Question

Air is escaping from a spherical balloon at the rate of 2𝑐𝑚3 per minute. How fast is the radius shrinking when the volume is 36𝜋 𝑐𝑚3 ?

Find the rate of change of the area 𝐴, of the circle with respect to its circumference C, 𝑖. 𝑒 𝑑𝐴 𝑑�

Answer 1

for a spherical balloon;

volume, $V= \frac {4}{3}\pi r^3 ...(i)$

area, $A= 4\pi r^2 ...(ii)$

circumference,$C=2\pi r ...(iii)$

a) given $\frac{dv}{dt}=-2 cm^3/min$ the negative sign to denote reduction in volume with time and $V= 36\pi cm^3$ we have;

$\frac {dv}{dt}=4\pi r^3 \frac {dr}{dt} from (i)$

$\therefore \frac{dr}{dt}= \frac{1}{4\pi r^2} \frac {dv}{dt}...(iv)$

when $V=36\pi\implies \frac {4}{3}\pi r^3 =36\pi$

$r^3= \frac {3*36}{4}=27$

$\implies r=3cm$

$\therefore \frac{dr}{dt}=\frac {1}{4\pi (3)^2}cm/min =\frac{-1}{18\pi}cm/min$

the radius is shrinking at the rate of $\frac{1}{18\pi}cm/min$

b) from (iii) $r= \frac{c}{2\pi}...(v)$

now (v) in (ii) gives;

$A= 4\pi (\frac {c}{2\pi})^2 = \frac {4\pi c^2}{4\pi ^2}= \frac 1\pi c^2 ...(vi)$

$\therefore \frac {dA}{dC}= \frac {2C}{\pi}$

when $V=36\pi cm^3, r= 3cm$  [from part (a)]

and thus $C=6\pi cm$

$\implies \frac {dA}{dC}= \frac {2(6\pi)} {\pi} (when c=6\pi)$

=12