Answer:
[tex]\frac{dh}{dt}=-\frac{1}{2\pi}cm/min[/tex]
Step-by-step explanation:
From similar triangles, see diagram in attachment
[tex]\frac{r}{4}=\frac{h}{16}[/tex]
We solve for [tex]r[/tex] to obtain,
[tex]r=\frac{h}{4}[/tex]
The formula for calculating the volume of a cone is
[tex]V=\frac{1}{3}\pi r^2h[/tex]
We substitute the value of [tex]r=\frac{h}{4}[/tex] to obtain,
[tex]V=\frac{1}{3}\pi (\frac{h}{4})^2h[/tex]
This implies that,
[tex]V=\frac{1}{48}\pi h^3[/tex]
We now differentiate both sides with respect to [tex]t[/tex] to get,
[tex]\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}[/tex]
We were given that water is drained out of the tank at a rate of [tex]2cm^3/min[/tex]
This implies that [tex]\frac{dV}{dt}=-2cm^3/min[/tex].
Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means [tex]h=8cm[/tex].
We substitute this values to obtain,
[tex]-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}[/tex]
[tex]\Rightarrow -2=4\pi \frac{dh}{dt}[/tex]
[tex]\Rightarrow -1=2\pi \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}=-\frac{1}{2\pi}[/tex]
