Struggling with related rates I'm trying to understand how to deal with related rates and here is my problem.
When a rocket is 2 km high it is moving vertically at 300 km/hr. At this time, how fast is the angle of elevation increasing as seen by a person on the ground 5km from the launchpad?
la picture
So we have
$$\frac{dy}{dt}=300 km/hr$$
$$y=2 km$$
$$x=5km$$
And we have to find $$\frac{d\Theta}{dt}=?$$
My solution is next:
$$\sin(\Theta)=\frac{y}{\sqrt{29}}$$ where $\sqrt{29}$ is the hypotenuse
Then i take derivative of this thing $$\cos(\Theta)*\frac{d\Theta}{dt} = \frac{\frac{dy}{dt}}{\sqrt{29}}$$
We know, that $\cos(\Theta)$ is $\frac{5}{\sqrt{29}}$ and $\frac{dy}{dt}$ is $300$ so that we have
$$\frac{5}{\sqrt{29}}*\frac{d\Theta}{dt}=\frac{300}{\sqrt{29}}$$
so that
$$\frac{d\Theta}{dt}=\frac{300}{\sqrt{29}}*\frac{\sqrt{29}}{5}$$
$$\frac{d\Theta}{dt}=60$$
But it doesn't look like the right answer.
Could you explain what did I do wrong, please?
 A: We have
$$\tan(\theta)=\frac{y}{5}$$
and by differentiation
$$(1+\tan^2(\theta))\frac{d \theta}{dt}=\frac{1}{5}\frac{dy}{dt}$$
thus
$$\frac{d\theta}{dt}=\frac{300}{5(1+(\frac{2}{5})^2)}$$
A: A modification on your solution that works: 
$\sin(\theta)=\frac{y}{\sqrt{5^2+y^2}} \\\  
$ 
Differentiating both sides gives 
$ \theta' cos(\theta)=\frac{y' \sqrt{5^2+y^2}-y \cdot \frac{1}{2} \cdot \frac{2y y'}{\sqrt{5^2+y^2}}}{(\sqrt{5^2+y^2})^2} \\\ 
$
So now you can replace the expression that represents the hyp with $\sqrt{29}$
So you can write 
$\theta' \cos(\theta)=\frac{y' \sqrt{29}-y \cdot \frac{1}{2} \frac{2 y y'}{\sqrt{29}}}{(\sqrt{29})^2}$
We can also make all other substitutions:
$\theta' \frac{5}{\sqrt{29}}=\frac{300 \sqrt{29}-2 \cdot \frac{1}{2} \frac{2 (2) (300)}{\sqrt{29}}}{(\sqrt{29})^2}$
Solving for $\theta'$ gives:
$\theta'=\frac{\sqrt{29}}{5} \cdot \frac{300 \sqrt{29}-\frac{1200}{\sqrt{29}}}{29} $
Distribute the $\sqrt{29}$ on top there and the $5$ there on bottom:
$\theta'=\frac{300(29)-1200}{5(29)}$
So now the simplified answer is:
$\theta'=\frac{1500}{29} \frac{rad}{hr}$
And yes I know this is not the easiest way... I just wanted to show what the op's way would look like if it was done correctly. 
