Answer to Question #294737 in Calculus for ZARUL

Question

The area of a square is increasing at the rate of 28cm² s^-1. Find the increasing rate of the length of a side,x when the area of the square is 49cm² .

Answer 1

Solution;

If the side of a square x;

Area,$A=x^2$

$\frac{dA}{dt}=2x\frac{dx}{dt}$

Make $\frac{dx}{dt}$ subject of the formula;

$\frac{dx}{dt}=\frac{1}{2x}\frac{dA}{dt}$

But;

$A=x^2=49$ therefore; $x=7cm$

$\frac{dA}{dt}=28cm^2s^{-1}$

$\frac{dx}{dt}=\frac{1}{2×7cm}(28cm^2s^{-1})$

$\frac{dx}{dt}=2cms^{-1}$

The rate of increase of the length is 2cm/s.