MHT CET · Maths · Application of Derivatives
The sides of an equilateral triangle are increasing at the rate of \(2 \mathrm{~cm} / \mathrm{sec}\). The rate at which the area increases, when side is \(10 \mathrm{~cm}\), is
- A \(\frac{\sqrt{3}}{10} \mathrm{~cm}^2 / \mathrm{sec}\)
- B \(\frac{10}{\sqrt{3}} \mathrm{~cm}^2 / \mathrm{sec}\)
- C \(\sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
- D \(10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
Answer & Solution
Correct Answer
(D) \(10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(A=\frac{\sqrt{3}}{4} S^2\)
\(\Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \cdot 2 S \frac{\mathrm{d} s}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \times 2 \times 10 \times\) \(2 \mathrm{~cm}^2 / \mathrm{sec}\)
\(\Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
\(\Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \cdot 2 S \frac{\mathrm{d} s}{\mathrm{~d} t}=\frac{\sqrt{3}}{4} \times 2 \times 10 \times\) \(2 \mathrm{~cm}^2 / \mathrm{sec}\)
\(\Rightarrow \frac{\mathrm{d} A}{\mathrm{~d} t}=10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}\)
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