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ashleynguyenx3 Group Title

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
  • 2 years ago

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  1. asnaseer Group Title
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    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?

    • 2 years ago
  2. ashleynguyenx3 Group Title
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    100?

    • 2 years ago
  3. asnaseer Group Title
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    no - what is the area of a square if its side length is equal to x?

    • 2 years ago
  4. ashleynguyenx3 Group Title
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    I don't get it

    • 2 years ago
  5. asnaseer Group Title
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    |dw:1350248149705:dw|

    • 2 years ago
  6. ashleynguyenx3 Group Title
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    Oh, A=x^2

    • 2 years ago
  7. asnaseer Group Title
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    :)

    • 2 years ago
  8. asnaseer Group Title
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    good, so now you just need to differentiate both sides with respect to time (t). Do you know how to do that?

    • 2 years ago
  9. ashleynguyenx3 Group Title
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    Derivative, right?

    • 2 years ago
  10. asnaseer Group Title
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    yes

    • 2 years ago
  11. ashleynguyenx3 Group Title
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    dA/dt=2x(dx/dt)

    • 2 years ago
  12. asnaseer Group Title
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    perfect! almost there now...

    • 2 years ago
  13. asnaseer Group Title
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    now, when the area is 150 cm^2, then what would the side length of the square equal?

    • 2 years ago
  14. asnaseer Group Title
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    i.e. what is the length of the side of a square if its area is 150 cm^2?

    • 2 years ago
  15. ashleynguyenx3 Group Title
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    sq rt of 150?

    • 2 years ago
  16. asnaseer Group Title
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    yes

    • 2 years ago
  17. ashleynguyenx3 Group Title
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    So you would plug in x and dx/dt to get dA/dt?

    • 2 years ago
  18. asnaseer Group Title
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    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:\[\frac{dx}{dt}=10\]

    • 2 years ago
  19. asnaseer Group Title
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    yes - you have it now - well done! :)

    • 2 years ago
  20. ashleynguyenx3 Group Title
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    Yay, thank you! :)

    • 2 years ago
  21. asnaseer Group Title
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    yw :)

    • 2 years ago
  22. ashleynguyenx3 Group Title
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    I would ask for more help, but it's okay.

    • 2 years ago
  23. asnaseer Group Title
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    ?

    • 2 years ago
  24. ashleynguyenx3 Group Title
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    I don't really understand related rates.

    • 2 years ago
  25. asnaseer Group Title
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    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.

    • 2 years ago
  26. asnaseer Group Title
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    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.

    • 2 years ago
  27. ashleynguyenx3 Group Title
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    Well, I'm just doing this packet. I understand some, but not all. http://tutorial.math.lamar.edu/ProblemsNS/CalcI/RelatedRates.aspx

    • 2 years ago
  28. asnaseer Group Title
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    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.

    • 2 years ago
  29. ashleynguyenx3 Group Title
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    Okay, thanks again :D

    • 2 years ago
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