Question

If the side of an equilateral triangle increases at the rate of \(\sqrt{3} \text{cm/sec}\), then the rate of change of increase of its area when the side is \(12 \text{cm}\) is ____

• A. \(18 \text{cm}^2/\text{sec}\)
• B. \(10 \text{cm}^2/\text{sec}\)
• C. \(12 \text{cm}^2/\text{sec}\)
• D. \(3\sqrt{3} \text{cm}^2/\text{sec}\)

Hint:

Write area in terms of the side and differentiate with respect to time.

Answer 1

The Correct Option is A

Step 1: Plug in first:
$A = \frac{\sqrt3}{4}s^2$, and $s$ grows by $\sqrt3$ each second, so $s = s_0 + \sqrt3 t$.

Step 2: Differentiate:
$\frac{dA}{dt} = \frac{\sqrt3}{4}\cdot 2s\cdot\sqrt3 = \frac{3s}{2}$. At $s = 12$, this is $18$ cm$^2$/s.

Final Answer:
The rate is $18$ cm$^2$/s, option (A). \[ \boxed{18\ \text{cm}^2/\text{sec}} \]