Question

Hola brainlic :) !
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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

★ Question Given :

• ➲ 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?

★ Required Solution :

✯ Assumption Needed :

• ➲ S { Horizontal distance of plane from radar system }

• ➲ X { Actual Direct Distance of plane from radar system }

✯ According to Question :

• ➦ Plane moving in horizontal direct at constant speed = ds / dt = 540

✯ Applying Pythagoras Theroum :

• ➝ X² = S² + 1²

• ➝ X² = S² + 1 ____ eq (1)

✯ Now , Differentiating Expression :

• ➝ 2x (dx / dt) = 2s (ds / dt) + 0

• ➝ x(dx / dt) = s(ds / dt)

• ➝ x(dx / dt) = 540 S

• ➝ x(dx / dt) = 540 √ x² - 1

✯ Where , S = 5

• ➝ 5(dx / dt) = 540 √ 25 - 1

• ➝ 5(dx / dt) = 540 √24

• ➝ dx / dt = 108 √ 24

• ➝ dx / dt = 529 .1 mi / h

★ Therefore :

• ➦ The rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station is 529 .1 mi / h

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Answer 2

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

$\red{\texttt{speed of plane }}= 540$

$\blue{\texttt{let 'x' }} = actual \: direct \: distance$

$\blue{\texttt{let 'y'}}= horizont.ally \: distance \: from \: the \: radar \: station$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\frac{dx}{dt} ≈529.1mi \: /hour$

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

$\sf{\underline{\overline{\pink{ Thanks!!}}}}$