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Question

A television camera is positioned 4000ft from the base of a 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. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600ft/s when it has risen 3000ft.

 

How fast is the distance from the television camera to the rocket changing at that moment?

 

If the television is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Answer 1

Ok this is what we know.

 

We are 4000 ft from the base.  That is a constant.  We also know the rocket has risen 3000 ft. 

 

4000² + 3000²  = c²   Therefore c = 5000 (Pythagorean)

 

 

Now let's look at our changes:

 

it is vertically changing at 6000 ft/s so that means we will have a dy/dt.  It is also changing in the hypotenuse because it is rising vertically. 

 

So we will again use this formula and derive.  We know the X direction is constant

 

so we get 4000² + y² = c²  We will derive

 

0 + 2y dy/dt = 2c dc/dt   We are solving for dc/dt

 

2 (600) (3000) = 2(5000) dc/dt

 

dc/dt = 360