The area of the square increases at the rate of $0.5 \mathrm{~cm}^{2} / \mathrm{sec}$. The rate at which its…
The area of the square increases at the rate of $0.5 \mathrm{~cm}^{2} / \mathrm{sec}$. The rate at which its
perimeter is increasing when the side of the square is $10 \mathrm{~cm}$ long, is
$0 \cdot 3 \mathrm{~cm} / \mathrm{sec}$
$0 \cdot 1 \mathrm{~cm} / \mathrm{sec}$
$0 \cdot 2 \mathrm{~cm} / \mathrm{sec}$
$0 \cdot 4 \mathrm{~cm} / \mathrm{sec}$
Solution
Let $x$ be the side of the square. Then area $A=x^{2}$
$\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow 0.5=2(10) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}$
$\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=0.025$
Now perimeter $=4 \mathrm{x}=\mathrm{p}$
$\therefore \frac{\mathrm{dp}}{\mathrm{dt}}=4 \frac{\mathrm{dx}}{\mathrm{dt}}=4(0.025)=0.1 \mathrm{~cm} / \mathrm{sec} .$