# Get Answers to all your Questions

### 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.

View full answer