrt{1-(\frac{1}{3} \sqrt{3}/2)^2}} 
\frac{1}{3} \cdot \frac{1}{2} \cdot 12 \pi 
= \sqrt{\frac{36}{33}} 2 \pi 
= \sqrt{\frac{9}{33}} 4\pi
= \sqrt{\frac{3}{11}} 4\pi
$$
Alternative without $\arcsin$:
$$
\sin(\alpha) = \frac{r}{l} \sin(\theta)
$$
Differentiating
$$
\cos(\alpha) \dot{\alpha} = \frac{r}{l} \cos(\theta) \dot{\theta} \iff \\
\dot{\alpha} 
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\cos(\alpha)}
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-\sin(\alpha)^2}}
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-(\frac{r}{l} \sin(\theta))^2}}
$$
