Given:
altitude, x = 1 mile
speed, v = 560 mi/h
distance from the station, x = 4 mi
Solution:
To find the rate,
[tex]\frac{dx}{dt} = 0[/tex]
Now, from the right angle triangle in fig 1.
Applying pythagoras theorem:
[tex]h^{2}=x^{2} + y^{2}[/tex]
differentiating the above eqn w.r.t 't' :
[tex]2h\frac{dh}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex] (1)
Now, putting values in eqn (1):
[tex]2h\frac{dh}{dt} = 2\times 1\times 0 + 2y\frac{dy}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{y}{h}\frac{dy}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}[/tex]
[tex]\frac{dh}{dt} = 140\sqrt{4^2 - 1}[/tex]
The rate at which distance from plane to station is increasing is:
[tex]\frac{dh}{dt} = 542.22 mph[/tex]