To determine

To find: The rate of change of the distance from the television camera to the rocket when height of the rocket is 3000 ft.

Answer

The rate of change of the distance from the television camera to the rocket is dldt=360  ft/s.

Explanation

Given:

A television camera is positioned at 4000 ft from the base of the rocket launching pad.

The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight.

The rocket rises vertically at a speed of 600 ft/s.

Formula used:

(1). Chain rule: dydx=dydu⋅dudx

(2). Pythagorean Theorem.

Calculation:

Let us suppose that A be the position of the rocket and B be the position of the television camera when the rocket is at 3000 ft height from the television camera.

And let y be the height of the rocket from the rocket launching pad, l be the distance between the rocket and television camera, and θ be the angle of elevation between the rocket and television at any time t, which is shown in Figure1.

Since the height of the rocket changes with time, so the angle of elevation and the distance between the rocket and the television camera also changes with time t.

Since dydt=600  ft/s .

Obtain dldt  when y=3000 ft.

Now by using the Pythagorean Theorem in ΔBCA.

l2=40002+y2 .

Now find the value of l when y=3000  ft.

l=40002+30002=16000000+9000000=25000000=5000  ft

Differentiate l2=40002+y2 with respect to the time t.

ddt[l2]=ddt[40002+y2]2ldldt=2ydydt                                                     [∵dydx=dydu⋅dudx]ldldt=ydydtdldt=yldydt

Substitutes y=3000, l=5000 and dydt=600 in dldt.

dldt=30005000(600)=3×6005=18005=360​  ft/s

Therefore, the rate of change of the distance from the television camera to the rocket is dldt=360  ft/s.

To determine

To find: The rate of change of the angle of elevation when height of the rocket is 3000 ft.

Answer

The rate of change of the angle of elevation is dθdt=0.096  rad/s.

Explanation

Formula used:

(1). Chain rule: dydx=dydu⋅dudx

Calculation:

By part (a)

Since dydt=600  ft/s.

Obtain dθdt when y=3000  ft.

Now in the ΔBCA.

tanθ=ACCBtanθ=y4000y=4000tanθ

Differentiate y with respect to the time t.

ddt(y)=ddt(4000tanθ)dydt=4000(sec2θ)dθdtdθdt=14000(sec2θ)dydtdθdt=cos2θ4000dydt

When y=3000, then l=5000

cosθ=CBAB=40005000=45

Substitutes cosθ=45 and dydt=600 in dθdt.

dθdt=(45)40002(600)=16×6004000×25=16×640×25=961000  rad/s

On further simplification, the rate of change in the angle of elevation is as follows.

dθdt=0.096  rad/s.

Therefore, the rate of change of the angle of elevation is dθdt=0.096  rad/s.