Posted by **Nel** on Saturday, November 6, 2010 at 8:02pm.

Two roads intersect at right angles. At a certain moment, one bicyclist is 8 miles due NORTH of the intersection traveling TOWARDS the intersection at a rate of 16 miles/hour. A second bicyclist is 10 miles due WEST of the intersection traveling AWAY from the intersection at 12 miles/hour.

(a) How fast is distance between the two bicyclists changing at this moment?

- AP CALC -
**Reiny**, Saturday, November 6, 2010 at 8:22pm
make a diagram, let the distance the north biker is from the intersection be y miles, let the distance the west biker is from the intersection be x miles

Let the distance between them be D

D^2 = x^2 + y^2

2D dD/dt = 2x dx/dt + 2y dy/dt

given:

dx/dt = 12 mph

dy/dt = -16 mph (y is decreasing)

find dD/dt when x=10 and y=8

then D^2 = 100+64

D = √164

dD/dt =(2(10)(12) + 2(8)(-16))/(2√164) = .....

you finish the arithmetic

- AP CALC -
**Nel**, Saturday, November 6, 2010 at 8:55pm
My final answer was -8/square root of 164. Is that correct?

- AP CALC -
**Jonathan**, Thursday, December 4, 2014 at 10:28pm
I believe the correct answer is 0.635 but i got 0.625.

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