Related Trigonometric Rates A port and a radar station are $3 \text{ mi}$ apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of $13 \frac{\text{mi}}{\text{hr}}$. Find the rate of change of the tracking angle $\theta$ between the shore and the line between the radar station and the ship at $12:30 \text{ PM}$, assuming the ship maintains its speed and course.
The following graphic represents this scenario:
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Let $\dfrac{ds}{dt} = 13 \frac{\text{mi}}{\text{hr}}$.
The ship is given to move from $A \to C$ at $\dfrac{ds}{dt}$ departing at noon and arriving at $12:30 \text{ PM}$ such that: 
$$s = \dfrac{13}{2}$$
Let the segment $\overline{CB} = x$.
By the law of cosines:
$$x = \sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}$$
Then apply the law of sines to find that:
$$\dfrac{\sin{\left( \frac{\pi}{4} \right)}}{x} = \dfrac{\sin{\theta}}{s}$$
Through substitution and manipulation of the equality its found that:
$$\dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} = \sin{\theta}$$
Take the derivative with respect to time of both sides to find that:
$$
\\ \begin{align*} 
\\ \frac{d}{dt}\left[ \arcsin \left( \dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} \right) \right] &= \frac{d}{dt}\left[ \theta \right]
\\
\\ \dfrac{\dfrac{1}{\sqrt{2}\sqrt{s^2-3\sqrt{2}s+9}} - \dfrac{s^2}{\sqrt{2}{(s^2-3\sqrt{2}s+9)}^{3/2}}}{\sqrt{1-\dfrac{s^2}{2(s^2-3\sqrt{2}s+9)}}}\cdot\frac{ds}{dt} &= \frac{d\theta}{dt}
\\ \end{align*}
$$
Which Wolfram evaluates to be $-4.5193 \frac{\text{rad}}{\text{hr}}$ as seen here.
This doesn't appear to be correct, however.
I'm uncertain of the error I've made, so if you could explicitly state at what point I've made an error in my approach, that would be very helpful. All that I'm certain of is that this is not the expected answer, but my approach appears proper to me (though it is clearly not).
 A: let the distance between the ship and the radar station be $y.$ so we have $$y^2 = s^2 + 3^2 - 2 \times 3 \times s\cos 45^\circ  = s^2 + 9 -3\sqrt 2s$$ differencing gives $$2y\frac{dy}{dt} = 2s\frac{ds}{st} - 3\sqrt 2 \frac{ds}{dt} \tag 1$$ now we put the data $$\frac{ds}{dt} = 13, s = 6.5, y = \sqrt{6.5^2 + 9 - 3\sqrt 2 \times 6.5}= 4.865$$
putting all these in $(1),$ we get $$2\times 4.865\frac{dy}{dt}=13\times 13-3\sqrt 2 \times 13=113.845\to \frac{dy}{dt} = 11.700\, \text{ mile/hour} $$
$\bf edit:$ if you need $\frac{d\theta}{dt},$  we can use $$\frac{s}{\sin \theta} = \frac{y}{\sin 45^\circ}\to\sin 45^\circ \frac{ds}{dt} = y\cos \theta \frac{d\theta}{dt} + \sin \theta \frac{dy}{dt} \tag 2$$  we will first compute $\theta.$
we have $$\cos \theta = \frac{-\frac{6.5}{\sqrt 2} + 3}{y} = -0.328, \sin \theta = \frac{6.5}{\sqrt 2 y}=0.944$$
now subbing $$13\frac{\sqrt 2}2=4.865\times (-0.328)\frac{d\theta}{dt}+0.944\times 11.700 $$
i get $$\frac{d\theta}{dt} = 1.165\, radian /hour = 66.787^\circ/hour$$
A: hint: Let $u = BC \Rightarrow \cos \theta = \dfrac{3^3+u^2-s^2}{2\cdot 3\cdot u}$, then using quotient rule to find $\theta '$, and also first find $s',u', u, s, \sin \theta$ at $12:30$ pm. To find $u'$ you need to use the law of cosine again using $u =\sqrt{3^2+s^2-2\cdot 3\cdot s\cdot \cos (45^{\circ})}$.
