Question

A plane flying horizontally at an altitude of 1 mile and a speed of 510 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 miles away from the station. (Round your answer to the nearest whole number.) mi/h

Answer 1

Answer:

442 miles

Step-by-step explanation:

Given

To properly solve this question, I illustrate some given parameters using attached image

From the image, apply Pythagoras theorem

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

Differentiate w.r.t time (t)

$2x\frac{dx}{dt} + 0 = 2y\frac{dy}{dt}$

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

Divide both sides by 2

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

From the question, we have that the plan travels are 510mi/h.

This implies that:

$\frac{dx}{dt} = 510mi/h$

So, we then calculate the value of x when the distance (y) is 2mi i.e.:

$y = 2mi$

Apply Pythagoras theorem

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

$x^2 + 1^2 = 2^2$

$x^2 + 1 = 4$

$x^2 = 4-1$

$x^2 = 3$

$x = \sqrt 3$

So, the expression becomes:

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

$\sqrt 3 * 510 = 2* \frac{dy}{dt}$

$\frac{\sqrt 3 * 510}{2} = \frac{dy}{dt}$

$\sqrt 3 * 255 = \frac{dy}{dt}$

$\frac{dy}{dt} = 255\sqrt 3$

$\frac{dy}{dt} = 255 * 1.7321$

$\frac{dy}{dt} = 441.655$

$\frac{dy}{dt} = 442$

Hence, the distance is 442 miles