A television camera is positioned 4 km from the base of a rocket launching pad. In order to keep the rocket in focus as it takes off, the camera must be programmed with the distance to the rocket and how that distance is changing. If we assume that the rocket rises vertically at a speed of 200 km/hr, how fast is the distance from the camera to the rocket changing when the rocket has risen 3 km

120 km/hr

Explanation:

Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.

Now, at any instant, D² = d² + L²

= d² + 4²

= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.

differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have

d(D²)/dt = d(d² + 16)/dt

2DdD/dt = 2d[d(d)/dt]

dD/dt = 2d[d(d)/dt] ÷ 2D

Now d(d)/dt = vertical speed of rocket = 200 km/hr

dD/dt = 200d/D  [D = √(d² + 16)]

dD/dt = 200d/[√d² + 16]

Now substituting d = 3 km, the distance the rocket has risen into the equation, we have

dD/dt = 200(3)/[√(3² + 16)]

dD/dt = 600/[√(9 + 16)]

dD/dt = 600/√25

dD/dt = 600/5

dD/dt = 120 km/hr

So,  the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.

answer: decibel

explanation:

nah.   i'm just sleepy.