Question

Water is being drained from a container which has the shape of an inverted right circular cone. The
container has a radius of 5.00 inches at the top and a height of 9.00 inches. At the instant when the
water in the container is 8.00 inches deep, the surface level is falling at a rate of 2 in./sec. Find the rate
at which water is being drained from the container.

Answer 1

@RyanRyan Is the contain full to the top when the draining starts?

Answer 2

yes

Answer 3

??????????

Answer 4

my thoughts exactly

Answer 5

@RyanRyan

Answer 6

thanks

Answer 7

no clue,im guessing 124in 3/s

Answer 8

\[\bf \frac{ 10 }{ b }=\frac{ 9 }{ h } \implies 10h = 9b \implies 10\frac{ dh }{ dt }=9\frac{ db }{ dt }\]Now notice that 'b' here is the "base" of the similar triangle that is formed at the top as the water level drops and the 'h' is the height of this similar triangle. Because the two triangles are similar, I was able to come up with this equality of ratios. Differentiating both sides, I got 5 dh/dt = 9 db/dt. Now the volume of the amount of water that has drained is given by:\[\bf V_{drained} = 75 \pi - \frac{ b^2h \pi }{ 12 }\]Differentiating both sides we get: \[\bf \frac{ dV_{drained} }{ dt }=-\frac{2hb \pi}{12} \frac{ db }{ dt }-\frac{\pi b^2}{12}\frac{ dh }{ dt }\]We know that the surface level drops at 2 in/sec when the height of the similar triangle is 8 in, this implies that the height of this similar triangle drops at a rate of -2 in/sec, hence dh/dt = -2 in/sec when h = 8. We also know that at this point, the base of the similar triangle is given by:\[\bf \frac{ b }{ 10 }=\frac{ 8 }{ 9 } \implies b \approx 8.889\]Now we need to find db/dt. Going all the way up in this post and looking at 10 dh/dt = 9 db/dt, we can solve for db/dt by plugging in -2 for dh/dt:\[\bf 10(-2)=9\frac{ db }{ dt } \implies \frac{ db }{ dt }=-\frac{ 20 }{ 9 }\]Now we have everything to find dV/dt. Just plug in b = 8.889, h = 8, dh/dt = -2, db/dt = -20/9 and solve for dV/dt:\[\bf \frac{ dV_{drained} }{ dt }=-\frac{ 2(8)(8.889) \pi }{ 12 }\left(-\frac{ 20 }{ 9 }\right)-\frac{ \pi (8.889)^2 }{12 }(-2) \approx 124.114\]
Hence the water is being drained at a rate of 124.114 in^3/sec, i.e.
\[\bf \frac{dV_{drained}}{dt} \approx -124.114 \ in^3/sec\]