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Question

(A) Let the side of the equilateral triangle be $x$ and its area be $A$.Given that the rate of change of the side is $\frac{dx}{dt} = 4 	ext{ cm/sec}

Vedclass · Accepted Answer

$28 \sqrt{3} 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(A) Let the side of the equilateral triangle be $x$ and its area be $A$.Given that the rate of change of the side is $\frac{dx}{dt} = 4 	ext{ cm/sec}$.The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt}$$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot x \cdot \frac{dx}{dt}$Substituting the given values $x = 14 	ext{ cm}$ and $\frac{dx}{dt} = 4 	ext{ cm/sec}$:$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 14 \cdot 4$$\frac{dA}{dt} = \sqrt{3} \cdot 7 \cdot 4$$\frac{dA}{dt} = 28 \sqrt{3} 	ext{ cm}^2/	ext{sec}$.Thus,the area is increasing at the rate of $28 \sqrt{3} 	ext{ cm}^2/	ext{sec}$.

The sides of an equilateral triangle are increasing at the rate of $4 \text{ cm/sec}$. Find the rate at which its area is increasing when the side is $14 \text{ cm}$.

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