Expert: Abe Mantell - 1/7/2004

Question
Hi! The question is: A radar station is 2000 ft from the launch site of a rocket.  if the rocket is launched vertically at the rate of 500 ft/sec, how fast is the distance between the radar station and the rocket changing 10 seconds later?
Thanks so much!

Answer
Hello Melanie,

Let x=the distance from the rocket to the radar station,
which will be the hypotenuse of a right triangle.  So we
can use the Pythagorean Thrm. to get a relationship we need.
The vertical height of the rocket is 500t ft, where t is
the time measured in seconds.  Thus, we get:
2000^2 + (500t)^2 = x^2, now solving for x and taking
the positive square root (since the distance must be +):
x=sqrt(4000000 + 250000t^2), now take the derivative with
respect to 't' ==>
dx/dt= (1/2)(4000000 + 250000t^2)^(-1/2) * (500000t),
which reduces to dx/dt=500t/sqrt(t^2 + 16),
now let t=10 to get dx/dt at 10 seconds later:
==> dx/dt= 2500/sqrt(29) = 464.24 ft/sec

OK?

Abe