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

$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{square},A\left(t\right)\mathrm{is}\mathrm{increased}$$$$\mathrm{at}\mathrm{a}\mathrm{rate}\mathrm{equal}\mathrm{to}\mathrm{its}\mathrm{perimeter}$$$$\mathrm{write}\mathrm{a}\mathrm{differential}\mathrm{equation}\mathrm{that}A\left(t\right)$$$$\mathrm{satisfy},\mathrm{starting}\mathrm{from}\frac{dA}{dt}=$$

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

$$\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{p}$$$$\frac{\mathrm{dA}}{\mathrm{dt}}=4\sqrt{\mathrm{A}}\Rightarrow \frac{1}{4}\int \frac{\mathrm{dA}}{\sqrt{\mathrm{A}}}=\int \mathrm{dt}$$$$\frac{1}{2}\sqrt{\mathrm{A}}=\mathrm{t}+\mathrm{C}$$

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

$$\mathrm{thanks}$$

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

$$\mathrm{Amusing},\mathrm{area}\mathrm{dependent}\mathrm{upon}\mathrm{time}.\mathrm{A}\mathrm{bacteria}\mathrm{probably}$$

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

$$A=a^2$$$$P=4a=4\sqrt{A}$$$$\frac{dA}{dt}=cP=k\sqrt{A}$$

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