Question #181111

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�



1
Expert's answer
2021-04-14T12:39:26-0400

for a spherical balloon;

volume, V=43πr3...(i)V= \frac {4}{3}\pi r^3 ...(i)

area, A=4πr2...(ii)A= 4\pi r^2 ...(ii)

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

a) given dvdt=2cm3/min\frac{dv}{dt}=-2 cm^3/min the negative sign to denote reduction in volume with time and V=36πcm3V= 36\pi cm^3 we have;

dvdt=4πr3drdtfrom(i)\frac {dv}{dt}=4\pi r^3 \frac {dr}{dt} from (i)

drdt=14πr2dvdt...(iv)\therefore \frac{dr}{dt}= \frac{1}{4\pi r^2} \frac {dv}{dt}...(iv)

when V=36π    43πr3=36πV=36\pi\implies \frac {4}{3}\pi r^3 =36\pi

r3=3364=27r^3= \frac {3*36}{4}=27

    r=3cm\implies r=3cm

drdt=14π(3)2cm/min=118πcm/min\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 118πcm/min\frac{1}{18\pi}cm/min

b) from (iii) r=c2π...(v)r= \frac{c}{2\pi}...(v)

now (v) in (ii) gives;

A=4π(c2π)2=4πc24π2=1πc2...(vi)A= 4\pi (\frac {c}{2\pi})^2 = \frac {4\pi c^2}{4\pi ^2}= \frac 1\pi c^2 ...(vi)


dAdC=2Cπ\therefore \frac {dA}{dC}= \frac {2C}{\pi}


when V=36πcm3,r=3cmV=36\pi cm^3, r= 3cm  [from part (a)]

and thus C=6πcmC=6\pi cm

    dAdC=2(6π)π(whenc=6π)\implies \frac {dA}{dC}= \frac {2(6\pi)} {\pi} (when c=6\pi)

=12







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