Question 1025620
We launch a rocket from 5km away and track the ascent with a telescope. The rocket ascends at a rate of 500 m/s. How fast is the angle between the telescope and the ground increasing 30 seconds into the launch? 
 The answer:
 Let P,V and T be the points of rocket launch, current position of rocket at time t and telescope. Then ΔPVT is a right triangle with angle P as right triangle. 
 We have PV=500t and PT=5km=5000m. Let ∠VTP=θ at time t. 
 Then, 
 tanθ=500t/5000=t/10. 
 Differentiating w.r.t. t we get 
 sec˛θ dθ/dt=1/10 
 ⇒dθ/dt=cos˛θ/10 
 At time t=30, PV=500x30=15000m, 
 cos˛θ=5000˛/(5000˛+15000˛)=1/10 
 Therefore, rate of change of angle dθ/dt at t=30s=(1/10)*(1/10)=1/100=0.01 
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 could someone explain to me how the whole PVT thing works?
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PVT is a right triangle, with the right angle at P (given).
PT is the distance from the telescope to the launch site 5000 meters ((given).
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 I don't understand what's happening there, how are we able to do this? 
 Why was tan chosen?
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Tan is the ratio of the altitude of the rocket to the 5000 m distance from the launch site.
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 Where did the ratio of 500t and 5000 come from?
At time t, the altitude is 500t meters. 5000 is constant, the given distance to the launch site.