Question #329560

Water is being poured at the rate of 15cm3/sec into an inverted cone with a diameter of 3m and a 

height of 6m. At what rate is the surface rising just as the tank is filled?


1
Expert's answer
2022-04-18T00:27:21-0400

Let u=15 cm3/sec, D=3 m, H=6 mu=15\space cm^3/sec,\space D=3\space m,\space H=6\space m

The volume of the cone: V=13πd24h=πd2h12,V=\frac{1}{3}\frac{\pi d^2}{4}h=\frac{\pi d^2h}{12},

where d - current diameter of the surface,

h - current height of the water cone.

Current water cone and the cone containing this water are the similar objects with a similarity factor of k = h / H.

d=kD=hDHV=π12(DH)2h3d=kD=h\frac{D}{H}\Rarr V=\frac{\pi}{12}(\frac{D}{H})^2h^3

u=dVdt=π12(DH)23h2dhdt=π(Dh2H)2dhdtdhdt=uπ(2HDh)2u=\frac{dV}{dt}=\frac{\pi}{12}(\frac{D}{H})^2\cdot3h^2\frac{dh}{dt}=\pi(\frac{Dh}{2H})^2\frac{dh}{dt}\Rarr\\ \Rarr\frac{dh}{dt}=\frac{u}{\pi}(\frac{2H}{Dh})^2

At the point in time when the cone is almost filled h = H:

dhdt(h=H)=uπ(2D)2153.14(26)20.53 (m/sec)\frac{dh}{dt}(h=H)=\frac{u}{\pi}(\frac{2}{D})^2\approx\frac{15}{3.14}(\frac{2}{6})^2\approx0.53\space(m/sec)


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