Im unsure how to do this question after attempting it a few times. Please help!
A Container is in the form of a right circular cone of height 16 and base radius 4 is held vertex downward and filled with liquid. If the liquid leaks our from the vertex at a rate of 4 find the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out.
Thank.
The change in height is the volume of the leak divided by the surface area at the water level.
The volume of water in the cone is $\alpha r^3$
When it is full it is $64\alpha$
and when it is half full, it is $32\alpha.$
And the radius of the surface of the water is $32^{\frac 13}$
The surface area is $\pi 32^\frac 23$
$\frac {4}{\pi 32^\frac 23} = \frac 1{\pi\sqrt[3]{16}}$
But you wanted to use calculus
$V = \frac 13 \pi r^2 h\\ h = 4r$
$V = \frac 13 \pi (4r^3) = \frac 13 \pi \frac {h^3}{16}\\ \frac {dv}{dt}=4=\pi {h^2}{16} \frac {dh}{dt}$
We need to find $h$ when the tank is half full.
When the tank is full, $h = 16$
$V = \frac 13 \pi (4r^3) = \frac 13 \pi 256$
and when it is half full
$V = \frac 13 \pi (4r^3) = \frac 13 \pi 128$
$h = \frac {16}{2^\frac 13} = 2^{11}{3}$
$2^2=\pi \frac {2^{\frac {22}{3}}}{2^4} \frac {dh}{dt}\\ \frac {dh}{dt} = \frac {2^6}{\pi2^\frac {22}{3}} = \frac 1{\pi 2^{\frac 43}}$