AP EAMCET · Maths · Application of Derivatives
The sides of an equilateral triangle are increasing at the rate of \(2 \mathrm{~cm} \mathrm{~s}^{-1}\). How fast does its area increase when its side is \(10 \mathrm{~cm}\) ?
- A \(10 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}\)
- B \(5 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}\)
- C \(\sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}\)
- D \(2 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(10 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
Area of an equilateral triangle is, \(A=\frac{\sqrt{3} a^2}{4}\) Differentiating with respect to time we get, \(\Rightarrow \quad \frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a \cdot \frac{d a}{d t}\) Here, \(\frac{d a}{d t}=2 \mathrm{~cm}^{-1}\) when \(a=10 \mathrm{~cm}\) So,…
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