Question

A plane flying horizontally at an altitude of 1 mi and a speed of 560 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.)

Answer 1

Given:

altitude, x = 1 mile

speed, v = 560 mi/h

distance from the station, x = 4 mi

Solution:

To find the rate,

$\frac{dx}{dt} = 0$

Now, from the right angle triangle in fig 1.

Applying pythagoras theorem:

$h^{2}=x^{2} + y^{2}$

differentiating the above eqn w.r.t 't' :

$2h\frac{dh}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$                  (1)

Now, putting values in eqn (1):

$2h\frac{dh}{dt} = 2\times 1\times 0 + 2y\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{y}{h}\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}$

$\frac{dh}{dt} = 140\sqrt{4^2 - 1}$

The rate at which distance from plane to station is increasing is:

$\frac{dh}{dt} = 542.22 mph$