A plane flying horizontally at an altitude of 1 mi and a speed of 540

1.	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» er: \textbf{ANSWER.}ANSWER. \red{\texttt{speed of plane }}speed of plane = 540 mi/h \blue{\texttt{let 'x' }}let ’x’ = actual direct distance from the radar station \blue{\texttt{let 'y'}}let ’y’ = horizontally distance from the radar station \blue{\textbf{then, we have to find,}}then, we have to find, \FRAC{dx}{DT} \: \: ,when \: x = 5dtdx,whenx=5 \blue{\texttt{speed of plane is CONSTANT}}speed of plane is constant so, i.e \: \frac{dx}{dt} = 540i.edtdx=540 \blue{\textbf{using pythagoras,}}using pythagoras, {x}^{2} = {y}^{2} + {1}^{2}x2=y2+12 therefore.. \: \: \: \: {x}^{2} = {y}^{2} + 1...(1)therefore..x2=y2+1...(1) \blue{\textbf{Differentiating Implicitly wrt..'t' we get, }}Differentiating Implicitly wrt..’t’ we get, 2x \frac{dx}{dt} = 2x \frac{dy}{dt} + 02xdtdx=2xdtdy+0 x \frac{dx}{dt} = y \frac{dy}{dt}xdtdx=ydtdy x \frac{dx}{dt} = 540sxdtdx=540s x \frac{dx}{dt} = 540\sqrt{ {x} ^{2} - 1}...(using(1) \: )xdtdx=540x2−1...(using(1)) \red{\texttt{when x =5 ,}}when x =5 , 5 \frac{dx}{dt} = 540 \sqrt{25 - 1}5dtdx=54025−1 5 \frac{dx}{dt} = 540 \sqrt{24}5dtdx=54024 \frac{dx}{dt} = 108 \sqrt{24}dtdx=10824 \frac{dx}{dt} ≈529.1mi \: /hourdtdx≈529.1mi/hour \textbf{so, RATE}so, rate ≈\textbf{529.1 mi/h}529.1 mi/h

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»

er:

\textbf{ANSWER.}ANSWER.

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

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

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

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

\FRAC{dx}{DT} \: \: ,when \: x = 5dtdx,whenx=5

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

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

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

{x}^{2} = {y}^{2} + {1}^{2}x2=y2+12

therefore.. \: \: \: \: {x}^{2} = {y}^{2} + 1...(1)therefore..x2=y2+1...(1)

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

2x \frac{dx}{dt} = 2x \frac{dy}{dt} + 02xdtdx=2xdtdy+0

x \frac{dx}{dt} = y \frac{dy}{dt}xdtdx=ydtdy

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

x \frac{dx}{dt} = 540\sqrt{ {x} ^{2} - 1}...(using(1) \: )xdtdx=540x2−1...(using(1))

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

5 \frac{dx}{dt} = 540 \sqrt{25 - 1}5dtdx=54025−1

5 \frac{dx}{dt} = 540 \sqrt{24}5dtdx=54024

\frac{dx}{dt} = 108 \sqrt{24}dtdx=10824

\frac{dx}{dt} ≈529.1mi \: /hourdtdx≈529.1mi/hour

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