at a distance of 4000ft from the launch site a spectator is observing a rocket being launched if the rocket lifts off virtically and is rising at a speed of 600ft/s when it is at an altitiude 3000ft,the distance between the rocket and the spectator changing at that instant at the rate :

21 Points
6 years ago
By applying Pythagoras theoremLet`s assume that rocket is at Y ft. Above ground and dist. from observer is X.Then,X^2=Y^2+5000Diff. it2xdx/dt=2ydy/dt+0Basically you have to find dx/dtTherefore,Putting x=5000ft.,Y=3000ft.&dy/dt=600ft/sYou will get dx/dt=360ft/s
13 Points
10 months ago
let us consider:
distance of rocket at 3000 feet = x
distance of rocket from the ground = y
as we all know the famous pythagorous theorem: a^2 + b^2 = c^2
this is going to be applied here.
so A.T.G.C our equation is as under:
x^2 + y^2 = 5000
now actually we have to differentiate this equation
so differentiating it on both sides with respect to t, we get
2x.dx/dt = 2y.dy/dt + 0   --------------------------- 1
and we are given that dy/dt = 600(as it is rate od change of rocket’s disttance from the ground i.e its speed)
and we have x and y.
so putting the values in our eq 1:
we get
2(5000) dx/dt = 2(3000) (600)
by simplification we get:
dx/dt = 360
so  ,the distance between the rocket and the spectator is changing at that instant at the rate of 360feet per second