Bubble Gum

Geometry Level 1

If the diameter of her bubble doubles, what happens to the thickness of the bubble's skin?

Assume that the bubbles are spheres and that the thickness of the gummy-skin is uniform.

The thickness is quartered The thickness stays the same The thickness doubles It depends on how big exactly the bubble is before it doubles The thickness is halved

5 solutions

Gabe Smith
Oct 14, 2015

The Surface Area of a Sphere is $4\pi r^2.$ When the diameter doubles, the radius doubles, so the surface area quadruples, since $2^2 = 4.$

Since there is a constant amount of gum equally distributed over the surface area, when the surface area quadruples, the thickness of the gum will be quartered.

The bubble is NOT a perfect sphere !!!

Rajdeep Bharati - 5 years, 7 months ago

It is approximately a perfect sphere, and so can be modelled as such.

Jake Lai - 5 years, 7 months ago

Saket Sinha
Oct 15, 2015

As volume of gum used is constant, VOLUME IS CONSERVED. Therefore Area * Thickness(t) = Constant

4π (d/2)^{2} (t) =4π (2d/2)^{2 } (t`)

=> t` = t/4

The folmula you used is wrong In sphere radius changes at every surface So we use integration to calculate the volume of sphere not just multiply the surface area and thikness

Alok Pathak - 5 years, 8 months ago

You are right. The volume of gum is not calculated by simply multiplying the thickness by the surface area. The problem with this approach is that the gum "area" is not only the surface area. The gum also contain a inner area. You can imagine it as one little sphere inside a larger one. The little sphere contains air, and the space between the little sphere surface and the bigger sphere is filled up with gum. The argument presented by Saket Sinha that the gum volume is conserved is right. But in that case the volume of gum should be calculated as the volume of the larger sphere minus the volume of the small sphere of air contained inside it. Initially (before blowing more air into the bubble gum) the volume of air is 4\3•pi•r^3, where 'r' is the initial radius of the air bubble. The initial gum volume is then 4/3•pi•(r+T)^3 - 4/3•pi•r^3 , where T is the initial gum thickness. After blowing, the radius of the air bubble will be twice the initial one, thus the final volume of air will be 4/3•pi•R^3, where R=2r. The new volume of gum is 4/3•pi•(R+t)^3 - 4/3•pi•R^3, where 't' is the new thickness. As Saket said, the new gum volume is the same as the initial one, so equalling both of them we get (R+t)^3 - R^3 = (r+T)^3 - r^3. This leads us to a cubic equation. It is not a "friendly" result. Expanding both sides we get that the new thickness is given by:

 t = (T^3 + 3r•T^2 + 3T•r^2 + 8r^3)^(1/3) - 2r

This means that the new thickness depends on the initial values of the radius and thickness.

Leonardo Bohac - 5 years, 8 months ago

Yes, you're right. We must not take approximate values in Mathematics. Integration must be used.

Rajdeep Bharati - 5 years, 7 months ago

But I think that in the question,we have given a certain point to compare their Thickness. Also volume is conserved at every instant. So we can easily compare two points where r = d/2 and where r= d.

Saket Sinha - 5 years, 8 months ago

Ariosvaldo da Silva
Oct 15, 2015

12.5664 x r² But, if the diameter is doubled we have 12.5664 x 2r²

How did you get to this result?

Leonardo Bohac - 5 years, 8 months ago

Lu Chee Ket
Oct 23, 2015

V = 4 Pi R^2 t = 4 Pi (2 R)^2 (t/ 4) OR V = Pi (D)^2 t = Pi (2 D)^2 (t/ 4)

With same volume of bubble gum, thickness t becomes (1/ 4) t.

Wee Xian Bin
Oct 15, 2015

Based on similarity the (multiplication factor of a nD property of a (n+m)D object) * (multiplication factor of a distinct mD property of a (n+m)D object) = 1. If diameter is doubled, the surface area would be quadrupled, and hence for the same volume of balloon skin the thickness would be quartered.

Bubble Gum

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