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Differential Calculus> at a distance of 4000ft from the launch s...

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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 :

vivek , 10 Years ago

Grade 6

2 Answers

Bhaskar bhadani

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

Last Activity: 8 Years ago

muhammad hassan

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}

Last Activity: 3 Years ago

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