Water is being poured into a conical container at a rate of 15 cm³/sec. The cone points directly down, and it has a height of 45 cm and a base radius of 15 cm. How fast is the water level rising when the water is 6 cm deep?

To solve this problem, we need to relate the rate at which the water level \( h \) in the conical container is rising to the rate at which the volume of water \( V \) is increasing. We use related rates and the formula for the volume of a cone.

### Given:

- **Rate at which water is poured into the container:** \( \frac{dV}{dt} = 15 \text{ cm}^3/\text{sec} \)
- **Height of the cone:** \( H = 45 \text{ cm} \)
- **Radius of the cone:** \( R = 15 \text{ cm} \)
- **Height of water at the given moment:** \( h = 6 \text{ cm} \)

### Find:

- **Rate at which the water level is rising:** \( \frac{dh}{dt} \) when \( h = 6 \text{ cm} \).

### Solution:

1. **Volume of the cone filled with water:**

   The volume \( V \) of water in a cone with height \( h \) and base radius \( r \) is given by:

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

   Since the water forms a smaller cone similar to the full cone, the radius \( r \) and height \( h \) of the water are proportional to the radius and height of the full cone. Hence,

   \[
   \frac{r}{R} = \frac{h}{H} \implies r = \frac{R}{H} h = \frac{15}{45}h = \frac{h}{3}
   \]

   Thus, \( r = \frac{h}{3} \).

   Substituting \( r = \frac{h}{3} \) into the volume formula:

   \[
   V = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h
   \]

   \[
   V = \frac{1}{3} \pi \frac{h^3}{9}
   \]

   \[
   V = \frac{\pi}{27} h^3
   \]

2. **Differentiate the volume with respect to time \( t \):**

   \[
   \frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt}
   \]

   \[
   \frac{dV}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt}
   \]

   We need to find \( \frac{dh}{dt} \) when \( h = 6 \text{ cm} \) and \( \frac{dV}{dt} = 15 \text{ cm}^3/\text{sec} \).

3. **Substitute the known values:**

   \[
   15 = \frac{\pi}{9} (6)^2 \frac{dh}{dt}
   \]

   \[
   15 = \frac{\pi}{9} \cdot 36 \frac{dh}{dt}
   \]

   \[
   15 = 4\pi \frac{dh}{dt}
   \]

   \[
   \frac{dh}{dt} = \frac{15}{4\pi}
   \]

4. **Calculate the value:**

   \[
   \frac{dh}{dt} \approx \frac{15}{4 \cdot 3.1416} \approx \frac{15}{12.5664} \approx 1.194 \text{ cm/sec}
   \]

Thus, the water level is rising at approximately \( \frac{15}{4\pi} \approx 1.194 \) cm/sec when the water is 6 cm deep.

Would you like a detailed explanation on any part of the solution or any additional questions? Here are some related questions you might find interesting:

1. How would the rate change if the radius of the cone were different?
2. What would happen to the rate if the water was being poured faster?
3. How can the time it takes to fill the cone be calculated?
4. What if the container was a cylinder instead of a cone?
5. How does the relationship between the height and radius affect the water level rise?
6. What would be the effect of a different cone height while keeping the radius constant?
7. How would the problem change if the cone were inverted (pointing upwards)?
8. How can we determine the rate at which the volume of water increases at a specific time?

**Tip:** When solving related rates problems, always relate all the relevant variables through known formulas and differentiate them with respect to time. Keep track of all the units for consistency.

Learn how to solve a related rates problem involving a conical container filling with water at a given rate. Understand the application of volume formulas and proportional relationships in calculus. Suitable for high school students studying calculus and related rates problems.

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