![volume~=\frac{4}{3}h^3](https://studentforums.biz/cgi-bin/mimetex.cgi?volume~=\frac{4}{3}h^3)
Differentiating :
![\frac{\mathrm{dV} }{\mathrm{d} h}=4h^2](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{\mathrm{dV} }{\mathrm{d} h}=4h^2)
The rate of change of volume with respect to time is equal to the water input minus the water loss :
![\frac{\mathrm{dV} }{\mathrm{d} t}=20-kh^2](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{\mathrm{dV} }{\mathrm{d} t}=20-kh^2)
-kh
2 represents the loss via leakage
![\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} t}](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} t})
![\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} V} = \frac{\mathrm{dh} }{\mathrm{d} t}<br />](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{\mathrm{dV} }{\mathrm{d} t}\times \frac{\mathrm{dh} }{\mathrm{d} V} = \frac{\mathrm{dh} }{\mathrm{d} t}<br />)
![\frac{\mathrm{dh} }{\mathrm{d} t}=\frac{20-kh^2}{4h^2}](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{\mathrm{dh} }{\mathrm{d} t}=\frac{20-kh^2}{4h^2})
Simplify to :
![\frac{5}{h^2}-\frac{k}{4}](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{5}{h^2}-\frac{k}{4})
![When~h=1,\frac{\mathrm{dh} }{\mathrm{d} t}=4.95](https://studentforums.biz/cgi-bin/mimetex.cgi?When~h=1,\frac{\mathrm{dh} }{\mathrm{d} t}=4.95)
Input h=1 into above equation and find k. You will see the end solution is the equation given in the paper.