Question

If the length of the diagonal of a square is increasing at the rate of \( 0.1 \) cm/secWhat is the rate of increase of its area when the side is \( \frac{15}{\sqrt{2}} \) cm?

• A. 3 cm\(^2\)/sec
• B. 0.15 cm\(^2\)/sec
• C. 1.5 cm\(^2\)/sec
• D. \( 3\sqrt{2} \) cm\(^2\)/sec

Hint:

Always relate variables first (like \( d = s\sqrt{2} \))Then differentiate to connect rates and substitute values carefully.

Answer 1

The Correct Option is C

Step 1: Relate diagonal and side.
For a square:
\[ d = s\sqrt{2} \]

Step 2: Differentiate with respect to time.
\[ \frac{dd}{dt} = \sqrt{2}\frac{ds}{dt} \]

Step 3: Substitute given rate.
\[ 0.1 = \sqrt{2}\frac{ds}{dt} \]
\[ \frac{ds}{dt} = \frac{0.1}{\sqrt{2}} \]

Step 4: Write area function.
\[ A = s^2 \]

Step 5: Differentiate area.
\[ \frac{dA}{dt} = 2s\frac{ds}{dt} \]

Step 6: Substitute values.
\[ s = \frac{15}{\sqrt{2}}, \quad \frac{ds}{dt} = \frac{0.1}{\sqrt{2}} \]
\[ \frac{dA}{dt} = 2 \cdot \frac{15}{\sqrt{2}} \cdot \frac{0.1}{\sqrt{2}} \]
\[ = 30 \cdot \frac{0.1}{2} \]
\[ = 1.5 \]

Step 7: Final conclusion.
\[ \boxed{1.5 \text{ cm}^2/\text{sec}} \]