Rates of change question involving water leaking out of a hemispherical tank

Question

Water is leaking out the bottom of a hemispherical tank of radius 9 feet at a rate of 4 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height h is 8 ​feet? Note​: the volume of a segment of height h in a hemisphere of radius r is $$πh^2[r− \frac {h}3]$$

Answer 1

The height at time $t$ is $h$=8ft. At a very small time interval after that $t+dt$, the new height is $h-dh$. Note the minus sign, since the height decreases. The volume change is $$dv=\pi h^2\left[r− \frac {h}3\right]-\pi (h-dh)^2\left[r− \frac {h-dh}3\right]$$ The rate of change in the volume is equal to 4ft$^3$/hr=$dv/dt$. If you ignore terms in $dh^2$ and $dh^3$, which are small, you get $$\frac{dv}{dt}=-\pi h(h-2r)\frac{dh}{dt}$$ From here $$\frac{dh}{dt}=\frac{1}{\pi(2r-h)h}\frac{dv}{dt}$$

Answer 2

Let the level of water after 1 hour be $h$.

so change in volume in one hour is :

$π8^2[9− \frac {8}3] - πh^2[9− \frac {h}3] = 4$, (the difference in volume of water stored in tank)

you can take positive value of $h$ after solving the above equation and your change in level is $8-h$ feet per hour.