Water is filowing into right circular conical vessel, 10 inch deep and 10 inch in diameter at the rate of 4 (inch)`""^(3)` /min, How fast is the water-level rising when the water is 8 inch deep?

Options

A. `(pi)/(4)` inch/min
B. `4pi` inch/min
C. `(1)/(4pi)` inch/min
D. `(1)/(2pi)` inch/min

Answer

To solve the problem step by step, we will follow the mathematical principles and relationships involved in the situation described.

### Step-by-Step Solution:

1. **Understand the Geometry of the Cone**:
- The conical vessel has a height \( H = 10 \) inches and a diameter \( D = 10 \) inches. Therefore, the radius \( R \) is:
\[ R = \frac{D}{2} = \frac{10}{2} = 5 \text{ inches} \]

2. **Establish the Relationship Between Radius and Height**:
- The relationship between the radius \( r \) of the water surface and the height \( h \) of the water in the cone can be derived from similar triangles:
\[ \frac{r}{h} = \frac{R}{H} = \frac{5}{10} = \frac{1}{2} \]
- Thus, we can express \( r \) in terms of \( h \):
\[ r = \frac{h}{2} \]

3. **Volume of the Cone**:
- The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
- Substituting \( r = \frac{h}{2} \) into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{\pi}{12} h^3 \]

4. **Differentiate Volume with Respect to Time**:
- We need to find how fast the height \( h \) is changing with respect to time \( t \) when the water is 8 inches deep. We differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} \]

5. **Substituting Known Values**:
- We know that water is flowing into the cone at a rate of \( \frac{dV}{dt} = 4 \) cubic inches per minute. We will substitute \( h = 8 \) inches into the equation:
\[ 4 = \frac{\pi}{4} (8^2) \frac{dh}{dt} \]
- Simplifying:
\[ 4 = \frac{\pi}{4} \cdot 64 \frac{dh}{dt} \]
\[ 4 = 16\pi \frac{dh}{dt} \]

6. **Solve for \( \frac{dh}{dt} \)**:
- Rearranging the equation to find \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{4}{16\pi} = \frac{1}{4\pi} \]

### Final Answer:
The rate at which the water level is rising when the water is 8 inches deep is:
\[ \frac{dh}{dt} = \frac{1}{4\pi} \text{ inches per minute} \]