Answer
(a) The altitude $y = 1~mile$
The velocity is $\frac{dx}{dt} = 500~mi/h$
The distance $D$ between the plane and the station is two miles.
(b) The unknown is $\frac{dD}{dt}$
(c) We can see a sketch below.
(d) $x^2+y^2 = D^2$
(e) The rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station is $~~433~mi/h$
Work Step by Step
(a) The altitude $y = 1~mile$
The velocity is $\frac{dx}{dt} = 500~mi/h$
The distance $D$ between the plane and the station is two miles.
(b) The unknown is $\frac{dD}{dt}$
(c) We can see a sketch below.
(d) $x^2+y^2 = D^2$
(e) We can find $x$ when $D = 2~miles$:
$x^2+y^2 = D^2$
$x^2 = D^2-y^2$
$x = \sqrt{D^2-y^2}$
$x = \sqrt{(2)^2-(1)^2}$
$x = \sqrt{3}~miles$
$2x~\frac{dx}{dt}+2y~\frac{dy}{dt} = 2D~\frac{dD}{dT}$
$\frac{dD}{dt} = \frac{x~\frac{dx}{dt}+y~\frac{dy}{dt}}{D}$
$\frac{dD}{dt} = \frac{(\sqrt{3})(500)+(1)(0)}{2}$
$\frac{dD}{dt} = 433~mi/h$
The rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station is $~~433~mi/h$
