Asked by Nate
A rocket fired straight in the air is being tracked by a radar station 3 miles from the launching pad. If the rocket is travelling at 2 miles per second, how fast is the distance between the rocket and the tracking station changing at the moment when the rocket is 4 miles in the air?
D^2=9+y^2
D^2=9+y^2
Answers
Answered by
Steve
I assume you meant fired straight up in the air. In that case, at time t seconds, the rocket has gone 2t miles.
So, the distance D is found, as you showed, by
D^2 = 9+y^2 = 9+4t^2
when the rocket is 4 miles up, t=2, and we have a 3-4-5 triangle, so D=5.
2D dD/dt = 8t
when t=2, we have
2(5) dD/dt = 8(2)
dD/dt = 8/5
So, the distance D is found, as you showed, by
D^2 = 9+y^2 = 9+4t^2
when the rocket is 4 miles up, t=2, and we have a 3-4-5 triangle, so D=5.
2D dD/dt = 8t
when t=2, we have
2(5) dD/dt = 8(2)
dD/dt = 8/5
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