Question

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 1

Answer:

$-\frac{1}{2\pi}$ cm³/min  

Step-by-step explanation:

Let h = height of water and r = radius of its surface  

$V = \frac{ 1}{ 3}\pi r^{2}h$

At all times,  

$\frac{r }{h } = \frac{4 }{16 } = \frac{ 1}{4 }$

So, $r = \frac{h }{4 }$  

$V = \frac{ 1}{ 3}\pi (\frac{h }{ 4})^{2}h$

$V = \frac{ 1}{ 48}\pi h^{3}$

$\frac{\text{d}V}{\text{d}h} = \frac{ 1}{ 16}\pi h^{2}$

We want $\frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}V}{\text{d}t}$ = -2 cm³/min  

$-2 = \frac{ 1}{ 16}\pi h^{2} \frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}h}{\text{d}t} = -\frac{2}{\frac{ 1}{ 16}\pi h^{2} }$

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi h^{2}}$

When h = 8  cm

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi\times8^{2}}$

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{64\pi}$

$\frac{\text{d}h}{\text{d}t} = -\frac{1}{2\pi}$ cm³/min

The depth of the water is decreasing at a rate of $\frac{1}{2\pi}$ cm³/min

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