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  • The radius of a right circular cylinder increases at the rate of 0.1 cm} /min and the height decreases at the rate of 0.2 cm / min. The rate of change of the volume of the cylinder in <img alt

The radius of a right circular cylinder increases at the rate of 0.1 cm} /min and the height decreases at the rate of 0.2 cm / min. The rate of change of the volume of the cylinder in cm^{3} / min, when the radius is 2 cm and the height is 3 cm is

 

Option: 1

-2 \pi


Option: 2

-\frac{8 \pi}{5}


Option: 3

-\frac{3 \pi}{5}


Option: 4

\frac{2 \pi}{5}


Answers (1)

best_answer

Given, 

V=\pi r^2 h

Differentiating both sides 

\begin{aligned} & \frac{d V}{d t}=\pi\left(r^2 \frac{d h}{d t}+2 r \frac{d r}{d t} h\right)=\pi r\left(r \frac{d h}{d t}+2 h \frac{d r}{d t}\right) \\ \\& \frac{d r}{d t}=\frac{1}{10} \quad \text { and } \quad \frac{d h}{d t}=-\frac{2}{10} \end{aligned}

\frac{d V}{d t}=\pi r\left(r\left(-\frac{2}{10}\right)+2 h\left(\frac{1}{10}\right)\right)=\frac{\pi r}{5}(-r+h)

Thus, when r=2\: \text{and} \: h = 3

        \frac{d V}{d t}=\frac{\pi(2)}{5}(-2+3)=\frac{2 \pi}{5}

Posted by

sudhir.kumar

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