Q94 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
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$3 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}$
- Correct answerB$18 \mathrm{~cm}^2 / \mathrm{sec}$
- C$12 \mathrm{~cm}^2 / \mathrm{sec}$
- D$10 \mathrm{~cm}^2 / \mathrm{sec}$
Solution
$A = \dfrac{\sqrt{3}}{4}a^2$
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \cdot a \cdot \dfrac{da}{dt}$
Substituting $a = 12$ cm and $\dfrac{da}{dt} = \sqrt{3}$ cm/sec:
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 12 \times \sqrt{3} = \dfrac{\sqrt{3} \times 12 \times \sqrt{3}}{2} = \dfrac{36}{2} = 18$ cm$^2$/sec
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \cdot a \cdot \dfrac{da}{dt}$
Substituting $a = 12$ cm and $\dfrac{da}{dt} = \sqrt{3}$ cm/sec:
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 12 \times \sqrt{3} = \dfrac{\sqrt{3} \times 12 \times \sqrt{3}}{2} = \dfrac{36}{2} = 18$ cm$^2$/sec
Hence, the correct answer is option: 'B'.