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VITEEE 2019 Paper

Section: Mathematics

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Question : 94 of 125

Marks: +1, -0

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

-2π
$-\frac{8\pi}{5}$
$-\frac{3\pi}{5}$
$\frac{2\pi}{5}$

Solution:

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} \left( -r + h \right)

Thus, when r = 2 and h = 3,

\frac{dV}{dt} = \frac{\pi (2)}{5} (-2+3) = \frac{2\pi}{5}

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