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Question Number 115706 by Rio Michael last updated on 27/Sep/20

the area of a square, A (t) is increased

at a rate equal to its perimeter

write a differential equation that A (t)

satisfy, starting from \frac{d A}{d t} =

Commented by Dwaipayan Shikari last updated on 27/Sep/20

\frac{dA}{dt} = p

\frac{dA}{dt} = 4 \sqrt{A} \Rightarrow \frac{1}{4} \int \frac{dA}{\sqrt{A}} = \int dt

\frac{1}{2} \sqrt{A} = t + C

Commented by Rio Michael last updated on 27/Sep/20

thanks

Commented by Dwaipayan Shikari last updated on 27/Sep/20

Amusing, area dependent upon time . A bacteria probably

Answered by mr W last updated on 27/Sep/20

A = a^{2}

P = 4 a = 4 \sqrt{A}

\frac{d A}{d t} = c P = k \sqrt{A}

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