Question

A plane flying horizontally at an altitude of 3 mi and a speed of 460 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 4 mi away from the station (Round your answer to the nearest whole number.) 368 X mi/h Enhanced Feedback Please try again. Keep in mind that distance - (altitude)2 + (horizontal distance)? (or y = x + n ). Differentiate with respect to con both sides of the equation, using the Chain Rule, to solve for the given speed of the plane is x.

Answer 1

Answer:

$\frac{dy}{dt}=304mi/h$

Step-by-step explanation:

From the question we are told that:

Height of Plane $h=3mi$

Speed $\frac{dx}{dt}=460mi/h$

Distance from station $d=4mi$

Generally the equation for The Pythagoras Theorem is is mathematically given by

$x^2+3^2=y^2$

For y=d

$x^2+3^2=d^2$

$x^2+3^2=4^2$

$x=\sqrt{7}$

Therefore

$x^2+3^2=y^2$

Differentiating with respect to time t we have

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

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

$\frac{dy}{dt}=\frac{\sqrt{7}}{4} *460$

$\frac{dy}{dt}=304.2614008mi/h$

$\frac{dy}{dt}=304mi/h$