Question

a rocket travels vertically at a speed of 400 km/hr. the rocket is tracked through a telescope by an observer located 9 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

Answer:

The rate at which the angle between the telescope and the ground is increasing 3 min after lift-off is 7.48radians per hour.

Step-by-step explanation:

We are given that

A rocket travels vertically at a speed of 400 km/hr. the rocket is tracked through a telescope by an observer located 9 km from the launching pad

Let theta be the angle between the telescope and the ground and the h be the height of the rocket above the ground.

Then

tanθ = $\frac{h}{9}$

Now we differentiate the trigonometric function with respect to time

$\frac{dtan(theta)}{dt}$ = $\frac{d}{dt}$($\frac{h}{9}$)

$sec^2(theta)\frac{d(theta)}{dt}$ = $\frac{1}{9}\frac{dh}{dt}$

$\frac{d(theta)}{dt} = \frac{1}{9sec^2(theta)}\frac{dh}{dt}$ ..................(1)

Now as we know that the rocket is travelling upwards with a speed of 400 km/hr so the distance it will cover in three minutes will be

400*$\frac{3}{60} = 20km$

tan(theta) = $\frac{20}{9}$

$sec^2(theta) = 1 + tan^2(theta)\\ = 1 + (\frac{20}{9})^2$

       = 1 + $\frac{400}{81}$

       =  $\frac{481}{81}$

upon substituting this value in the equation (1) we get

$\frac{dtheta}{dt} = \frac{1}{9*\frac{481}{81} } *400$

          = $\frac{9*400}{481}$

          = $\frac{3600}{481}$

$\frac{dtheta}{dt} = 7.484radians$/hour

Therefore the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off is 7.48radians per hour.