Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 1/2 cubic feet per second. Note: The volume of a cylinder is V=(pi)(r^2)(h).

Question

A. Find the volume, V, of the water remaining in the small tank as a function of time.

B. How long does it take for the small tank to completely empty?

C. Let z be the depth of the water in the large tank, which is initially empty. Compute dz/dt.

D. What fraction of the total amount of water is in the large tank at time t=6?

Answer 1

the small tank has a volume of 96π ft^3

its depth decreases at a rate of (1/2)/(16π) = 1/(32π) ft/s

So,

A: v(t) = 96π - 1/2 t
B: duh
C: the large tank's area is 4 times as big, so its depth increases 1/4 as fast, or 1/8 ft/s.
D: 6/(192π)

Answer 2

The large tank's area or volume is not 4 times as big though.....