Question

A plane flying horizontally at an altitude of 1 mi and a speed of 490 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station. (Round your answer to the nearest whole number.) mi/h

Answer 1

Answer: 438.3 mph

Explanation:

Let z be the distance from the plane to the station. You should draw a right triangle

for the diagram with z on the hypotenuse, 1 on the vertical side, and x on the horizontal

side. The 1 never changes, but x changes with time.

Find the attached file for the solution.

Answer 2

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 424 mi/h

Explanation:

Given;

vertical position of the plane, h = 1 mi

when plane is 2 mi away from the station, this position and vertical position forms a right - angled triangle.

Let the vertical position = y = 1 mi

Let the 2 mi position = hypotenuse = p

Let the remaining side of the triangle, which is horizontal = x

x² = p² - y²

$x = \sqrt{p^2 -y^2} = \sqrt{2^2 -1^2} = \sqrt{3}$

Again;

p² = x² + y²

the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station, can be determined by differentiating P with respect to time t.

$p^2.\frac{dp}{dt} = x^2.\frac{dx}{dt} +y^2.\frac{dy}{dt} \\\\2p.\frac{dp}{dt} = 2x.\frac{dx}{dt} + 2y.\frac{dy}{dt}\\\\2p.\frac{dp}{dt} = 2x.\frac{dx}{dt} + 0\\\\\frac{dp}{dt} = \frac{2x}{2p} .\frac{dx}{dt} \\\\\frac{dp}{dt} = \frac{x}{p} .\frac{dx}{dt} = \frac{\sqrt{3}}{2} . 490(\frac{mi}{h}) = 424 \ mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 424 mi/h