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(\rm km)/(\rm h)\right)t`

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

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

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

`\tan\theta=(y(t))/(13\,\rm km)`

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

`\tan\theta=(30\,\rm km)/(13\,\rm km)\implies\theta\approx66.57^\circ`

Differentiating both sides of the equation above gives

`\sec^2\theta(\mathrm d\theta)/(\mathrm dt)=\frac1{13\,\rm km}(\mathrm dy(t))/(\mathrm dt)`

and substituting the angle
`\theta` found above, and
`(\mathrm dy(t))/(\mathrm dt)=800(\rm km)/(\rm h)`, we get

`\sec^2\theta66.57^\circ(\mathrm d\theta)/(\mathrm dt)=\frac1{13\,\mathrm km}\left(800(\rm km)/(\rm h)\right)`

`\implies\boxed{(\mathrm d\theta)/(\mathrm dt)\approx9.73(\rm rad)/(\rm h)}`