# Math Help - Rate of Change

1. ## Rate of Change

Can someone assist me with this one:

A person obeserves an airplane directly overhead which moves at 500ft/sec at an altitue of 8000ft. The plane maintains constant altitude and horizontal velocity. What is the rate of change of the distance between the obeserver and the airplane after 12 seconds?

2. Did you draw a triangle. We can use Pythagoras.

dx/dt=500, y=8000, dy/dt=0 because it remains constant.

After 12 seconds, the plane will have travelled 6000 feet.

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

$D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$

By Pythagoras, $D=\sqrt{8000^{2}+6000^{2}}=10,000$

$10,000\frac{dD}{dt}=6000(500)+8000(0)$

Solve for dD/dt.