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: Understanding the Concept:
The area of an equilateral triangle of side \(s\) is \(A = \frac{\sqrt3}{4}s^2\). Both \(A\) and \(s\) change with time, so use the chain rule.

Step 2: Key Formula or Approach:
\(\frac{dA}{dt} = \frac{\sqrt3}{4}\cdot 2s\cdot\frac{ds}{dt} = \frac{\sqrt3}{2}s\frac{ds}{dt}\).

Step 3: Detailed Explanation:
Given \(\frac{ds}{dt} = \sqrt3\) cm/s and \(s = 12\) cm:
\[ \frac{dA}{dt} = \frac{\sqrt3}{2} \times 12 \times \sqrt3 = \frac{3 \times 12}{2} = 18\ \text{cm}^2/\text{s} \]
Option D (\(3\sqrt3\)) would be obtained by forgetting the factor \(s\), and option C would be \(12\), from using only \(\frac{ds}{dt}\) and \(s\) but not the \(\sqrt3\cdot\sqrt3 = 3\) factor.

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