a juice is shaped so that the volume of liquid it contains is givens by the formula: v=1/4πh2.where h cm is the depth of the liquid in the glass. juice is poured into the glass at a rate of 5cm3/s. at what rate the depth of juice increasing in cm/s when the depth of the juice is 3cm.

Given the volume formula for the liquid in the glass:

\[
V = \frac{1}{4} \pi h^2
\]

where \(V\) is the volume in cubic centimeters (cm³) and \(h\) is the depth of the liquid in centimeters (cm).

The juice is being poured into the glass at a rate of 5 cm³/s. This is given by \(\frac{dV}{dt} = 5\) cm³/s. We need to find the rate at which the depth of the juice (\(h\)) is increasing, denoted by \(\frac{dh}{dt}\), when the depth \(h = 3\) cm.

### Step 1: Differentiate the Volume Formula with Respect to Time

We differentiate both sides of the volume formula with respect to time \(t\):

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{4} \pi h^2 \right)
\]

Applying the chain rule:

\[
\frac{dV}{dt} = \frac{1}{4} \pi \cdot 2h \cdot \frac{dh}{dt}
\]

Simplifying:

\[
\frac{dV}{dt} = \frac{1}{2} \pi h \frac{dh}{dt}
\]

### Step 2: Solve for \(\frac{dh}{dt}\)

Rearrange the equation to solve for \(\frac{dh}{dt}\):

\[
\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{1}{2} \pi h}
\]

Substitute \(\frac{dV}{dt} = 5\) cm³/s and \(h = 3\) cm:

\[
\frac{dh}{dt} = \frac{5}{\frac{1}{2} \pi \cdot 3}
\]

\[
\frac{dh}{dt} = \frac{5}{\frac{3\pi}{2}}
\]

\[
\frac{dh}{dt} = \frac{5 \times 2}{3\pi}
\]

\[
\frac{dh}{dt} = \frac{10}{3\pi}
\]

### Step 3: Calculate the Numerical Value

Now, calculate the numerical value:

\[
\frac{dh}{dt} \approx \frac{10}{3 \times 3.1416}
\]

\[
\frac{dh}{dt} \approx \frac{10}{9.4248}
\]

\[
\frac{dh}{dt} \approx 1.061 \, \text{cm/s}
\]

### Conclusion

The rate at which the depth of the juice is increasing when the depth of the juice is 3 cm is approximately **1.061 cm/s**.

---

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

Here are 5 related questions that may interest you:

1. How would the rate change if the pouring rate was different?
2. How would this formula change if the shape of the container were different?
3. What is the volume of the juice when the depth is 3 cm?
4. How does the rate of increase of depth change as the depth increases?
5. What if the glass has a maximum height—how long would it take to fill it?

**Tip:** When dealing with rates of change, always remember to use the chain rule effectively to differentiate composite functions!

Discover how to calculate the rate at which the depth of juice increases in a glass when the depth of the juice is 3cm. Using calculus and the chain rule, find the solution step-by-step.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation