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The sides of a square are increasing at a rate of 10 cm/sec. How fast is the area enclosed by the square increasing when the area is 150 cm^2.
 one year ago
 one year ago
The sides of a square are increasing at a rate of 10 cm/sec. How fast is the area enclosed by the square increasing when the area is 150 cm^2.
 one year ago
 one year ago

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asnaseerBest ResponseYou've already chosen the best response.1
start by assigning a variable to represent the side length of the square. lets call it x. now, what would be the area of the square in terms of x?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
no  what is the area of a square if its side length is equal to x?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1350248149705:dw
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
good, so now you just need to differentiate both sides with respect to time (t). Do you know how to do that?
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
Derivative, right?
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
dA/dt=2x(dx/dt)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
perfect! almost there now...
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
now, when the area is 150 cm^2, then what would the side length of the square equal?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
i.e. what is the length of the side of a square if its area is 150 cm^2?
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
So you would plug in x and dx/dt to get dA/dt?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
now, in the equation you derived, you got:\[\frac{dA}{dt}=2x\frac{dx}{dt}\]Your questions tells you how fast the side length is increasing, so this gives you the value for:\[\frac{dx}{dt}=10\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  you have it now  well done! :)
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
Yay, thank you! :)
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
I would ask for more help, but it's okay.
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
I don't really understand related rates.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
What you should do is find a specific example and then post that as a question. Otherwise it becomes difficult to explain in general terms.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
For example, in this question, the rate of increase of side length is RELATED to the rate of increase of area by the formula you just derived.
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
Well, I'm just doing this packet. I understand some, but not all. http://tutorial.math.lamar.edu/ProblemsNS/CalcI/RelatedRates.aspx
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
In that case I would advise you to pick one that you don't understand and post it as a question. If you also include what parts of it you DO understand then it becomes easier for others to focus on the parts that you are having difficulties with.
 one year ago

ashleynguyenx3Best ResponseYou've already chosen the best response.1
Okay, thanks again :D
 one year ago
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