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Question Number 145337 by physicstutes last updated on 04/Jul/21
The Area of a square,A(t),is increased at a rate  equal to its perimeter ,A(t) satisfies the differential  equation (dA/dt)=  A. 4A         B. 2A         C.−4(√A)        D. 4(√A)
$$\mathrm{The}\:\mathrm{Area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square},{A}\left({t}\right),\mathrm{is}\:\mathrm{increased}\:\mathrm{at}\:\mathrm{a}\:\mathrm{rate} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{its}\:\mathrm{perimeter}\:,{A}\left({t}\right)\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\frac{{dA}}{{dt}}= \\ $$$$\mathrm{A}.\:\mathrm{4}{A}\:\:\:\:\:\:\:\:\:\mathrm{B}.\:\mathrm{2}{A}\:\:\:\:\:\:\:\:\:\mathrm{C}.−\mathrm{4}\sqrt{{A}}\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{4}\sqrt{{A}} \\ $$
Answered by gsk2684 last updated on 06/Jul/21
(d/dt)(x^2 )=4x⇒(dA/dt)=4(√A)
$$\frac{{d}}{{dt}}\left({x}^{\mathrm{2}} \right)=\mathrm{4}{x}\Rightarrow\frac{{dA}}{{dt}}=\mathrm{4}\sqrt{{A}} \\ $$

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