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?
Answers
Answer:
\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
★ 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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