A) \[-2\pi \]
B) \[-\frac{8\pi }{5}\]
C) \[-\frac{3\pi }{5}\]
D) \[\frac{2\pi }{5}\]
Correct Answer: D
Solution :
[d] Given \[V=\pi {{r}^{2}}h.\] Differentiating both sides, we get \[\frac{dV}{dt}=\pi \left( {{r}^{2}}\frac{dh}{dt}+2r\frac{dr}{dt}h \right)=\pi r\left( r\frac{dh}{dt}+2h\frac{dr}{dt} \right)\]\[\frac{dr}{dt}=\frac{1}{10}\] and \[\frac{dh}{dt}=\frac{2}{10}\] \[\frac{dV}{dt}=\pi r\left( r\left( -\frac{2}{10} \right)+2h\left( \frac{1}{10} \right) \right)=\frac{\pi r}{5}(-r+h)\]Thus, when r=2 and h=3. \[\frac{dV}{dt}=\frac{\pi (2)}{5}(-2+3)=\frac{2\pi }{5}\]
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