Answer to Question #329560 in Calculus for Keila

Question

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?

Answer 1

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

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

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.

$d=kD=h\frac{D}{H}\Rarr V=\frac{\pi}{12}(\frac{D}{H})^2h^3$

$u=\frac{dV}{dt}=\frac{\pi}{12}(\frac{D}{H})^2\cdot3h^2\frac{dh}{dt}=\pi(\frac{Dh}{2H})^2\frac{dh}{dt}\Rarr\\ \Rarr\frac{dh}{dt}=\frac{u}{\pi}(\frac{2H}{Dh})^2$

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

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