Water droplet coalescence - prove S < S1 + S2

[bioinformaticsgirl]

Homework Statement

There are two water droplets with the radii R1 and R2, volumes V1 and V2, and surface areas S1 and S2. Assume that R1 is not equal to R2.

Upon coalescence, two droplets form a "joint" droplet with the volume V=V1+V2.

Prove analytically that the surface area of the "joint" droplet S < S1 + S2.

Homework Equations

surface area of a sphere, volume of a sphere, and radius of a sphere.

The Attempt at a Solution

So far I have, but not sure how to factor this out and turn into the inequality proof:

S < S1 + S2
4piR^2 < 4piR1^2 + 4piR2^2

V = V1 + V2
4/3piR^3 = 4/3piR1^3 + 4/3piR2^3

 

[drvrm]

There are two water droplets with the radii R1 and R2, volumes V1 and V2, and surface areas S1 and S2. Assume that R1 is not equal to R2.

Upon coalescence, two droplets form a "joint" droplet with the volume V=V1+V2.

is there one more condition related to surface tension?

 

[bioinformaticsgirl]

Assumed that pressure is equivalent

 

[drvrm]

Assumed that pressure is equivalent

Is it wise assumption?

i do not think so - as any water droplet is having a pressure related to surface tension and radius r . you must use it

 

[Chestermiller]

Is it wise assumption?

i do not think so - as any water droplet is having a pressure related to surface tension and radius r . you must use it

What has the surface tension got to do with it if the shape is a sphere and the final volume is the sum of the two initial volumes?

 

[Chestermiller]

This is really a math problem, and I'm moving it to the Introductory Math Forum.

Chet

 

[drvrm]

What has the surface tension got to do with it if the shape is a sphere and the final volume is the sum of the two initial volumes?

actually i remember that droplets or bubbles are in equilibrium because of an excess pressure inside it which is due to the surface stretched to make the shape - its related to (2.surface tension/radius)so i thought that can give a relationship to get out of the jam'

 

[Chestermiller]

actually i remember that droplets or bubbles are in equilibrium because of an excess pressure inside it which is due to the surface stretched to make the shape its related to (surface tension/radius)so i thought that can give a relationship to get out of the jam'

I don't see where there is a "jam." It's strictly a mathematics problem.

 

[Chestermiller]

Homework Statement

There are two water droplets with the radii R1 and R2, volumes V1 and V2, and surface areas S1 and S2. Assume that R1 is not equal to R2.

Upon coalescence, two droplets form a "joint" droplet with the volume V=V1+V2.

Prove analytically that the surface area of the "joint" droplet S < S1 + S2.

Homework Equations

surface area of a sphere, volume of a sphere, and radius of a sphere.

The Attempt at a Solution

So far I have, but not sure how to factor this out and turn into the inequality proof:

S < S1 + S2
4piR^2 < 4piR1^2 + 4piR2^2

V = V1 + V2
4/3piR^3 = 4/3piR1^3 + 4/3piR2^3

Just substitute R from the volume equation for R in the inequality, and determine if it is satisfied.

 

[drvrm]

I don't see where there is a "jam." It's strictly a mathematics problem.

well ,i thought its related to idea of surface tension- it might have helped

 

[SammyS]

well ,i thought its related to idea of surface tension- it might have helped

But the problem does not have a thing about surface tension in it, so it doesn't help. It just adds confusion.

 

[haruspex]

S < S1 + S2
4piR^2 < 4piR1^2 + 4piR2^2

V = V1 + V2
4/3piR^3 = 4/3piR1^3 + 4/3piR2^3

Bear in mind that the second is what you are given and the first is what you are trying to show, but we can cross that bridge later.
Can you simplify those two equations?
Can you combine them to eliminate a variable?

 

[drvrm]

But the problem does not have a thing about surface tension in it, so it doesn't help. It just adds confusion.

actually liquid drops are made/exist in spherical shape due to surface tension property- in general one can not have water drops of different radius possible . they may flow out - it may not form or stay in spherical shape- for a drop to exist the pressure inside must be greater than outside pressure by an amount 2.T/r-
but the present problem may be a purely 'mathematical' one -as @Chestermiller opined.

