# Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm^2?

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Antoine Share
Apr 6, 2015

The Area, A is increasing by $48$ $c {m}^{2} {s}^{-} 1$

Area, A = ${s}^{2}$
$s$ is the side length

$\implies \frac{\mathrm{dA}}{\mathrm{dt}} = 2 s \cdot \frac{\mathrm{ds}}{\mathrm{dt}}$$\text{ {[Chain rule](http://socratic.org/calculus/basic-differentiation-rules/chain-rule)}}$

$\frac{\mathrm{ds}}{\mathrm{dt}} = 6 \text{ cm/s}$

When A = 16 $c {m}^{2} \implies 16 = {s}^{2}$

$\implies s = \sqrt{16} \implies s = 4 c m$

Hence,

$\implies \frac{\mathrm{dA}}{\mathrm{dt}} = 2 \times 4 \times 6 = 48 c {m}^{2} {s}^{-} 1$

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Gió Share
Apr 6, 2015

Consider the side of the square as a function of time $l \left(t\right)$ so:

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