Math CentralQuandaries & Queries


Question from Houston, a student:

Bazooka Joe is blowing a spherical bubble gum bubble. Let V be the volume in the bubble, R the inside of the bubble, and T the thickness of the bubble. V, T, and R are functions of time t.

(a) Write a formula for V in terms of T and R. Hint: Draw a picture
(b) Assume that the amount of bubble gum in the bubble is not changing. What is V'(t)?
(c) After 5 minutes of blowing a bubble gum bubble, the bubble is 3ft in diameter and .01 feet thick. If the inside radius of the bubble is expanding at a rate of .5 feet per minute, how fast is the thickness changing? Hint: Remember that the volume of gum in the bubble does not change over time.

I am so confused!!!

Hi Houston,

If R is the radius of the inside of the bubble and T is the thickness of the gum that forms the skin of the bubble then the radius of the bubble is R + T. Thus the volume of the bubble is

V = 4/3 π (R + T)3

and hence

V'(t) = 4/3 π × 3 (R + T)2 (R' + T') = 4 π (R + T)2 (R' + T')

The volume of air VA in the bubble is

VA = 4/3 π R3

and hence the volume of gum (the skin of the bubble) is

V - VA = 4/3 π (R + T)3 - 4/3 π R3

But this is a constant so its derivative is zero. Hence

4 π (R + T)2 (R' + T') - 4 π R2 R' = 0 or 4 π (R + T)2 (R' + T') = 4 π R2 R'


V'(t) = 4 π (R + T)2 (R' + T') = 4 π R2 R'

Now try part (c).


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