Question 1025620: Can someone explain the answer to this related rates problem to me, I don't understand it.
The question:
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? I don't understand what's happening there, how are we able to do this?
Why was tan chosen?
Where did the ratio of 500t and 5000 come from?
I am so confused here.....
Please help
Thank you
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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
====================================
could someone explain to me how the whole PVT thing works?
--
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).
-----
I don't understand what's happening there, how are we able to do this?
Why was tan chosen?
--
Tan is the ratio of the altitude of the rocket to the 5000 m distance from the launch site.
------
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.
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