COMEDK · Maths · 27. Application of Derivatives
The side of an equilateral triangle expands at the rate of \(\sqrt{3} \mathrm{~cm} / \mathrm{sec}\). When the side is \(12 \mathrm{~cm}\), the rate of increase of its area is
- A \(3 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
- B \(18 \mathrm{~cm}^2 / \mathrm{sec}\)
- C \(12 \mathrm{~cm}^2 / \mathrm{sec}\)
- D \(10 \mathrm{~cm}^2 / \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(18 \mathrm{~cm}^2 / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(A = \dfrac{\sqrt{3}}{4}a^2\) \(\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \cdot a \cdot \dfrac{da}{dt}\) Substituting \(a = 12\) cm and \(\dfrac{da}{dt} = \sqrt{3}\) cm/sec:…
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