Question

A plane flying horizontally at an altitude of 1 mi and a speed of 540 mi/h passes directly over a radar station. How do you find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station?

Answer 1

$\textbf{ANSWER.}$

$\red{\texttt{speed of plane }}$= 540 mi/h

$\blue{\texttt{let 'x' }}$ = actual direct distance from the radar station

$\blue{\texttt{let 'y'}}$= horizontally distance from the radar station

$\blue{\textbf{then, we have to find,}}$

$\frac{dx}{dt} \: \: ,when \: x = 5$

$\blue{\texttt{speed of plane is constant}}$so,

$i.e \: \frac{dx}{dt} = 540$

$\blue{\textbf{using pythagoras,}}$

${x}^{2} = {y}^{2} + {1}^{2}$

$therefore.. \: \: \: \: {x}^{2} = {y}^{2} + 1.......(1)$

$\blue{\textbf{Differentiating Implicitly wrt..'t' we get, }}$

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

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

$x \frac{dx}{dt} = 540s$

$x \frac{dx}{dt} = 540\sqrt{ {x} ^{2} - 1} .........(using(1) \: )$

$\red{\texttt{when x =5 ,}}$

$5 \frac{dx}{dt} = 540 \sqrt{25 - 1}$

$5 \frac{dx}{dt} = 540 \sqrt{24}$

$\frac{dx}{dt} = 108 \sqrt{24}$

$\frac{dx}{dt} ≈529.1mi \: /hour$

$\textbf{so, rate}$ ≈$\textbf{529.1 mi/h}$

Answer 2

see the attachment above .