Question #294737

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 .


1
Expert's answer
2022-02-08T08:16:35-0500

Solution;

If the side of a square x;

Area,A=x2A=x^2

dAdt=2xdxdt\frac{dA}{dt}=2x\frac{dx}{dt}

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

dxdt=12xdAdt\frac{dx}{dt}=\frac{1}{2x}\frac{dA}{dt}

But;

A=x2=49A=x^2=49 therefore; x=7cmx=7cm

dAdt=28cm2s1\frac{dA}{dt}=28cm^2s^{-1}

dxdt=12×7cm(28cm2s1)\frac{dx}{dt}=\frac{1}{2×7cm}(28cm^2s^{-1})

dxdt=2cms1\frac{dx}{dt}=2cms^{-1}

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


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