 

[haruspex]

actually liquid drops are made/exist in spherical shape due to surface tension property- in general one can not have water drops of different radius possible . they may flow out - it may not form or stay in spherical shape- for a drop to exist the pressure inside must be greater than outside pressure by an amount 2.T/r-
but the present problem may be a purely 'mathematical' one -as @Chestermiller opined.

Your determination to make surface tension relevant somehow reminds me of a classic BBCTV sitcom.
Basil Fawlty: What are you a doctor of?
Hotel Guest: Paediatrics
Fawlty: Feet?
Guest: Children
Fawlty: Well, children have feet, don't they?

 

[Mark44]

but the present problem may be a purely 'mathematical' one

That's my opinion, as well. The droplets assume a spherical shape because of surface tension, but that's all you need to say about surface tension. The problem is strictly a matter of geometry, comparing the volumes and surface areas of droplets before and after they coalesce.

Bringing up surface tension (several times) is really nothing more than a distraction in this problem.

 

[bioinformaticsgirl]

Yes, I believe this boils down to a math problem for me at this point. I understand the concepts but it's been a long time for me since I have been in school so I'm still brushing up on my math. I'll spend some time working on understanding solving the inequality and substitution. Thank you all for the input. If anyone can direct me to what I should brush up on the math part to solve this. I have my algebra/calc book out reading inequalities now.

 

[bioinformaticsgirl]

Just substitute R from the volume equation for R in the inequality, and determine if it is satisfied.

So I take the V equation and solve it for R, then plug that in for R into the S equation. Got it. Working on it now...

 

[bioinformaticsgirl]

So I have the following, and got a bit stuck.

V = V1 + V2

V = 4/3piR^3

Substitute
4/3piR^3 = 4/3piR1^3 + 4/3piR2^3 (R, R1 and R2 are variables)
Reduce:
R^3 = R1^3 + R2^3

S < S1 + S2

a = 4piR^2 (surface area formula, substitute to have both equations for R)

4piR^2 < 4piR1^2 + 4piR2^2
R^2 < R1^2 + R2^2

Is this correct so far, and now I am stuck with how to put the R^3 in for the R^2 in the second equation.

Thank you.

 

[Chestermiller]

So I have the following, and got a bit stuck.

V = V1 + V2

V = 4/3piR^3

Substitute
4/3piR^3 = 4/3piR1^3 + 4/3piR2^3 (R, R1 and R2 are variables)
Reduce:
R^3 = R1^3 + R2^3

S < S1 + S2

a = 4piR^2 (surface area formula, substitute to have both equations for R)

4piR^2 < 4piR1^2 + 4piR2^2
R^2 < R1^2 + R2^2

Is this correct so far, and now I am stuck with how to put the R^3 in for the R^2 in the second equation.

Thank you.

$$R^2=(R^3)^{2/3}$$

 

[bioinformaticsgirl]

Got it. I think I'm having trouble with the distributive properties of exponents with the plus sign its not distributive. So when I plug that in, how do I distribute the exponent here?

R^3 = R1^3 + R2^3
convert to R^2
(R^3)2/3 = R^2
so
R^2 = (R1^3 + R2^3)^2/3 ? is this right

R^2 < R1^2 + R2^2
(R1^3 + R2^3)^2/3 < R1^2 + R2^2
0 < R1^2 + R2^2 - (R1^3 + R2^3)^2/3 ? need to simplify

 

[Chestermiller]

Got it. I think I'm having trouble with the distributive properties of exponents with the plus sign its not distributive. So when I plug that in, how do I distribute the exponent here?

R^3 = R1^3 + R2^3
convert to R^2
(R^3)2/3 = R^2
so
R^2 = (R1^3 + R2^3)^2/3 ? is this right

R^2 < R1^2 + R2^2
(R1^3 + R2^3)^2/3 < R1^2 + R2^2

If A<B, then A³<B³. So cube both sides of the inequality, and show us what you get.

