Question

The radius 'r' of a cylinder increases with time at a constant rate of 2m/sec and its height decreases
with time at constant rate of 3 m/sec. The rate of change of volume of cylinder at a time when radius
and height of cylinder are 3m and 2m respectively will be :​

Answer 1

Given:

The radius 'r' of a cylinder increases with time at a constant rate of 2m/sec and its height decreases

with time at constant rate of 3 m/sec.

To find:

Rate of change of volume

Calculation:

$\therefore \: V = \pi {r}^{2} h$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{\dfrac{d( {r}^{2} h)}{dt} \bigg \}$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{h\dfrac{d( {r}^{2} )}{dt} + {r}^{2} \dfrac{d(h)}{dt} \bigg \}$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{2rh\dfrac{d(r)}{dt} + {r}^{2} \dfrac{d(h)}{dt} \bigg \}$

Putting available values ;

$= > \: \dfrac{dV}{dt} = \pi \bigg \{2(3)(2)\dfrac{d(r)}{dt} + {(3)}^{2} \dfrac{d(h)}{dt} \bigg \}$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{2(3)(2)(2) + {(3)}^{2} ( - 3) \bigg \}$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{24 + (- 27)\bigg \}$

$= > \: \dfrac{dV}{dt} = \pi \bigg \{ - 3\bigg \}$

$= > \: \dfrac{dV}{dt} = - 3\pi \: \: {m}^{3} {s}^{ - 1}$

So, final answer is:

$\boxed{ \sf{ \red{ \: \dfrac{dV}{dt} = - 3\pi \: \: {m}^{3} {s}^{ - 1} }}}$

Answer 2

Answer:

• -3 pie m^3/sec

Explanation:

hope it is useful