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July 30, 2014

July 30, 2014

Posted by **laquisha** on Monday, November 17, 2008 at 9:46pm.

- calculus -
**Reiny**, Tuesday, November 18, 2008 at 12:21amDraw a right-angled triangle ABC, B the right angle, and the vertical AB = 17 feet

Somewhere between B and C draw a vertical DE = 6 feet, the height of the woman

let BD = x and DC = y, the length of her shadow

given: dx/dt = 6 ft/s

find: dy/dt

(actually d(x+y)/dt, I'll come back to that later)

by similar triangles 6/y = 17/(x+y)

.

.

.

11y = 6x

then 11dy/dt = 6dx/dt

dy/dt = 6(6)/11 = 36/11 ft/s

so no matter where she is, the length of her shadow is increasing at 36/11 ft/s

but the woman herself is moving at 6 ft/sec

so her shadow is moving at 6+36/11 or

102/11 ft/s

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