A juice glass is shaped so that the volume of liquid it contains is given by the formula: v=1/4pieh2. where h is the depth of the liquid in the glass. Juice is poured into the glass at a rate of 5cm3/. At what rate is the depth of juice increasing in cm/s when the depth of the juice is 3cm?

Let's clarify and correct the interpretation of the formula and steps to find the rate at which the depth of juice is increasing.

### Given:
- The volume of liquid in the glass: \( v = \frac{1}{4} \pi h^2 \).
- The rate at which juice is poured into the glass: \( \frac{dv}{dt} = 5 \) cm³/s.
- We need to find the rate at which the depth \( h \) is increasing when \( h = 3 \) cm.

### Steps to Solve:

1. **Differentiate the volume formula with respect to time \( t \)**:
   \[
   v = \frac{1}{4} \pi h^2
   \]
   Differentiating both sides with respect to \( t \):
   \[
   \frac{dv}{dt} = \frac{1}{4} \pi \cdot 2h \cdot \frac{dh}{dt}
   \]
   Simplifying:
   \[
   \frac{dv}{dt} = \frac{1}{2} \pi h \frac{dh}{dt}
   \]

2. **Substitute the given values**:
   - \( \frac{dv}{dt} = 5 \) cm³/s.
   - \( h = 3 \) cm.
   
   Substituting these values into the differentiated equation:
   \[
   5 = \frac{1}{2} \pi \cdot 3 \cdot \frac{dh}{dt}
   \]

3. **Solve for \( \frac{dh}{dt} \)**:
   \[
   5 = \frac{3\pi}{2} \cdot \frac{dh}{dt}
   \]
   \[
   \frac{dh}{dt} = \frac{10}{3\pi} \text{ cm/s}
   \]

### Conclusion:
The depth of the juice is increasing at a rate of \( \frac{10}{3\pi} \) cm/s when the depth of the juice is 3 cm.

If you'd like a more detailed breakdown of any part, feel free to ask!

Here are some related questions you might find interesting:
1. How would the rate of height change if the shape of the glass changed (e.g., cylindrical or conical)?
2. What would the depth increase rate be if the pouring rate was different, say 10 cm³/s?
3. Can we generalize the formula for other volumes like spheres or cones?
4. How does the rate of height change vary as the depth \( h \) increases?
5. What practical applications could this type of rate-of-change problem have in engineering?

**Tip:** Always double-check the interpretation of the given formulas to ensure they match the physical scenario being described.

Explore how to find the rate at which the depth of juice is increasing in a glass using calculus and related rates. Get a step-by-step solution involving the volume formula v = (1/4)πh^2 and differentiation with respect to time. Suitable for advanced high school or college students.

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