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a spherical balloon is inflated so that its volume is increasing at the rate of 6ft^3/minute. how fast is the diameter of the balloon increasing when the radius is 1 feet? reminder: volume of a sphere= 4/3 (pi)r^3

- Thread starter skivail
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a spherical balloon is inflated so that its volume is increasing at the rate of 6ft^3/minute. how fast is the diameter of the balloon increasing when the radius is 1 feet? reminder: volume of a sphere= 4/3 (pi)r^3

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Tide

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-When simplified you get pi/6(D(t))^3.

-Differentiate and you get V'= pi/6[3(D(t))^2] [D'(t)] (implicit diff.)

-plug in the numbers--remember any variable with the ' means it's a rate, a unit over time

- I got 3.8197 ft/min. but if you have an answer key go with that.

I really hope this helps! Good luck on your test!

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