Solution to Exercise 8.2-4


© 1999 by Karl Hahn 		KCT logo

The problem is: A rocket is launched vertically and travels at 100 meters per second. A tracking radar is 500 meters from the launch site. When the rocket is 800 meters high, how fast must the radar antenna slew (in radians per second) in order to track the rocket?

You should be getting pretty good at answering questions 1 and 2 by now. Is it clear to you that time, t, is the independent variables in this problem and that rocket height, h, and and the angle that the radar looks up, θ, are the dependent variables? And since the radar does not get any nearer or farther from the launch site, it should be clear too that the distance, x, between the radar and the launch site is constant.

In this problem it pays to draw a diagram (did you?). Here is my diagram of the problem (not drawn to scale).

Passing quickly to question 3, there are no hidden variables here. All of them are shown in the diagram.

As to question 4, the relationship between the two dependent variables, h and θ, is gathered by analyzing the right triangle in the diagram. Since  tan(θ)  is opposite over adjacent, it follows that

              h
   tan(θ)  =   
              x
and 
   x tan(θ)  =  h
Question 5 asks you to take the derivative of each relationship you have among dependent variables. Remembering that x is a constant here and h and θ are dependent variables, you have


   x sec2(θ)


 dθ
     =
 dt

  dh
    
  dt
Question 6 asks you to figure out what the problem is asking for and solve it. Since θ is the radar's "look" angle and the problem is asking how fast the radar must slew, it should be clear that the problem is asking you for  dθ/dt.  Solving the above for that gives:

   1 dh
       
   x dt


 cos2(θ)  =


  dθ
    
  dt
But you can make this even simpler by remembering that 

   sec2(θ)  =  1 + tan2(θ)
and that 
              h
   tan(θ)  =   
              x
If you substitute these facts before you solve for  dθ/dt,  you find that 
   1 dh      1           dθ
                      =    
   x dt         h2       dt
          1  +    
                x2
This looks rather nasty, but multiply out the x2, then cancel the  1/x  and you have 
   dh    x        dθ
               =    
   dt x2 + h2     dt
Question 7 has you plug in the numbers. We have  x = 500 meters,   h = 800 meters,  and  dh/dt = 100 meters/second.  When I put them into the solution-equation, I get 
   dθ
       =  0.056179775 radians/sec  =  3.218864018 degrees/sec
   dt
What happens to the radar's slew rate as the rocket gets higher according to the solution equation above? Is that what you'd expect it to do based upon your intuition about tracking a rocket in a vertical trajectory?

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