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Using differentials to estimate the amount of tin

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm, and height 12cm, if the tin is 0.04 cm thick.

    The way I did this was to subtract the Max volume by the Enclosed volume.

    Max Volume is pi(4)(12)
    Enclosed Volume is pi ( 8 - 2(0.04) )/2 *12

    I got about 12cm^3


    How do I do this with differentials ?


    V = pi*r^2*h

    dv = 2pi*rh * dr + pi*r^2 dh

    Then what?

    I think dr = ( 8 - (8 - 2*0.4) ) / 2 = 0.4
    Then what is dh ?

    Do I have to compare it to the Total volume somehow?
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Mark44

    Staff: Mentor

    It seems to me that dh = 2* 0.4, since you have two lids (top and bottom) of thickness 0.4 cm each.
    No, your expression for dV should do the trick.
     
  4. Oct 18, 2009 #3
    No, the answer in the book says that its 16cm^3.

    The first method I did, subtracting the max volume by the volume with radius 1/2( 8 - 2*0.04) was close because the radius will be 3.6, from the answer(16cm^3) the radius has to be close to 3.5.

    I am not sure how to work with differentials with this problem though.
     
  5. Oct 18, 2009 #4

    Mark44

    Staff: Mentor

    Using dh = .08, I get
    dV [itex]\approx~4\pi(24(.04) + 4(.08))~\approx~16.08495[/itex] cm^3
    If I use dh = .04, I get an answer that is about 14.07 cm^3

    My answer using dh = .08 is a lot closer to your book's answer.
     
  6. Oct 18, 2009 #5

    Sorry for my miscalculation, thanks a million, Mark.
     
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