Question 1130066
given:

{{{10}}} feet across the top => {{{d=10ft}}} => {{{r=5ft}}}
and {{{12 }}}feet deep=> {{{h=12ft}}}

Water is flowing into the tank at a rate of {{{dV/dt=20 ft^3/min}}} 
the water is {{{4}}} feet deep=> {{{h[1]=4ft}}}

find: {{{dh/dt}}}

{{{r/h=5/12}}}

{{{12r=5h}}}

{{{r=5h/12}}}


{{{V=(1/3)pi*r^2*h}}}


{{{V=(1/3)pi*(5h/12)^2*h}}}


{{{V=(1/3)pi*(25h^2/144)*h}}}


{{{V=(1/3)pi*(25/144)h^3}}}



{{{dV/dt=(1/3)pi*(25/144)*3h^2(dh/dt)}}}


{{{dV/dt=(1/cross(3))pi*(25/144)*cross(3)h^2(dh/dt)}}}


{{{dV/dt=pi*(25/144)*h^2(dh/dt)}}}


{{{dV/dt=(25*h^2*pi)/144(dh/dt)}}}..........multiply by recip.


{{{(dh/dt)=144/(25*h^2*pi)(dV/dt)}}}..........since {{{h=4}}}, and {{{dV/dt=20 }}} we have


{{{(dh/dt)=144/(25*4^2*pi)(20)}}}


{{{(dh/dt)=cross(144)9/(cross(25)5*cross(16)*pi)(cross(20)4)}}}


{{{(dh/dt)=9/(5*pi)(4)}}}


{{{(dh/dt)=36/(5*pi) (ft/min)}}}