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how-fast-is-the-area-of-a-rectangle-changing-if-one-side-is-10-cm-long-and-increasing-at-a-rate-of-2-cm-s-and-the-other-side-is-8-cm-long-and-is-decreasing-at-a-rate-of-3-cm-s-

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Question Number 132211 by benjo_mathlover last updated on 12/Feb/21
how fast is the area of a rectangle changing if one side is 10 cm long and increasing at a rate of 2 cm/s and the other side is 8 cm long and is decreasing at a rate of 3 cm/s
$$ \\ $$$$\mathrm{how}\:\mathrm{fast}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\: \\ $$$$\mathrm{rectangle}\:\mathrm{changing}\:\mathrm{if}\:\mathrm{one}\:\mathrm{side}\:\mathrm{is}\:\mathrm{10} \\ $$$$\:\mathrm{cm}\:\mathrm{long}\:\mathrm{and}\:\mathrm{increasing}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{2}\:\mathrm{cm}/\mathrm{s}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{side}\:\mathrm{is}\: \\ $$$$\mathrm{8}\:\mathrm{cm}\:\mathrm{long}\:\mathrm{and}\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{at}\: \\ $$$$\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{3}\:\mathrm{cm}/\mathrm{s} \\ $$
Answered by ajfour last updated on 12/Feb/21
A=xy (dA/dt)=y((dx/dt))+x((dy/dt)) =(8cm)(((2cm)/s))+(10cm)(−((3cm)/s)) ⇒ (dA/dt)=−((14cm^2 )/s) ⇒ Area is decreasing at a rate of ((14cm^2 )/s).
$${A}={xy} \\ $$$$\frac{{dA}}{{dt}}={y}\left(\frac{{dx}}{{dt}}\right)+{x}\left(\frac{{dy}}{{dt}}\right) \\ $$$$\:\:\:=\left(\mathrm{8}{cm}\right)\left(\frac{\mathrm{2}{cm}}{{s}}\right)+\left(\mathrm{10}{cm}\right)\left(−\frac{\mathrm{3}{cm}}{{s}}\right) \\ $$$$\Rightarrow\:\frac{{dA}}{{dt}}=−\frac{\mathrm{14}{cm}^{\mathrm{2}} }{{s}} \\ $$$$\Rightarrow\:\:{Area}\:{is}\:{decreasing}\:{at}\:{a}\:{rate} \\ $$$${of}\:\:\frac{\mathrm{14}{cm}^{\mathrm{2}} }{{s}}. \\ $$

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