Chet

 

[bioinformaticsgirl]

(R1^3 + R2^3)^2/3 < R1^2 + R2^2

((R1³ + R2³)^2/3)³ < (R1² + R2²)³

0 < (R1² + R2²)³ - ((R1³ + R2³)^2/3)³

0 < (R1² + R2²) - ((R1³ + R2³)^2/3) factor out 3's?

 

[Chestermiller]

(R1^3 + R2^3)^2/3 < R1^2 + R2^2

((R1³ + R2³)^2/3)³ < (R1² + R2²)³

With regard to the left-hand side of the inequality $((x)^a)^b=(x)^{ab}$

 

[bioinformaticsgirl]

((R1³ + R2³)^2/3)³ < (R1² + R2²)³

((R1³ + R2³)^6/3 < (R1² + R2²)³

(R1³ + R2³)² < (R1² + R2²)³

0 < (R1² + R2²)³ - (R1³ + R2³)²

 

[Chestermiller]

((R1³ + R2³)^2/3)³ < (R1² + R2²)³

((R1³ + R2³)^6/3 < (R1² + R2²)³

(R1³ + R2³)² < (R1² + R2²)³

0 < (R1² + R2²)³ - (R1³ + R2³)²

Good. Now expand the two terms out.

 

[bioinformaticsgirl]

I used the foil method to expand the monomials (did this part on paper but did not include here):

Expand the first part to:

R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶

Expand the second part to:

R1⁶ + 2R1³R2³ + R2⁶

Back to the equation with those expanded

0 < R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶ - (R1⁶ + 2R1³R2³ + R2⁶)

simplify

0 < 3R1⁴ + 3R2⁴R1² - 2R1³R2³

 

[Chestermiller]

I used the foil method to expand the monomials (did this part on paper but did not include here):

Expand the first part to:

R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶

Expand the second part to:

R1⁶ + 2R1³R2³ + R2⁶

Back to the equation with those expanded

0 < R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶ - (R1⁶ + 2R1³R2³ + R2⁶)

simplify

0 < 3R1⁴ + 3R2⁴R1² - 2R1³R2³

OK. The first term in your last equation is missing a factor of $R_2^2$. Next, divide by $R_1^2R_2^2$. What do you get?

 

[bioinformaticsgirl]

Found my factoring mistake. Now I see the common divisor.

0 < R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶ - R1⁶ - 2R1³R2³ - R2⁶
...
0 < 3R1² + 3R2² - 2R1R2

 

[haruspex]

Found my factoring mistake. Now I see the common divisor.

0 < R1⁶ + 3R1⁴R2² + 3R2⁴R1² + R2⁶ - R1⁶ - 2R1³R2³ - R2⁶
...
0 < 3R1² + 3R2² - 2R1R2

Good. Can you see a way to express the right hand side as a sum of terms, each of which is guaranteed non-negative (and at least one is nonzero).

 

[bioinformaticsgirl]

Using the square of a binomial where (u + v)² = u² - 2uv +v² I would think I can find the sum of terms with this, but I'm not totally sure what to do with those 3's.

I rewrite into standard form:
0 < 3R1² - 2R1R2 + 3R2²

 

[bioinformaticsgirl]

Oh wait. those are for a single variable x, whereas I essentially have an x and a y... Not sure which rule to use here.

0 < (R1 - R2)^2 would give me something close but those 3's I'm not sure what to do with.

 

[haruspex]

Oh wait. those are for a single variable x, whereas I essentially have an x and a y... Not sure which rule to use here.

0 < (R1 - R2)^2 would give me something close but those 3's I'm not sure what to do with.

What's the obvious way to get 3R₁²+3R₂² from R₁²+R₂² ?

 

[Chestermiller]

Using the square of a binomial where (u + v)² = u² - 2uv +v² I would think I can find the sum of terms with this, but I'm not totally sure what to do with those 3's.

I rewrite into standard form:
0 < 3R1² - 2R1R2 + 3R2²

Suppose you wrote $3R_1^2+3R_2^2-2R_1R_2=2R_1^2+2R_2^2+(R_1^2-2R_1R_2+R_2^2)$
Would that help?