Hello, McDiesel!

Two airplanes are flying away from an airport at a right angle to each other.

The distance between them is 470 miles.

Plane $\displaystyle X$ is 270 miles from the airport.

Plane $\displaystyle Y$ is traveling at a speed of 520 mi/hr.

The distance between them is increasing by 660 mi/hr.

How fast is plane $\displaystyle X$ traveling? Code:

Q *
: | *
y | *
: | *
C * * z
: | * *
___| * 470 *
20√370| * *
: | * *
: | * *
- * - - - - - - - - * - - - - - *
A 270 B x P

The airport is at $\displaystyle A.$

Plane $\displaystyle X$ is at $\displaystyle B\!:\;AB = 270$

Plane $\displaystyle Y$ is at $\displaystyle C\!:\;BC = 470.$

In right triangle $\displaystyle CAB\!:\;\;AC^2 + 270^2 \:=\:470^2 \quad\Rightarrow\quad AC \:=\:20\sqrt{370}$

In a certain time, plane $\displaystyle X$ flew $\displaystyle x$ miles from $\displaystyle B$ to $\displaystyle P.$

. . $\displaystyle AP \:=\:x+270$

In a certain time, plane $\displaystyle Y$ flew $\displaystyle y$ miles from $\displaystyle C$ to $\displaystyle Q\!:\;\;\tfrac{dy}{dt} = 520\text{ mph}$

. . $\displaystyle AQ \:=\:y + 20\sqrt{370}$

The distance between them is: .$\displaystyle z \:=\:PQ\:\text{ and }\:\tfrac{dz}{dt} = 660\text{ mph}$

From right triangle $\displaystyle QAP\!:\;z^2 \;=\;(x+270)^2 + (y + 20\sqrt{370})^2 $

Differentiate with respect to time: .$\displaystyle 2z\,\frac{dz}{dt} \;=\;2(x+270)\,\frac{dx}{dt} + 2(y + 20\sqrt{370})\,\frac{dy}{dt}$

. . and we have: .$\displaystyle (x+270)\,\frac{dx}{dt} \;=\;z\,\frac{dz}{dt} - (y + 20\sqrt{370})\,\frac{dy}{dt}$

At that instant in question: . $\displaystyle \begin{Bmatrix}x &=& 0 \\ y&=&0 \\ z &=& 470 \\ \frac{dy}{dt} &=& 520 \\ \\[-4mm] \frac{dz}{dt} &=& 660 \end{Bmatrix}$

Substitute those values and solve for $\displaystyle \frac{dx}{dt}$