related rate change problem

Thanks to both Mufasa and Ron Gordon, I now understand when I difficultly came from, misunderstanding of the term vertically.

Problem:

A rocket takes off vertically from the ground. 2000 ft. away, a camera captures it image. The rocket lift-off vertically to the position equation $s=50t^2$. Find the rate of change of the angle of the camera at 10 seconds after lift-off.

I am having trouble solving this problem. Also, a bit unsure if I fully understand the language used in the problem.

I know that the rocket lift off vertically, and I know the graph of the position function $s$.

I know to use the right half of the graph, otherwise time would being going backwards.

However, how to give the angle is puzzling me, from the camera's reference point. I know how to use the arc length to find the angle, but only for circles.

Now, I understand that vertically means a straight line, but how does one solve it if it doesn't follow the strict meaning of vertically, and insteads follows a parabola?

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the rocket is travelling vertically (i.e. in a straight line). Therefore there is a right angled triangle formed between where the camera is, where the rocket is, and the point from where the rocket took off. –  Mufasa Sep 19 '13 at 8:15
@Mufasa so Vertically, means it follows a path of x="constant", I was a bit confused by the term because I thought it was another way to say upward. –  yiyi Sep 19 '13 at 8:17
@Mufasa, how would you find the change in angle if it followed the parabola path? –  yiyi Sep 19 '13 at 8:18

The angle of the camera $\theta$ and the height $s$ of the rocket are related as follows:

$$\tan{\theta}= \frac{s(t)}{2000}$$

so that

$$\sec^2{\theta} \frac{d\theta}{dt} = \frac{1}{2000} \frac{ds}{dt}$$

Note that $\sec^2{\theta}=1+\tan^2{\theta}$ and we have

$$\frac{d\theta}{dt} = \frac{(ds/dt)/2000}{1+s^2/(2000)^2}$$

Now use $s(t)=50 t^2$, $ds/dt=100 t$, to get

$$\frac{d\theta}{dt} = \frac{t/20}{1+(t^4/1600)}$$

At $t=10$, I get that

$$d\theta/dt = \frac{1/2}{1+(10000/1600)} \approx 0.06897$$

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Great answer, that is with the rocket following a straight line upwards and hen downwards; how does one solve it when it following the parabola? –  yiyi Sep 19 '13 at 8:41
@yiyi; not sure what you mean, "follow the parabola." The height is a quadratic in time, so a plot of this gives a parabola over time. But this is linear motion, not 2D parabolic notion like that of a projectile. –  Ron Gordon Sep 19 '13 at 8:57
Thanks, the last sentence cleared up my confusion about the proper way to understand the question. I just realized the difference between linear motion and how it can give a non-linear graph over time. Wonderful! –  yiyi Sep 19 '13 at 9:39