An Indian chef is taking tetrahedron shaped samosas to deep fry. When the samosas are in the frying process, the length of each edge of each samosa increases by 1 millimeter per minute. Find the rate of change of a samosa's volume if the edge length is 54 millimeters. Round your answer to the nearest hundredth along with using the unit of cubic millimeters per minute.

To find the rate of change of a samosa's volume during the frying process, we can use the formula for the volume of a tetrahedron and apply related rates. Here's the step-by-step solution:

### Step 1: Volume of a Tetrahedron
The volume \( V \) of a regular tetrahedron with edge length \( a \) is given by:
\[
V = \frac{a^3 \sqrt{2}}{12}
\]

### Step 2: Differentiate with Respect to Time
To find the rate of change of the volume with respect to time, we differentiate the volume formula with respect to time \( t \). We use the chain rule:
\[
\frac{dV}{dt} = \frac{dV}{da} \cdot \frac{da}{dt}
\]

First, differentiate \( V \) with respect to \( a \):
\[
\frac{dV}{da} = \frac{d}{da} \left( \frac{a^3 \sqrt{2}}{12} \right) = \frac{3a^2 \sqrt{2}}{12} = \frac{a^2 \sqrt{2}}{4}
\]

Now, since \( \frac{da}{dt} \) is the rate at which the edge length increases, and it is given as 1 millimeter per minute:
\[
\frac{dV}{dt} = \frac{a^2 \sqrt{2}}{4} \cdot 1
\]
\[
\frac{dV}{dt} = \frac{a^2 \sqrt{2}}{4}
\]

### Step 3: Substitute the Given Edge Length
Substitute \( a = 54 \) millimeters into the equation:
\[
\frac{dV}{dt} = \frac{(54)^2 \sqrt{2}}{4}
\]
\[
\frac{dV}{dt} = \frac{2916 \sqrt{2}}{4}
\]
\[
\frac{dV}{dt} = 729 \sqrt{2}
\]

### Step 4: Calculate the Numerical Value
Now, compute the numerical value:
\[
\frac{dV}{dt} \approx 729 \times 1.4142 = 1031.29 \text{ cubic millimeters per minute}
\]

Thus, the rate of change of the samosa's volume when the edge length is 54 millimeters is approximately **1031.29 cubic millimeters per minute**.

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Do you want details on any step or have any questions?

Here are 8 related questions to explore further:
1. How would the rate of change of volume differ if the edge length increased by 2 mm per minute?
2. What is the formula for the surface area of a tetrahedron, and how would you calculate the rate of change of the surface area?
3. How does the volume of a tetrahedron compare to that of a cube with the same edge length?
4. If the edge length was given in centimeters instead of millimeters, how would that affect the rate of change of volume?
5. How would you find the rate of change of the edge length if the volume's rate of change was given?
6. What happens to the rate of change of volume as the edge length increases?
7. How would you calculate the rate of change of the height of the tetrahedron?
8. How does the formula for the volume of a tetrahedron change if it's not a regular tetrahedron?

**Tip:** When dealing with related rates problems, always identify the known rates and relationships between variables before differentiating.

Learn how to calculate the rate of change of volume for tetrahedron-shaped samosas as their edge length increases over time. Detailed step-by-step solution using differentiation and the volume formula for a tetrahedron. Suitable for advanced high school students and anyone interested in related rates problems.

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