A conical paper cup 20 cm across the top and 15 cm deep is full of water. The cup springs a leak at the bottom and losses water at 5 cu. cm per minute.
How fast is the water level dropping at the instant when the water is exactly 7.5 cm deep ?

Options

A. `(1)/(pi)cm//min`
B. `(1)/(5pi)cm//min`
C. `(1)/(2pi)cm//min`
D. `(2)/(3pi)cm//min`

Answer

To solve the problem step by step, we will follow the process of relating the volume of the cone to the height of the water and then finding the rate at which the height is changing.

### Step 1: Understand the dimensions of the cone
The cone has a diameter of 20 cm, which gives a radius \( R = \frac{20}{2} = 10 \) cm. The height of the cone is \( H = 15 \) cm.

### Step 2: Set up the relationship between the radius and height
Let \( h \) be the height of the water in the cone at any time \( t \), and let \( r \) be the radius of the water surface at height \( h \). By the similarity of triangles, we can write:
\[ \frac{H}{R} = \frac{h}{r} \]
Substituting the known values:
\[ \frac{15}{10} = \frac{h}{r} \]
This simplifies to:
\[ r = \frac{2}{3}h \]

### Step 3: Write the volume of water in the cone
The volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = \frac{2}{3}h \):
\[ V = \frac{1}{3} \pi \left(\frac{2}{3}h\right)^2 h = \frac{1}{3} \pi \frac{4}{9} h^2 h = \frac{4}{27} \pi h^3 \]

### Step 4: Differentiate the volume with respect to time
We need to find \( \frac{dh}{dt} \) when \( h = 7.5 \) cm. We know that the volume is decreasing at a rate of \( \frac{dV}{dt} = -5 \) cm³/min (negative because the volume is decreasing). Using the chain rule:
\[ \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \]
First, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \frac{d}{dh}\left(\frac{4}{27} \pi h^3\right) = \frac{4}{27} \pi \cdot 3h^2 = \frac{4\pi}{9} h^2 \]

### Step 5: Substitute known values
Now we substitute \( h = 7.5 \) cm into the derivative:
\[ \frac{dV}{dh} = \frac{4\pi}{9} (7.5)^2 = \frac{4\pi}{9} \cdot 56.25 = \frac{225\pi}{9} = 25\pi \]
Now we can set up the equation:
\[ -5 = 25\pi \cdot \frac{dh}{dt} \]
Solving for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-5}{25\pi} = \frac{-1}{5\pi} \text{ cm/min} \]

### Final Answer
The water level is dropping at a rate of \( \frac{-1}{5\pi} \) cm/min when the water is exactly 7.5 cm deep.