# Thread: experiment/ word problem help

1. ## 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?

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

3. Originally Posted by emakarov
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?

4. Sorry, I meant D mm = 25.4 mm/in * D' in.

5. Originally Posted by emakarov
Sorry, I meant D mm = 25.4 mm/in * D' in.

How did you get this?

6. An inch has 25.4 millimeters.

7. Originally Posted by emakarov
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 $Dmm = \frac{25.4mm Din}{in}$

8. 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.

9. 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.

10. 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?

11. Originally Posted by Tweety
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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# aseed crystal of diameter D(mm) is placed in asolution of dissolved salt and new crystals are observed to nucleate(form) at aconstant rate r(crystals/min).experiments with seed crystals of different sizes show that the rate of nucleation varies with the s

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