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  • The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is: A. 10 cm2/s

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:
A. 10 cm^2/s
B. \sqrt3 cm^2/s
C. 10\sqrt3 cm^2/s
D. \frac{10}{3} cm^2/s

Answers (1)

Let x cm be the side of the equilateral triangle, then the area of the triangle is

\\A=\frac{\sqrt{3}}{4} x^{2}$ \\$\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{sec}$
Also, the rate of side increasing at instant of time t is
Differentiate area with respect to time t and get
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{d}\left(\frac{\sqrt{3}}{4} \mathrm{x}^{2}\right)}{\mathrm{dt}}$
Take the constants out and get,

\\ \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}$
Apply the derivative and get
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \times 2 \mathrm{x} \times \frac{\mathrm{dx}}{\mathrm{dt}}$
Substitute given value of \frac{\mathrm{dx}}{\mathrm{dt}}$ and get

\\\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{4} \times 2 \mathrm{x} \times 2$ \\$\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\sqrt{3} \mathrm{x}$

Now, put side \mathrm{x}=10 \mathrm{~cm}$
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}_{\mathrm{x}=10}}=\sqrt{3} \times 10$
Hence, the rate at which the area increases is 10 \sqrt 3 \mathrm{~cm}^{2} / \mathrm{s}$.
So the correct answer is option C.

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infoexpert22

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