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which is the constant rate in this question h, r or v which one

In the given problem, the constant rate refers to the rate at which the volume of water is being poured into the cup. This is given as \( \frac{dV}{dt} = 2 \, \text{cm}^3/\text{sec} \). 

Here’s the problem breakdown:

1. The cup is a right circular cone with:
   - Height (\(H\)) = 12 cm
   - Radius (\(R\)) = 3 cm

2. Water is poured into the cup at a constant rate of \(2 \, \text{cm}^3/\text{sec}\).

3. We need to find the rate at which the water level (height \(h\)) is rising when the depth of the water in the cup is 5 cm.

The volume of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]

To solve this, we will use related rates. Let’s go through the solution:

1. **Find the relationship between \(r\) and \(h\)**: 

Since the water forms a smaller cone similar to the full cone, the ratios of corresponding dimensions are equal:
\[ \frac{r}{h} = \frac{R}{H} \Rightarrow \frac{r}{h} = \frac{3}{12} = \frac{1}{4} \Rightarrow r = \frac{h}{4} \]

2. **Express the volume \(V\) in terms of \(h\)**:

\[ V = \frac{1}{3} \pi \left( \frac{h}{4} \right)^2 h = \frac{1}{3} \pi \frac{h^3}{16} = \frac{\pi h^3}{48} \]

3. **Differentiate \(V\) with respect to time \(t\)**:

\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi h^3}{48} \right) = \frac{\pi}{48} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{16} \frac{dh}{dt} \]

4. **Solve for \(\frac{dh}{dt}\)**:

Given \(\frac{dV}{dt} = 2 \, \text{cm}^3/\text{sec}\) and when \(h = 5 \, \text{cm}\):

\[ 2 = \frac{\pi (5)^2}{16} \frac{dh}{dt} \]

\[ 2 = \frac{25\pi}{16} \frac{dh}{dt} \]

\[ \frac{dh}{dt} = \frac{2 \times 16}{25\pi} \]

\[ \frac{dh}{dt} = \frac{32}{25\pi} \, \text{cm/sec} \]

Therefore, the rate at which the water level is rising when the depth of the water in the cup is 5 cm is \( \frac{32}{25\pi} \, \text{cm/sec} \).

Would you like further details or have any questions?

### Relative Questions:
1. How does the relationship between radius and height affect the volume formula for the cone?
2. What happens to the rate of change of height if the pouring rate of water changes?
3. How would you calculate the volume if the cone was not right circular but oblique?
4. What other real-life scenarios involve related rates problems?
5. How can you apply the concept of similar triangles in other geometric problems?

### Tip:
When solving related rates problems, always identify the given rates and the rates you need to find, and express all variables in terms of a single variable if possible to simplify differentiation.

A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 2 cm³/sec. What is the rate at which the water level is rising when the depth of the water in the cup is 5 cm? (The volume of a cone of height h and radius r is given by V = 1/3πr²h)

Calculate the rate at which the water level is rising when the depth is 5 cm in a conical cup. This problem involves related rates, the volume of a cone, and differentiation techniques. Water is being poured at 2 cm³/sec into a cup with a height of 12 cm and a radius of 3 cm.

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