# related rates

• Dec 12th 2007, 10:27 AM
asdf
related rates
Two boats are headed due north. Boat A is traveling at 25 ft/sec while Boat B is traveling at 15 ft/sec. Boat A is following a course which is 240 ft west of boat B. How fast is the distance between the boats changing when Boat A is 450 ft behind Boat B?
• Dec 12th 2007, 11:38 AM
galactus
Problems like this call for ol' Pythagoras.

The boats are 240 feet apart. That is constant. Therefore, dx/dt=0

One is travelling at 25 ft/sec and the other at 15 ft/sec. 25-15=10

dy/dt=10 ft/sec.

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

When x=240 and y=450, then D=510

Differentiate:

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

$(510)\frac{dD}{dt}=240(0)+(450)(10)$

$\frac{dD}{dt}=\frac{150}{17} \;\ ft/sec$

Check my figures. I have done this in a hurry.