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**Exotique** I just want to check if my reasoning and answer are correct. The question is:

A rocket is launched straight up, and it's altitude is given by $\displaystyle h=10t^2$ after $\displaystyle t$ seconds. You are on the ground 300m from the launch site watching the rocket going up. The line of sight from you to the rocket makes and angle $\displaystyle \theta$ with the horizontal. By how many radians per second is $\displaystyle \theta$ changing 10 seconds after the launch?

My reasoning is as follows:

Need to find $\displaystyle \frac{d\theta}{dt}$

$\displaystyle h=10t^2$, so $\displaystyle \frac{dh}{dt}=20t$.

$\displaystyle \tan\theta=\frac{h}{300}$, so $\displaystyle \frac{d\theta}{dh}=\frac{300}{h^2+90000}$.

Remembering $\displaystyle t=10$, $\displaystyle \frac{dh}{dt}=20t=200$

Also, since $\displaystyle h=10t^2$, $\displaystyle h=1000$ when $\displaystyle t=10$.

Therefore, $\displaystyle \frac{d\theta}{dh}=\frac{300}{h^2+90000}=\frac{300 }{91000}$

So using Chain Rule:

$\displaystyle \frac{d\theta}{dt}=\frac{d\theta}{dh}\cdot\frac{dh }{dt}$

$\displaystyle =\frac{300}{91000}\cdot200=\frac{60}{91}$

Any help will be much appreciated.