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Question

A water tank is in the shape of an inverted cone. The radius of the base is 8 meters and the height is 12 meters. The tank is being emptied for cleaning at the rate of 4met3/minute. Find the rate at which the water level will be decreasing, when the water is 6 meters deep.

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Solution

Let at any moment height and radius of a cone made by water is h and r respectively.
From similartiy principle of a triangle we have
OABC=ODBD
8r=12h
8h12=r
r=2h3
At time t, volume of water in a tank is,
V=13πr2h=13π(49h2)h
V=4π27h3
dVdt=4π27(3h2.dhdt)
dhdt=274π(3h2)dVdt=94πh2dVdt
But dVdt=4m3minute
dhdt=94πh2(4)
But we have h=6m
dhdt=9π(36)=14πm/minute
(dhdt)h=3=9π(36)=14πm/minute
Rate of decrease of the height of water tank is 14πmminute
884734_958719_ans_b6f5e7ac9a024b2a90b7430eafb8576d.PNG

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