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-dA-dt-

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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 (dA/dt) =

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

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

(dA/dt)=p (dA/dt)=4(√A) ⇒(1/4)∫ (dA/( (√A)))=∫dt (1/2)(√A) =t+C

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

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