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Exercise 2.1 · Q15

Q.The surface area of a spherical balloon is increasing at the rate of 22 cm2^2/sec. At what rate the volume of the balloon is increasing when radius of the balloon is 66 cm?

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Let r,S,Vr,S,V be the radius, surface area and volume: S=4πr2S=4\pi r^2, V=43πr3V=\dfrac43\pi r^3.

From S=4πr2S=4\pi r^2: dSdt=8πrdrdt⇒drdt=dS/dt8πr\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt} = \dfrac{dS/dt}{8\pi r}.

Given dSdt=2\dfrac{dS}{dt}=2; at r=6r=6: drdt=28π(6)=248π=124π\dfrac{dr}{dt}=\dfrac{2}{8\pi(6)}=\dfrac{2}{48\pi}=\dfrac{1}{24\pi}. …

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