Draining a sphere: height is not volume

A spherical tank with negligible thickness and internal radius aa cm contains water. At time tt s, the water surface is at a height xx cm above the lowest point of the tank and the volume of water in the tank, VV cm³, is given by V=13πx2(3a−x)V=\frac13\pi x^2(3a-x). Water flows from the tank, through an outlet at its lowest point, at a rate πkx\pi k\sqrt x cm³ s⁻¹, where kk is a positive constant.

(i) Show that (2ax−x2)dxdt=−kx(2ax-x^2)\frac{dx}{dt}=-k\sqrt x. marks

(ii) Find the general solution for tt in terms of x,a,kx,a,k. marks

(iii) Find the ratio T1:T2T_1:T_2, where T1T_1 is the time taken to empty the tank when initially it is completely full, and T2T_2 is the time taken to empty the tank when initially it is half full. marks

[2019/NYJC/Prelim/I/7]

Total marks: 9

Solution

(i)

Related rates

V=13πx2(3a−x)V=\frac13\pi x^2(3a-x) and water leaves at πkx\pi k\sqrt x cm³/s. Show that (2ax−x2)dxdt=−kx(2ax-x^2)\dfrac{dx}{dt}=-k\sqrt x.

Hint

  1. dVdt=dVdx⋅dxdt\frac{dV}{dt}=\frac{dV}{dx}\cdot\frac{dx}{dt}.

  2. The volume is decreasing.

Answer · (i)

(2ax−x2)x′=−kx(2ax-x^2)x\prime=-k\sqrt x.

(ii)

Separating variables

Find the general solution for tt in terms of xx, aa, kk.

Explore this idea

Hint

  1. Separate: 2ax−x2xdx=−k dt\frac{2ax-x^2}{\sqrt x}dx=-k\,dt.

Answer · (ii)

t=[C−(4a/3)x3/2+(2/5)x5/2]/kt=[C-(4a/3)x^{3/2}+(2/5)x^{5/2}]/k.

Explore the idea

Depth against time (a = 1, k = 1)

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 01
Read the solution at t0.5
Try an example
x at t = 0.5
0.528
Rate dx/dt there
-0.935
  • Solution that starts at the purple dot
Key idea

Depth changes rapidly near both the full and empty endpoints because the water surface is small. This is a depth rate; the volume outflow is πkx\pi k\sqrt x.

Ideas to try
  1. Start the solution at the top (x≈2x\approx2) and at x=1x=1: which empties faster per unit depth?
  2. Read the time when the depth reaches 0.
(iii)

Emptying times

Find the ratio T1:T2T_1:T_2 of the times to empty from full and from half full.

Hint

  1. Use t=0t=0 at the start: C=43ax3/2−25x5/2C=\frac43ax^{3/2}-\frac25x^{5/2} evaluated at the initial depth.

  2. The tank is empty at x=0x=0: T=CkT=\frac Ck.

Answer · (iii)

T1:T2=82:7T_1:T_2=8\sqrt2:7.