Question

The side of a square is increasing at the rate of 0.1 cm/s. The rate of increase of perimeter w.r.t. time is

Answer 1

We know that the perimeter of a square of side $\sf{a}$ is,

$\longrightarrow\sf{P=4a}$

So the change in perimeter with respect to time is,

$\longrightarrow\sf{\dfrac{dP}{dt}=\dfrac{d}{dt}\,(4a)}$

Since 4 is a constant we can take it out.

$\longrightarrow\sf{\dfrac{dP}{dt}=4\cdot\dfrac{da}{dt}}$

Thus, the rate of change of perimeter is 4 times that of side.

Here, rate of increase in side is,

$\longrightarrow\sf{\dfrac{da}{dt}=0.1\ cm\,s^{-1}}$

Hence the rate of increase in perimeter is,

$\longrightarrow\sf{\dfrac{dP}{dt}=4\cdot\dfrac{da}{dt}}$

$\longrightarrow\sf{\dfrac{dP}{dt}=4\times 0.1}$

$\longrightarrow\sf{\underline{\underline{\dfrac{dP}{dt}=0.4\ cm\,s^{-1}}}}$

Hence the answer is $\bf{0.4\ cm\ s^{-1}.}$