If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, is

Options

A. `3 cm^(2)//sec`
B. `(6)/(sqrt(2))cm^(2)//sec`
C. `3sqrt(2)cm^(2)//sec`
D. `6cm^(2)//sec`

Explanation

To solve the problem step by step, we will follow the mathematical reasoning outlined in the video transcript.

### Step 1: Understand the relationship between the diagonal and the side of the square.
The diagonal \( x \) of a square is related to its side \( a \) by the formula:
\[ x = a\sqrt{2} \]

### Step 2: Differentiate the diagonal with respect to time.
Given that the diagonal is increasing at a rate of \( \frac{dx}{dt} = 0.2 \) cm/s, we can differentiate the equation \( x = a\sqrt{2} \) with respect to time \( t \):
\[ \frac{dx}{dt} = \sqrt{2} \frac{da}{dt} \]

### Step 3: Find the side length when the diagonal is \( 30 \) cm.
We are given that the side \( a \) is \( \frac{30}{\sqrt{2}} \) cm. We can find the diagonal \( x \) at this side length:
\[ x = a\sqrt{2} = \left(\frac{30}{\sqrt{2}}\right) \sqrt{2} = 30 \text{ cm} \]

### Step 4: Find the area of the square.
The area \( A \) of the square is given by:
\[ A = a^2 \]
Substituting \( a = \frac{30}{\sqrt{2}} \):
\[ A = \left(\frac{30}{\sqrt{2}}\right)^2 = \frac{900}{2} = 450 \text{ cm}^2 \]

### Step 5: Differentiate the area with respect to time.
To find the rate of change of the area, we differentiate \( A = a^2 \):
\[ \frac{dA}{dt} = 2a \frac{da}{dt} \]

### Step 6: Substitute \( a \) and \( \frac{da}{dt} \).
From the previous steps, we know \( a = \frac{30}{\sqrt{2}} \) and we need to find \( \frac{da}{dt} \). We can express \( \frac{da}{dt} \) in terms of \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \sqrt{2} \frac{da}{dt} \implies \frac{da}{dt} = \frac{1}{\sqrt{2}} \frac{dx}{dt} = \frac{1}{\sqrt{2}} \times 0.2 = \frac{0.2}{\sqrt{2}} \text{ cm/s} \]

### Step 7: Calculate \( \frac{dA}{dt} \).
Now substitute \( a \) and \( \frac{da}{dt} \) into the area rate of change formula:
\[ \frac{dA}{dt} = 2 \left(\frac{30}{\sqrt{2}}\right) \left(\frac{0.2}{\sqrt{2}}\right) \]
\[ = 2 \times \frac{30 \times 0.2}{2} = 30 \times 0.2 = 6 \text{ cm}^2/\text{s} \]

### Final Answer:
The rate of increase of the area when the side is \( \frac{30}{\sqrt{2}} \) cm is \( 6 \text{ cm}^2/\text{s} \).

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