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The Radius of a Cylinder is Increasing at the Rate 2 Cm/Sec. and Its Altitude is Decreasing at the Rate of 3 Cm/Sec. Find the Rate of Change of Volume When Radius is 3 Cm and Altitude 5 Cm. - Mathematics

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Question

The radius of a cylinder is increasing at the rate 2 cm/sec. and its altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of volume when radius is 3 cm and altitude 5 cm.

Sum
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Solution

\[\text { Let r e the radius,h be the height and V be the volume of the cylinder at any time t. Then },\]
\[V = \pi r^2 h\]
\[ \Rightarrow \frac{dV}{dt} = 2\pi r h\frac{dr}{dt} + \pi r^2 \frac{dh}{dt}\]
\[ \Rightarrow \frac{dV}{dt} = \pi r\left( 2h\frac{dr}{dt} + r\frac{dh}{dt} \right)\]
\[ \Rightarrow \frac{dV}{dt} = \pi \times 3\left( 2 \times 5 \times 2 + 3 \times - 3 \right)\]
\[ \Rightarrow \frac{dV}{dt} = 3\pi\left( 20 - 9 \right)\]
\[ \Rightarrow \frac{dV}{dt} = 33\pi  \text{ cm}^3 /\sec\]

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Chapter 13: Derivative as a Rate Measurer - Exercise 13.2 [Page 20]

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RD Sharma Mathematics [English] Class 12
Chapter 13 Derivative as a Rate Measurer
Exercise 13.2 | Q 22 | Page 20

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