AIRPLANE Problem

**McDiesel** · November 7th 2008, 12:51 PM

Two airplanes (at the same height) are flying away from an airport at a right angle to each other.
The distance between them is 470 miles.
Plane X is 270 miles from the airport.
Plane Y is traveling at a speed of 520 mi/hr.
The distance between them is increasing by 660 mi/hr.

How fast is plane X traveling?

**Soroban** · November 7th 2008, 02:27 PM

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 $X$ is 270 miles from the airport.
Plane $Y$ is traveling at a speed of 520 mi/hr.
The distance between them is increasing by 660 mi/hr.

How fast is plane $X$ traveling?

Code:

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

The airport is at $A.$

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

Plane $Y$ is at $C\!:\;BC = 470.$
In right triangle $CAB\!:\;\;AC^2 + 270^2 \:=\:470^2 \quad\Rightarrow\quad AC \:=\:20\sqrt{370}$

In a certain time, plane $X$ flew $x$ miles from $B$ to $P.$
. . $AP \:=\:x+270$

In a certain time, plane $Y$ flew $y$ miles from $C$ to $Q\!:\;\;\tfrac{dy}{dt} = 520\text{ mph}$
. . $AQ \:=\:y + 20\sqrt{370}$

The distance between them is: . $z \:=\:PQ\:\text{ and }\:\tfrac{dz}{dt} = 660\text{ mph}$
From right triangle $QAP\!:\;z^2 \;=\;(x+270)^2 + (y + 20\sqrt{370})^2$

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

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

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

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

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