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iogann1982 [59]
4 years ago
12

A spherical balloon is inflated so that its volume is increasing at the rate of 2.7 ft3/min. How rapidly is the diameter of the

balloon increasing when the diameter is 1.3 feet
Mathematics
1 answer:
Mice21 [21]4 years ago
5 0
The formula for volume of a sphere is
V=\frac{4}{3}\pi r^{3}
\rightarrow\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}

If the diameter is 1.3, then r=\frac{1.3}{2}=0.65

We are given
\frac{dV}{dt}=2.7 and r=0.65
\rightarrow2.7=4\pi(0.65)^{2}\frac{dr}{dt}
\rightarrow\frac{dr}{dt}=\frac{2.7}{4\pi(0.65)^{2}}\approx0.5085\ ft/min

Change in diameter is twice change in radius, so we get 2*0.5085=1.017 ft/min
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