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We are asked to find how fast the water level is rising in a conical tank, given that water is flowing into the tank at a rate of \( 9 \, \text{ft}^3/\text{min} \). The tank is inverted, with a height of 10 feet and a base radius of 5 feet, and we need to determine the rate at which the water level is rising when the depth of the water is 6 feet.

### Step 1: Volume of a Cone
The volume \( V \) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the water's surface at time \( t \) and \( h \) is the depth of the water.

### Step 2: Relation between \( r \) and \( h \)
The problem gives us the dimensions of the tank: height \( H = 10 \, \text{ft} \) and radius \( R = 5 \, \text{ft} \). The radius and height of the water at any time are proportional, meaning:
\[
\frac{r}{h} = \frac{R}{H} = \frac{5}{10} = \frac{1}{2}
\]
Thus, \( r = \frac{h}{2} \).

### Step 3: Substitute \( r \) into the Volume Formula
Substituting \( r = \frac{h}{2} \) into the volume formula:
\[
V = \frac{1}{3} \pi \left( \frac{h}{2} \right)^2 h = \frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi}{12} h^3
\]

### Step 4: Differentiate Volume with Respect to Time
We are given that \( \frac{dV}{dt} = 9 \, \text{ft}^3/\text{min} \). To find how fast the water level is rising, we differentiate the volume equation with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}
\]

### Step 5: Solve for \( \frac{dh}{dt} \)
We know \( \frac{dV}{dt} = 9 \) and we are asked to find \( \frac{dh}{dt} \) when \( h = 6 \). Substituting these values into the equation:
\[
9 = \frac{\pi}{4} \cdot 6^2 \cdot \frac{dh}{dt}
\]
\[
9 = \frac{\pi}{4} \cdot 36 \cdot \frac{dh}{dt}
\]
\[
9 = 9\pi \cdot \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = \frac{9}{9\pi} = \frac{1}{\pi} \, \text{ft/min}
\]

Thus, the water level is rising at a rate of \( \frac{1}{\pi} \, \text{ft/min} \) when the water is 6 feet deep.

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Would you like more details, or do you have any other questions?

Here are some related questions to consider:
1. How would the rate change if the cone had a different base radius or height?
2. What happens to the water level rate when the depth is near the top of the cone?
3. How would the answer change if the water flow rate was variable?
4. What if the tank was cylindrical rather than conical? How would that affect the math?
5. Can this approach be applied to tanks of different shapes (e.g., pyramidal)?

**Tip:** In related rates problems, always try to express one variable in terms of another to reduce the number of variables before differentiating.

Water runs into a conical tank at the rate of 9 cubic feet per minute. The tank stands vertex down and has a height of 10 feet and a base radius of 5 feet. How fast is the water level rising when the water is 6 feet deep?

Solve a conical tank related rates problem where water is flowing in at 9 cubic feet per minute, and the water level is rising when it is 6 feet deep. Learn step-by-step how to use the volume of a cone, proportional relationships, and implicit differentiation.

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