Question

A rocket travels vertically at a speed of 800 km/hr. the rocket is tracked through a telescope by an observer located 13 km from the launching pad. find the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off. (round your answer to two decimal places.)

Answer 1

The rocket's altitude $y$ at time $t$ is

$y(t)=\left(800\dfrac{\rm km}{\rm h}\right)t$

so that after $t=3\,\mathrm{min}$, it will have traveled

$y=\left(800\dfrac{\rm km}{\rm h}\right)(3\,\mathrm{min})=40\,\mathrm{km}$

The angle of elevation $\theta$ at time $t$ is such that

$\tan\theta=\dfrac{y(t)}{13\,\rm km}$

At the moment when $y(t)=30\,\mathrm{km}$, this angle is

$\tan\theta=\dfrac{30\,\rm km}{13\,\rm km}\implies\theta\approx66.57^\circ$

Differentiating both sides of the equation above gives

$\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\rm km}\dfrac{\mathrm dy(t)}{\mathrm dt}$

and substituting the angle $\theta$ found above, and $\dfrac{\mathrm dy(t)}{\mathrm dt}=800\dfrac{\rm km}{\rm h}$, we get

$\sec^2\theta66.57^\circ\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\mathrm km}\left(800\dfrac{\rm km}{\rm h}\right)$

$\implies\boxed{\dfrac{\mathrm d\theta}{\mathrm dt}\approx9.73\dfrac{\rm rad}{\rm h}}$