document.write( "Question 1025620: Can someone explain the answer to this related rates problem to me, I don't understand it.\r
\n" ); document.write( "\n" ); document.write( "The question:
\n" ); document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "The answer:
\n" ); document.write( "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.
\n" ); document.write( "We have PV=500t and PT=5km=5000m. Let ∠VTP=θ at time t.
\n" ); document.write( "Then,
\n" ); document.write( "tanθ=500t/5000=t/10.
\n" ); document.write( "Differentiating w.r.t. t we get
\n" ); document.write( "sec˛θ dθ/dt=1/10
\n" ); document.write( "⇒dθ/dt=cos˛θ/10
\n" ); document.write( "At time t=30, PV=500x30=15000m,
\n" ); document.write( "cos˛θ=5000˛/(5000˛+15000˛)=1/10
\n" ); document.write( "Therefore, rate of change of angle dθ/dt at t=30s=(1/10)*(1/10)=1/100=0.01\r
\n" ); document.write( "\n" ); document.write( "====================================\r
\n" ); document.write( "\n" ); document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "Why was tan chosen?
\n" ); document.write( "Where did the ratio of 500t and 5000 come from?\r
\n" ); document.write( "\n" ); document.write( "I am so confused here.....\r
\n" ); document.write( "\n" ); document.write( "Please help
\n" ); document.write( "Thank you \n" ); document.write( "
Algebra.Com's Answer #640916 by Alan3354(69443)\"\" \"About 
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?
\n" ); document.write( " The answer:
\n" ); document.write( " 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.
\n" ); document.write( " We have PV=500t and PT=5km=5000m. Let ∠VTP=θ at time t.
\n" ); document.write( " Then,
\n" ); document.write( " tanθ=500t/5000=t/10.
\n" ); document.write( " Differentiating w.r.t. t we get
\n" ); document.write( " sec˛θ dθ/dt=1/10
\n" ); document.write( " ⇒dθ/dt=cos˛θ/10
\n" ); document.write( " At time t=30, PV=500x30=15000m,
\n" ); document.write( " cos˛θ=5000˛/(5000˛+15000˛)=1/10
\n" ); document.write( " Therefore, rate of change of angle dθ/dt at t=30s=(1/10)*(1/10)=1/100=0.01
\n" ); document.write( " ====================================
\n" ); document.write( " could someone explain to me how the whole PVT thing works?
\n" ); document.write( "--
\n" ); document.write( "PVT is a right triangle, with the right angle at P (given).
\n" ); document.write( "PT is the distance from the telescope to the launch site 5000 meters ((given).
\n" ); document.write( "-----
\n" ); document.write( " I don't understand what's happening there, how are we able to do this?
\n" ); document.write( " Why was tan chosen?
\n" ); document.write( "--
\n" ); document.write( "Tan is the ratio of the altitude of the rocket to the 5000 m distance from the launch site.
\n" ); document.write( "------
\n" ); document.write( " Where did the ratio of 500t and 5000 come from?
\n" ); document.write( "At time t, the altitude is 500t meters. 5000 is constant, the given distance to the launch site.
\n" ); document.write( " \n" ); document.write( "
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