Question

A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

Answer 1

Answer:

$250\sqrt 3mi/h$

Explanation:

Altitude,h=1 mi

Speed,v=$\frac{dx}{dt}=$500mi/h

We have to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y^2=x^2+h^2=x^2+1$

y=2 mi

$2^2=x^2+1$

$4-1=x^2$

$3=x^2$

$x=\sqrt 3$mi

Differentiate w.r. t

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

$y\frac{dy}{dt}=x\frac{dx}{dt}$

$\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}$

$\frac{dy}{dt}=\frac{\sqrt 3}{2}\times 500=250\sqrt 3mi/h$