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Question
If \[V = \frac{4}{3}\pi r^3\] , at what rate in cubic units is V increasing when r = 10 and \[\frac{dr}{dt} = 0 . 01\] ? _________________
Options
π
4π
40π
4π/3
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Solution
4π
\[\text { Given }:V = \frac{4}{3}\pi r^3 , r = 10 \text { and } \frac{dr}{dt} = 0 . 01\]
\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\]
\[ \Rightarrow \frac{dV}{dt} = 4\pi \left( 10 \right)^2 \times 0 . 01\]
\[ \Rightarrow \frac{dV}{dt} = 4\pi\]
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