Draining a sphere: height is not volume
A spherical tank with negligible thickness and internal radius cm contains water. At time s, the water surface is at a height cm above the lowest point of the tank and the volume of water in the tank, cm³, is given by . Water flows from the tank, through an outlet at its lowest point, at a rate cm³ s⁻¹, where is a positive constant.
(i) Show that . marks
(ii) Find the general solution for in terms of . marks
(iii) Find the ratio , where is the time taken to empty the tank when initially it is completely full, and is the time taken to empty the tank when initially it is half full. marks
[2019/NYJC/Prelim/I/7]
Solution
Related rates
and water leaves at cm³/s. Show that .
Hint
.
The volume is decreasing.
Water flows out, so .
Answer · (i)
.
Differentiate .
Chain rule
The volume decreases.
Cancel
As required.
Separating variables
Find the general solution for in terms of , , .
Hint
Separate: .
.
Answer · (ii)
.
Separate
For , divide by to separate the variables before integrating. The empty endpoint is handled by the integrated expression.
Integrate
Power rule.
Solve for
Explore the idea
Emptying times
Find the ratio of the times to empty from full and from half full.
Hint
Use at the start: evaluated at the initial depth.
The tank is empty at : .
The upper and lower halves hold equal volumes. Outflow is larger at greater depth, so the upper half drains faster. The full time is about 1.62 times the half-full time.
Answer · (iii)
.
Full:
Initially means , which fixes . At full height the depth derivative is singular, but the integrated emptying time is finite.
Half full:
By symmetry is half the full volume . Set , ; .
Ratio
At empty, gives . The common factors and cancel in the ratio.