1. ## Related Rates Problem

A hot air balloon is rising vertically from the ground. Let h denote its time-dependent height above the center of the earth, with the radius of the earthnormalized to 1, and let 0 denote the time dependent angle at the center of the earth bwtween the balloon and the furthest horizon point visible from it.When the balloon is at height h and rising with speed h', what is 0', the rate of change with respect to time? Your answer should be in terms of h and h'.

Assuming that the balloon leaves the groun dat t=0 with intial speed h'(0)=1, what is limit (as t goes towards 0) of 0'(t) according to your formula for 0'? Explain, if you can, why this makes sense in terms of the geometry of the problem and maybe your experience

2. ## Re: Related Rates Problem

A hot air balloon is rising vertically from the ground. Let h denote its time-dependent height above the center of the earth, with the radius of the earthnormalized to 1, and let 0 denote the time dependent angle at the center of the earth bwtween the balloon and the furthest horizon point visible from it.When the balloon is at height h and rising with speed h', what is 0', the rate of change with respect to time? Your answer should be in terms of h and h'.

Assuming that the balloon leaves the groun dat t=0 with intial speed h'(0)=1, what is limit (as t goes towards 0) of 0'(t) according to your formula for 0'? Explain, if you can, why this makes sense in terms of the geometry of the problem and maybe your experience
I assume you meant that angle to be $\theta$ ... diagram of my interpretation of the problem is attached.

$\theta = \text{arcsec}(h)$

derivative w/respect to time ...

$\theta' = \dfrac{h'}{h\sqrt{h^2-1}}$

I leave it for you do determine the requested limit ...

$\displaystyle \lim_{t \to 0} \dfrac{h'}{h\sqrt{h^2-1}} = \, ?$