the side of a square is increasing at a constant rate of 0.4cm/sec. In terms of the perimeter what is the rate change of the area of the square in cm²/sec

The side length of the square is increasing at a constant rate of 0.4 cm/sec. We want to find how the area of the square is changing with respect to time, but in terms of the **perimeter**.

### Step 1: Express the relevant formulas
- **Area of a square**: \( A = s^2 \), where \( s \) is the side length.
- **Perimeter of a square**: \( P = 4s \), where \( P \) is the perimeter and \( s \) is the side length.

We are given that the side length is increasing at a rate of:
\[
\frac{ds}{dt} = 0.4 \, \text{cm/sec}
\]

### Step 2: Relate the perimeter to the side length
Since \( P = 4s \), we can solve for \( s \) in terms of \( P \):
\[
s = \frac{P}{4}
\]

### Step 3: Differentiate the area with respect to time
The area of the square is \( A = s^2 \). To find the rate of change of the area, we take the derivative of the area with respect to time using the chain rule:
\[
\frac{dA}{dt} = 2s \frac{ds}{dt}
\]

### Step 4: Substitute \( s \) and \( \frac{ds}{dt} \)
From earlier, we know that \( s = \frac{P}{4} \) and \( \frac{ds}{dt} = 0.4 \, \text{cm/sec} \). Substitute these into the derivative of the area:
\[
\frac{dA}{dt} = 2 \left(\frac{P}{4}\right) \cdot 0.4
\]

### Step 5: Simplify the expression
Simplifying the right-hand side:
\[
\frac{dA}{dt} = 2 \cdot \frac{P}{4} \cdot 0.4 = \frac{P}{2} \cdot 0.4 = 0.2P
\]

Thus, the rate of change of the area of the square in terms of the perimeter is:
\[
\boxed{\frac{dA}{dt} = 0.2P \, \text{cm}^2/\text{sec}}
\]

### Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:
1. How would the result change if the side length of the square increased at a different rate?
2. What happens to the rate of change of the area if the perimeter increases at a constant rate?
3. How would you calculate the rate of change of the diagonal of the square?
4. Can you express the rate of change of the area in terms of the side length instead of the perimeter?
5. How would the approach change if we were working with a different shape, like a rectangle?

**Tip:** Whenever solving related rates problems, identify the fundamental relationships between the quantities involved, and carefully apply the chain rule for differentiation.

The side of a square is increasing at a constant rate of 0.4 cm/sec. In terms of the perimeter, what is the rate of change of the area of the square in cm²/sec?

Find the rate of change of the area of a square when its side length is increasing at 0.4 cm/sec. This problem involves geometry and related rates, using differentiation and the chain rule to express the result in terms of the perimeter. Suitable for students in Grades 10-12.

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