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Question 872343: My question has to do with the much used example, as it relates to scuba diving, of a "flexible sphere" (in this case a balloon) and how volume and diameter of that balloon decrease with depth. So, if at a certain depth (33 feet) the volume of the balloon is decreased by exactly half, does the diameter of the balloon also decrease by exactly half?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Say the volume of the sphere starts at 100 cubic feet. The radius of this sphere is...
V = (4/3)*pi*r^3
100 = (4/3)*pi*r^3
100(3/4) = pi*r^3
75 = pi*r^3
75/pi = r^3
r^3 = 75/pi
r = cuberoot(75/pi)
r = (75/pi)^(1/3)
r = 2.87941191148487
So the diameter is 2*2.87941191148487 = 5.75882382296973 feet
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Now let's say the volume gets cut in half to 50 cubic feet. Use the same formula to solve for r
V = (4/3)*pi*r^3
50 = (4/3)*pi*r^3
50(3/4) = pi*r^3
75 = pi*r^3
37.5/pi = r^3
r^3 = 37.5/pi
r = cuberoot(37.5/pi)
r = (37.5/pi)^(1/3)
r = 2.28539074867041
The diameter is now 2*2.28539074867041 = 4.57078149734082
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The diameter goes from 5.75882382296973 feet to 4.57078149734082 feet. Clearly the diameter has NOT been cut in half.
This example shows that the initial claim is false. You could disprove it using algebra and using an unknown in place of 100 cu ft, but this counterexample is enough to disprove it because the claim needs to be true for every single case. If it's wrong in one case (like shown above), then it's false overall.
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