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Mathematics 17 Online
OpenStudy (anonymous):

2nd question) Need to calculate rate of change of length of square whose area equal to 100m2, where rate of change dAdt=1m2s

OpenStudy (anonymous):

\[\frac{ dA }{ dt }=1 \frac{ m ^{2} }{ s }\]

OpenStudy (anonymous):

Need to find rate of change of this square's length

OpenStudy (anonymous):

Okay so remember the chain rule: \[ \frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt} \]Where \(x\) is the length of a side in this case.

OpenStudy (anonymous):

"rate of change of length of square" This means they want us to find: \[ \frac{dx}{dt} \]

OpenStudy (anonymous):

So first of all, can you dentify what \(dA/dx\) is?

OpenStudy (anonymous):

Consider what \(A(x)\) is first.

OpenStudy (anonymous):

A=L^2

OpenStudy (anonymous):

Well, \(x\) not \(L\) in this case.

OpenStudy (anonymous):

So what is \[ \frac{dA}{dx} \]?

OpenStudy (anonymous):

i dont know, what can be x?

OpenStudy (anonymous):

Well, we know \[ A=100m^2 \]And \[ A=x^2 \]

OpenStudy (anonymous):

\[\frac{ dA }{ dx }=\frac{ dx ^{2} }{ dx }=2x\]

OpenStudy (anonymous):

Right!

OpenStudy (anonymous):

so 20m/s is the answer?

OpenStudy (anonymous):

\[ 1\frac{m^2}{s}=2(10m)\frac{dx}{dt} \]

OpenStudy (anonymous):

It's not \(20mx/\)

OpenStudy (anonymous):

1/20?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

damn it), i made a mistake, thx for a second time :D

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