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Rate of increase of the surface area

  1. Feb 28, 2007 #1
    a spherical balloon is being inflated. find the rate of increase of the surface area (S=4 pie r squared) with respect to radius r when r is (A) 1 ft, (B) 2 ft, (C) 3ft.

    Here's what i did

    i found the derivative of s and i put 4 2R and then i plugged in the numbers in R
    and i got 8 ft sq/ft 16 ft sq/ft and 24 ft sq/ft

    except i'm not sure if im doing it right, although those are the right answers i didnt get the units along with them
     
    Last edited: Feb 28, 2007
  2. jcsd
  3. Feb 28, 2007 #2
    is there simple algebra involved in this calculus problem
     
  4. Feb 28, 2007 #3
    are you not given dr/dt or dv/dt?
     
  5. Mar 1, 2007 #4
    You've got a graph there, simply find the gradient of the slope, that's the rate of increase.

    (1,8)
    (2,16)
    (3,24)
     
    Last edited: Mar 1, 2007
  6. Mar 1, 2007 #5

    HallsofIvy

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    Give more detail about what you did (I, for one, don't know what "i put 4 2R" means!). What is the formula for surface area (I see you give that)? what is its derivative?

    Plastic Photon, since the problem asks for the rate of increas eof surface area with respect to r you don't need to know dr/dt.

    If the surface area is in square feet and the radius in feet, then the "rate of change of surface area with respect ot radius" is in (square feet)/feet or just feet!
     
    Last edited: Mar 1, 2007
  7. Mar 4, 2007 #6
    ok so ur saying i dont need to find the derivative of it, ok check this mate,
    the i put 4 2 R means thats the derivative of the surface area
     
  8. Mar 5, 2007 #7

    HallsofIvy

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    No, "i put 4 2 R" doesn't mean anything!
     
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