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Math Help - experiment/ word problem help

  1. #1
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    experiment/ word problem help

    2.6 A seed crystal of diameter D (mm) is placed in a solution of dissolved salt, and
    new crystals are observed to nucleate (form) at a constant rate   r (crystals \\\\   min^{-1}) .
    Experiments with seed crystals of differentt sizes show that the rate of nucleation
    varies with the seed crystal diameter as

     r(crystals\\ min^{-1})   = AD + BD^{2}

    where D is in mm, and where A and B are constants (which must be determined).

    (a) What are the units of the constants A and B? Suppose that a D = 1 mm
    seed crystal produces a nucleation rate of r = 210 crystals min��1, whereas a
    D = 2 mm seed crystal produces a nucleation rate of r = 440 crystals min��1.
    Use this information to evaluate A and B.

    (b) Calculate the crystal nucleation rate in crystals min��1 for a seed crystal di-
    ameter of 0.05 inch.

    (c) Derive a formula for r (crystals min��1) in terms of D (in). What are the
    values and units of the constants in the expression? Check the formula with
    the result of 2.6(b). Comment on your results.

    (d) Derive a formula for   r (crystals s^{-1}) . what are the values snd units of the constants in the expression.
    for part 'a', I got the units of A and B, as  crystals\\\\\\ min^{-1} mm^{-1} for A

     crystals\\\\\\\\ min^{-1} mm^{-2} for B

    so the mm would cancels out, with the D units, and the equation balances. is this correct?

    Also worked out A = 200 and B =10

    so equation would look something like this,

    r( crystals \\\\\ min^{-1} ) = \frac{200 Dmm\\\\\    crystals}{min\\\\ mm} + \frac{10Dmm^{2} \\\\\\crystals}{min\\\ mm^{2}}

    b) =  r = 200(1.27) + 10(1.27)^{2}

    c) ? Really Need help with part c. Do I just replace the units for diameter in my equation?
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  2. #2
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    r( crystals \\\\\ min^{-1} ) = \frac{200 Dmm\\\\\ crystals}{min\\\\ mm} + \frac{10Dmm^{2} \\\\\\crystals}{min\\\ mm^{2}}
    It should be 10D^2mm^{2}. Otherwise, you are correct so far.

    Concerning c), suppose that the diameter of the seed crystal in inches is D', while the diameter in millimeters is still D. Then D=25.5D'. So it suffices to substitute 25.5D' for D in your formula for r to express r through D'.
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  3. #3
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    Quote Originally Posted by emakarov View Post
    It should be 10D^2mm^{2}. Otherwise, you are correct so far.

    Concerning c), suppose that the diameter of the seed crystal in inches is D', while the diameter in millimeters is still D. Then D=25.5D'. So it suffices to substitute 25.5D' for D in your formula for r to express r through D'.
    Thank you, could you please explain how you got D = 25.5D' ? I don't see how the two equal each other?
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  4. #4
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    Sorry, I meant D mm = 25.4 mm/in * D' in.
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  5. #5
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    Quote Originally Posted by emakarov View Post
    Sorry, I meant D mm = 25.4 mm/in * D' in.


    How did you get this?
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  6. #6
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    An inch has 25.4 millimeters.
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  7. #7
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    Quote Originally Posted by emakarov View Post
    An inch has 25.4 millimeters.
    Thank you, but how does this relate to the question? I understand that I have to find an expression for Dmm in terms of inches, and than substitute this into my equation, but I dont understand how to do it from the expression you have come up with.

    why does   Dmm = \frac{25.4mm Din}{in}
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  8. #8
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    Question: How many millimeters are there in 5 inches?
    Answer: 25.4 mm per inch * 5 inches = 127 mm.

    Question: How many millimeters are there in D' inches?
    Answer: 25.4 mm/in * D' in = 25.4 * D' mm. Call the result D.

    r(D) = 200 * D + 10 * D^2. Also, D = 25.4 * D'. Therefore, r(D') = 200 * 25.4 * D' + 10 * (25.4 * D')^2.

    Edit.
    why does Dmm = \frac{25.4mm Din}{in}
    Note that D and D' are two different numbers.
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  9. #9
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    Thank you, I kind of get it now.

    I also tried doing part 'd', and need some help.

     r(crystals min^{-1}) = 200Dmm + 10D^{2}mm^{2}

    So we have to change from minutes to seconds, 1 minute = 60 seconds. Using your same method,

     r = \frac{crystals}{min}

     r'= \frac{crystals}{sec}

    not sure what to do from here.
    Last edited by Tweety; October 9th 2010 at 07:17 AM.
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  10. #10
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    Actually I have worked out part d, so ignore my previous post.


    can someone please clarify part 'c', for me as I still not able to grasp how to do it.

    if  1 inch = 25.4 mm

    than  Dmm = \frac{1}{25.4} \times D inch

    so substitute in  \frac{D(in)}{25.4} for Dmm

    Why is this method not correct?
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  11. #11
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    Quote Originally Posted by Tweety View Post
    if  1 inch = 25.4 mm

    than  Dmm = \frac{1}{25.4} \times D inch
    Conversion between two systems is always tricky for me because I can never remember whether one has to add or subtract, or whether to multiply or divide. For example, if f:\mathbb{R}\to\mathbb{R} is a function, the graph of f(x-5) is the graph of f(x) moved 5 units to the right. Similarly, I remember that we are supposed to get up earlier when the shift to summer time happens, but I can never remember whether the watch has to be adjusted forward or backward.

    Nevertheless, this issue is trivial when considered formally. Again, I prefer to use different symbols D and D' to avoid confusion. Let D be the size in millimeters and let D' be the same size in inches.

    Question: How many millimeters are there in D' inches?
    Answer: 25.4 mm/in * D' in = 25.4 * D' mm. Call the result D.
    So D = 25.4 * D', not D = 1/25.4 * D', as you write.
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