Mathematics, 12.10.2019 05:10 melissareid65

An airplane, flying at a constant speed of 360 mi/hr and climbing at a 30 degree angle, passes over a point p on the ground at an altitude of 21,120 ft. find the rate (in miles per hour) at which its distance from p is changing one minute later.

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Answer from: mahadharun

The point P is a point below the path of travel of the plane such as the

location of a house close to an airport.

Rate of change of the distance of the airplane from P is $\underline{\dfrac{1440\cdot \sqrt{19} }{19} \ \dfrac{km}{hr}}$

Reason:

The given parameters are;

The speed of the airplane = 360 mi/hr

The degree angle of the airplane = 30°

The height from which the airplane passes the point, P = 21,120 ft.

The distance the airplane travels per minute from the point in the x-direction is given as follows;

$D_x = \dfrac{360}{60} \times t \times cos (30^{\circ}) = 6 \cdot t \cdot cos (30^{\circ})$

The distance the airplane travels per minute from the point in the y-direction is given as follows;

$D_y = 4 + \dfrac{360}{60} \times t \times sin(30^{\circ}) = 4 + 6 \cdot t \cdot sin(30^{\circ})$

The magnitude of the distance, D, is given, by Pythagoras's theorem, as follows;

$D = \sqrt{D_x^2 + D_y^2}$ D^2 = D_x^2 + D_y^2}

Therefore;

D² = (6·t·cos(30°))² + (4 + 6·t·sin(30°))²

D² = 36·t²·cos²(30°) + 16 + 24·t·sin(30°) + 36·t²·sin²(30°)

D² = 36·t²·cos²(30°) + 36·t²·sin²(30°) + 16 + 24·t·sin(30°)

D² = 36·t²·(cos²(30°) + sin²(30°)) + 16 + 24·t·sin(30°)

D² = 36·t² + 16 + 24·t·sin(30°) = 36·t² + 24·t + 16

$\dfrac{d}{dt} D^2 = \dfrac{d}{dt} (36 \cdot t^2 + 24 \cdot t + 16)$

$2 \cdot D \cdot\dfrac{d}{dt} D = 72\cdot t + 24$

$\dfrac{dD}{dt} = \dfrac{72\cdot t + 24}{2 \cdot D}$

The distance covered in a minute, is given as follows;

D = √(6 × 1 × cos(30°))² + (4 + 6 × 1 × sin(30°))²) = 2·√19

The rate of change of the distance in a minute is therefore;

t = 1

$\dfrac{dD}{dt} = \dfrac{72\times 1 + 24}{2 \times 2 \times \sqrt{19} } = \dfrac{24 \cdot \sqrt{19} }{19}$

$\dfrac{dD}{dt} = \dfrac{24 \cdot \sqrt{19} }{19} \ km/min$

The rate of change of distance in miles per hour is found as follows;

$\dfrac{dD}{dt} \ in \ mi/hr = 60 \ min/hr\times \dfrac{24 \cdot \sqrt{19} }{19} \ km/min = \dfrac{60\cdot24 \cdot \sqrt{19} }{19} \ km/hr$

$\dfrac{dD}{dt} \ in \ mi/hr = \dfrac{60\cdot24 \cdot \sqrt{19} }{19} \ km/hr = \dfrac{1440\cdot \sqrt{19} }{19} \ km/hr$

The rate at which its distance from P is changing one minute later is therefore;

$\underline{\dfrac{1440\cdot \sqrt{19} }{19} \ \dfrac{km}{hr}}$

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Answer from: shyannmcmahon29991

Step-by-step explanation:

Given

velocity of airplane is 360 mi/hr

inclination of climbing $30^{\circ}$

Plane velocity can be divided in to x & y component

Velocity in x component is $360\cos30$

Velocity in y component is $360\sin30$

Thus distance traveled in x axis is $=360\cos30\times \frac{1}{60}$

=60cos30=51.96 miles

Distance travel in upward direction i.e. in Y axis

$=360\sin30\times \frac{1}{60}=30 miles$

so total altitude gain by plane is 30+4=34 miles

Horizontal distance traveled is 51.96

inclination w.r.t to point p

$tan\theta =\frac{34}{51.96}=0.654$

$\theta =33.18^{\circ}$

Initial inclination is $90^{\circ}$

therefore rate at which its distance is changing is $360\cos30=311.76 miles /hour$

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