Question

The volume of a sphere is increasing at the rate of π cm³/sec.The rate at which the radius is increasing is....... when the radius is 3 cm.Select correct option from the given options.
(a) 1/36 cm/sec
(b) 36 cm/sec
(c) 9 cm/sec
(d) 27 cm/sec

Answer 1

volume of sphere is increasing at the rate , $\frac{dV}{dt}=\pi cm^3/sec$......(1)

we have to find rate at which the radius is increasing.
we know, volume of sphere , $V=\frac{4}{3}\pi r^3$
now, differentiate both sides with respect to time,
$\frac{dV}{dt}=\frac{4}{3}\pi 3r^{3-1}\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

put equation (1) and r = 3cm

$\pi cm^3/sec=4\pi (3)^2\frac{dr}{dt}$

$1=4\times9\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{36} cm/sec$

hence, option (a) is correct

Answer 2

Dear Student,

Answer:Option A ( 1/36 cm/sec)

Solution:

Given that change in volume of a sphere with respect to time
$\frac{dv}{dt} = \pi \: {cm}^{3} {(sec)}^{ - 1}$
Volume of sphere
$v \: = \frac{4}{3} \pi {r}^{3} \\ \frac{dv}{dt} = \frac{4}{3} \pi(3 {r}^{2} )\frac{dr}{dt} \\ \\\frac{dv}{dt}= 4\pi {r}^{2}\frac{dr}{dt} \\ \\ \pi = 4\pi {3}^{2}\frac{dr}{dt}\\ \\ \frac{dr}{dt}= \frac{1}{36} cm/sec$

hope it helps you