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January 30, 2015

January 30, 2015

Posted by **Joshie** on Wednesday, October 6, 2010 at 5:09pm.

- calculus -
**Reiny**, Wednesday, October 6, 2010 at 5:27pmV = (4/3)πr^3

when V = 36 ...

36 = (4/3)πr^3

r^3 = 27/π

r = 2.04835

dV/dt = 4πr^2 dr/dt

1 = 4π(2.04835^2) dr/dt

dr/dt = .018966 ft^3/min

since diameter = 2r

d(diameter)/dt = .0379 ft^3 /min

- calculus -
**Damon**, Wednesday, October 6, 2010 at 5:35pmfirst find dr/dt. The diameter changes twice as fast as the radius.

v = (4/3) pi r^3

dv/dt = 4 pi r^2 dr/dt

so

1 = 4 pi (r^2) dr/dt

but r^3 = (3/4)(36)/pi

so r = 2.05 ft

so

1 = 4 pi (4.2) dr/dt

so

dr/dt = .019 ft/min

D = 2 r

dD/dt = 2 dr/dt = .038 ft/min

another way

rate of volume increase = surface area * dr/dt

1 = 4 pi r^2 * dr/dt

same old equation

- calculus -
**Jorge**, Thursday, November 8, 2012 at 5:37pmA spherical balloon is inflated so that its volume is increasing at the rate of 2.2 ft^3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?

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