Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer

Answer:[tex]-\frac{1}{2\pi}[/tex] cm³/min  Step-by-step explanation:Let h = height of water and r = radius of its surface  [tex]V = \frac{ 1}{ 3}\pi r^{2}h[/tex]At all times,  [tex]\frac{r }{h } = \frac{4 }{16 } = \frac{ 1}{4 }[/tex]So, [tex]r = \frac{h }{4 }[/tex]  [tex]V = \frac{ 1}{ 3}\pi (\frac{h }{ 4})^{2}h[/tex][tex]V = \frac{ 1}{ 48}\pi h^{3}[/tex][tex]\frac{\text{d}V}{\text{d}h} = \frac{ 1}{ 16}\pi h^{2}[/tex]We want [tex]\frac{\text{d}h}{\text{d}t}[/tex][tex]\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \frac{\text{d}h}{\text{d}t}[/tex][tex]\frac{\text{d}V}{\text{d}t}[/tex] = -2 cm³/min  [tex]-2 = \frac{ 1}{ 16}\pi h^{2} \frac{\text{d}h}{\text{d}t}[/tex][tex]\frac{\text{d}h}{\text{d}t} = -\frac{2}{\frac{ 1}{ 16}\pi h^{2} }[/tex][tex]\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi h^{2}}[/tex]When h = 8  cm[tex]\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi\times8^{2}}[/tex][tex]\frac{\text{d}h}{\text{d}t} = -\frac{32}{64\pi}[/tex][tex]\frac{\text{d}h}{\text{d}t} = -\frac{1}{2\pi}[/tex] cm³/minThe depth of the water is decreasing at a rate of [tex]\frac{1}{2\pi}[/tex] cm³/min.