Using implicit differentiation, it is found that the depth of the water is increasing at a rate of 0.19 feet/s.
The volume of a cone of radius r and height h is given by:
[tex]V = \frac{\pi r^2h}{3}[/tex]
It's rate of change, applying implicit differentiation, is given by:
[tex]\frac{dV}{dt} = \frac{1}{3}\left(2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}\right)[/tex]
The parameters are given as follows, considering the radius remains constant:
[tex]\frac{dV}{dt} = 20, r = 10, \frac{dr}{dt} = 0[/tex]
Hence:
[tex]\frac{dV}{dt} = \frac{1}{3}\left(2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}\right)[/tex]
[tex]20 = \frac{1}{3}\left(10^2\pi \frac{dh}{dt}\right)[/tex]
[tex]\frac{dh}{dt} = \frac{60}{100\pi}[/tex]
[tex]\frac{dh}{dt} = 0.19[/tex]
Hence, the depth of the water is increasing at a rate of 0.19 feet/s.
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