Let 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
a paper cup in the shape of a cone has a diameter of 10cm across the top, and is 8cm deep. if the cup is leaking out the bottom at 2pi cm^3/min, at what rate is thet area of the water surface ( only the top surface of the water) changing, when the cup has a the water depth of 5cm? exact solutions first; round to two decimal places
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