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 solve the problem of determining the rate of change of the volume of a tetrahedron-shaped samosa, we first need to understand the geometry and the rate at which the volume changes with respect to the edge length.

### Step 1: Formula for the volume of a tetrahedron
For a regular tetrahedron with edge length \( a \), the volume \( V \) is given by the formula:

\[
V = \frac{a^3}{6\sqrt{2}}.
\]

### Step 2: Differentiate the volume formula with respect to time
We are asked to find the rate of change of the volume, \( \frac{dV}{dt} \), when the edge length is changing. To do this, we first differentiate the volume formula with respect to time \( t \).

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{a^3}{6\sqrt{2}} \right).
\]

Using the chain rule:

\[
\frac{dV}{dt} = \frac{3a^2}{6\sqrt{2}} \cdot \frac{da}{dt}.
\]

Simplifying:

\[
\frac{dV}{dt} = \frac{a^2}{2\sqrt{2}} \cdot \frac{da}{dt}.
\]

### Step 3: Apply the given values
We are given:
- The rate at which the edge length is increasing: \( \frac{da}{dt} = 1 \) millimeter per minute.
- The edge length at the moment of interest: \( a = 54 \) millimeters.

Substituting \( a = 54 \) and \( \frac{da}{dt} = 1 \) into the equation:

\[
\frac{dV}{dt} = \frac{(54)^2}{2\sqrt{2}} \cdot 1.
\]

First, calculate \( 54^2 \):

\[
54^2 = 2916.
\]

Now, substitute this value into the equation:

\[
\frac{dV}{dt} = \frac{2916}{2\sqrt{2}}.
\]

### Step 4: Simplify and calculate
We can simplify this expression:

\[
\frac{dV}{dt} = \frac{2916}{2 \times 1.414} = \frac{2916}{2.828}.
\]

Now, performing the division:

\[
\frac{dV}{dt} \approx 1031.57 \text{ cubic millimeters per minute}.
\]

### Final Answer:
The rate of change of the samosa's volume is approximately **1031.57 cubic millimeters per minute** when the edge length is 54 millimeters.

---

Would you like more details or explanations on any part of this process? Here are five related questions that can further deepen your understanding:

1. How does the volume formula change if the shape is a cube instead of a tetrahedron?
2. How would the solution differ if the edge length increased at a different rate?
3. Can you find the surface area of the tetrahedron when the edge length is 54 mm?
4. What would be the rate of change of the surface area of the samosa?
5. How would the solution change if the samosa were a pyramid with a square base?

**Tip:** When dealing with geometric shapes, differentiating the volume or surface area formulas is key to finding rates of change. Always ensure your units remain consistent during the process!

Learn how to calculate the rate of change of volume for tetrahedron-shaped samosas as their edge length increases. This problem involves using the volume formula for tetrahedrons and applying differentiation to find the rate of change. The rate of change is calculated to be approximately 1031.57 cubic millimeters per minute when the edge length is 54 millimeters.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation