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-

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 \frac{d A}{d t} =

A . 4 A B . 2 A C . - 4 \sqrt{A} D . 4 \sqrt{A}

Answered by gsk2684 last updated on 06/Jul/21

\frac{d}{d t} (x^{2}) = 4 x \Rightarrow \frac{d A}{d t} = 4 \sqrt{A}

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