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July 28, 2014

July 28, 2014

Posted by **john** on Wednesday, June 22, 2011 at 11:00am.

- calculus ( related rates ) -
**Reiny**, Wednesday, June 22, 2011 at 12:39pmLet the radius of the water level be r cm

let the height of the water be h cm

by ratios of similar triangles (using a cross-section)

r/h = 5/8 ----> h = 8r/5

V= (1/3)πr^2h

= (1/3)πr^2(8r/5) = (8π/15)r^3

dV/dt = (8π/5)r^2 dr/dt

-2π = (8π/5)r^2 dr/dt

when r = 5

dr/dt = -1/20

A =πr^2

dA/dt = 2πr dr/dt

when r = 5

dA/dt = 2π(5)(-1/20) = -π/2 cm^2/min

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