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Question 256825: The formula relating the volume, V, of a baloon and its radius,r, as it is being inflated is: V(r)=4/3 pi r^3,where r is the radius in cm.
a) Determine the average rate of change in volume between r=2 and r=10cm.
b) Determine the average rate of change in volume between r=8 and r=10cm.
c) Determine the average rate of change in volume between r=9.99 and r=10
d) Calculate the value of 4 pi r ^2 for r=10. What do you notice?
Please help me :) Thanks.
Found 2 solutions by richwmiller, Alan3354:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! reminder :
do not submit similar problems.
And put them in the right category This problem did not belong in mixtures.
We ask very little of you.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! The formula relating the volume, V, of a baloon and its radius,r, as it is being inflated is: V(r)=4/3 pi r^3,where r is the radius in cm.
a) Determine the average rate of change in volume between r=2 and r=10cm.
At r=2, Vol = (32/3) pi cc =~ 33.51 cc
At r=10, Vol = (4000/3)pi cc =~4188.79 cc
Diff = (3968/3)pi cc
Avg rate of change = (3968/3)*pi/(10-2) =~ 519.41
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b) Determine the average rate of change in volume between r=8 and r=10cm.
At r=8, Vol = (2048/3)pi cc =~ 2144.66 cc
At r=10, Vol = (4000/3)pi cc =~ 4188.79 cc
Avg rate of change = (1952*pi/3)/(10-8) = ~ 1022
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c) Determine the average rate of change in volume between r=9.99 and r=10
At r = 9.99, Vol =~ 4176.24 cc
At r = 10, Vol =~ 4188.79 cc
Avg rate of change =~ 1255
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d) Calculate the value of 4 pi r ^2 for r=10. What do you notice?
4pi*r^2 = 4000pi =~ 1256.637
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As the difference between 10 and the other radius becomes smaller, the average rate of change becomes closer to 4pi*r^2 at r=10.
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Tho it might not be relevant, 4pi*r^2 is the surface area of the sphere.
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PS I did these on a calculator, but I recommend using Excel. Errors are less likely, and you can easily change the increments and see the effects.
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