Answer:
[tex]\dfrac{dh}{dt}=1.41\ ft/min[/tex]
Explanation:
Given that
h= 12 ft
r= 6 ft
h= 2 r
r=h/2
We know that volume of cone given as
[tex]V=\dfrac{1}{3}\pi r^2h[/tex]
Now by putting the values
r=h/2
[tex]V=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2h[/tex]
[tex]V=\dfrac{1}{12}\pi h^3[/tex]
Given that when h= 3 ft
[tex]\dfrac{dV}{dt}=10\ ft^3/min[/tex]
[tex]V=\dfrac{1}{12}\pi h^3[/tex]
[tex]\dfrac{dV}{dt}=\dfrac{\pi}{12}(3h^2)\times \dfrac{dh}{dt}[/tex]
By putting the values
[tex]10=\dfrac{\pi}{12}(3\times 3^2)\times \dfrac{dh}{dt}[/tex]
[tex]\dfrac{dh}{dt}=1.41\ ft/min[/tex]
So the rate at which the water level is rising is 1.41 ft/min
