A large container has the shape of a frustum of a cone with top radius 5 m, bottom radius 3m, and height 12m.
The container is being filled with water at the constant rate 4.9 m^3/min. At what rate is the level of water rising at the instant the water is 2 deep?
the finial answer is = 14.0 cm/min
can any one help me solving this question (step by step solution)
1 answer
a frustum is part of a cone with radius 5m for bigger cone:.R=5m H=12+h for smaller cone:r=3m and height h by comparison:5/12+h=3/h 5h=36+3h 2h=36 h=18m the volume of water in full cone will also increase by 4.9m3/s considering only the bigger cone alone since water has passed the smaller cone.assume water is at h0 and radius r0,by comparison,5/r0=30/h0 h0=6r0 V=1/3pir0^h0 by substituting h0=6r0 V=1/3pi(h0/6)^2h0=1/3pih0^3/36.by differentiating dV/dt=pih0^2/36*dh0/dt when h0=18+2=20m,4.9=22/7*20^2/36*dh0/dt 4.9=34.92dh0/dt dh0/dt=0.143m/s=14.3cm/s
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