Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3.8 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?

Answer 1

let d be the diameter of the baloon at any time t.
\[Volume~ V=\frac{ 4 }{3 }\pi \left( \frac{ d }{2 } \right)^3=\frac{ \pi }{ 6 }d^3\]
\[\frac{ dV }{dt }=\frac{ \pi }{6 }*3 d^2*\left( d \frac{ d }{dt } \right)\]
\[\frac{ dV }{dt }=3.8ft^3/\min.when~d=1.2 ~ft\]
\[3.8=\frac{ \pi }{ 2 }(1.2)^2\frac{ d }{dt }(d)\]
\[\frac{ d }{dt }\left( d \right)=\frac{( 3.8)(2) }{ 1.44 }=?\]

Answer 2

correction
\[\frac{ d }{ dt }d=\frac{ \left( 3.8 \right)\left( 2 \right) }{ 1.44\pi }=?\]