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 2 years 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.
 2 years 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.

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asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1start 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?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1no  what is the area of a square if its side length is equal to x?

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

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1Derivative, right?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1perfect! almost there now...

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

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

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1So you would plug in x and dx/dt to get dA/dt?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1now, 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:\[\frac{dx}{dt}=10\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1yes  you have it now  well done! :)

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1Yay, thank you! :)

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1I would ask for more help, but it's okay.

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1I don't really understand related rates.

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1For 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.

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1Well, I'm just doing this packet. I understand some, but not all. http://tutorial.math.lamar.edu/ProblemsNS/CalcI/RelatedRates.aspx

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1In 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.

ashleynguyenx3
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, thanks again :D
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