We can make use of the complying with layout to assist in the resolution:
The deepness of the water in the cone is stood for by $latex x$. After that, the distance $latex r$ of the cross-section of the water is offered by:
$$ tan( 30 ^ {circ} )= frac {r} {x} $$
$$ r= xtan( 30 ^ {circ} )$$
$$= frac {x} {sqrt {3}} $$
The quantity of water in the cone is offered by:
$$ V= frac {1} {3} pi r ^ 2x$$
$$= frac {1} {3} pi left( frac {x} {sqrt {3}} right) ^ 2x$$
$$= frac {1} {9} pi x ^ 3$$
Then, the by-product of the quantity relative to $latex x$ is:
$$ dfrac {dV} {dx} pi x ^ 2$$
From the concern, we understand that $latex dfrac {dV} {dt} =5$. Additionally, by the chain guideline, we have:
$$ dfrac {dV} {dt} =dfrac {dV} {dx} dfrac {dx} {dt} $$
Then,
$$ 5= frac {1} {3} pi x ^ 2 dfrac {dx} {dt} $$
$$ dfrac {dx} {dt} = frac {15} {pi x ^ 2} $$
When $latex x= 10$, we have:
$$ dfrac {dx} {dt} =frac {15} {pi (10 )^ 2} =frac {3} {20pi} $$
The price of adjustment of deepness when the deepness is 10 centimeters is $latex frac {3} {20pi} $ cm/s.