Applications Of Derivatives Question 224
Question: The radius of a right curular 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^{2} $ /min when the radius is 2 cm and the height is 3 cm is
Options:
A) $ -2\pi $
B) $ -\frac{8\pi }{5} $
C) $ -\frac{3\pi }{5} $
D) $ \frac{2\pi }{5} $
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Answer:
Correct Answer: D
Solution:
[d] Given $ V=\pi r^{2}h. $
Differentiating both sides, we get $ \frac{dV}{dt}=\pi ( r^{2}\frac{dh}{dt}+2r\frac{dr}{dt}h )=\pi r( r\frac{dh}{dt}+2h\frac{dr}{dt} ) $
$ \frac{dr}{dt}=\frac{1}{10} $ and $ \frac{dh}{dt}=\frac{2}{10} $
$ \frac{dV}{dt}=\pi r( r( -\frac{2}{10} )+2h( \frac{1}{10} ) )=\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} $
BETA
