KCET · Maths · Application of Derivatives
The sides of an equilateral triangle are increasing at the rate of \( 4 \mathrm{~cm} / \mathrm{sec} \). The rate at which its
area is increasing, when the side is \( 14 \mathrm{~cm} \)
- A \( 14 \sqrt{3} \mathrm{~cm}^{2} / \)
- B \( 14 \mathrm{~cm}^{2} / \mathrm{sec} \)
- C \( 10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec} \)
- D none
Answer & Solution
Correct Answer
(D) none
Step-by-step Solution
Detailed explanation
Given Options are not matching
\( \frac{d x}{d t}=4 \mathrm{~cm} / \mathrm{sec}, x=14 \mathrm{~cm} \)
\( A=\frac{\sqrt{3}}{4} x^{2} \)
\( \frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t} \)
\( =\frac{\sqrt{3}}{2} \cdot 14 \times 4 \)
\( =\sqrt{3} \cdot 7 \times 4 \)
\( =28 \sqrt{3} \)
\( \frac{d x}{d t}=4 \mathrm{~cm} / \mathrm{sec}, x=14 \mathrm{~cm} \)
\( A=\frac{\sqrt{3}}{4} x^{2} \)
\( \frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t} \)
\( =\frac{\sqrt{3}}{2} \cdot 14 \times 4 \)
\( =\sqrt{3} \cdot 7 \times 4 \)
\( =28 \sqrt{3} \)
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