Question

A balloon is pumped at the rate of a cm^3/min . the rate of increase of its

Answer 1

sorry i dont know because i am not good at this

Answer 2

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$\frac{dV}{dt} = a\:\: cm³/ min$

$V = (\frac{4}{3}) × πr³$

$\frac{dV}{dt} = (\frac{4}{3}) × π3r² (\frac{dr}{dt})$

$a = 4πr²(\frac{dr}{dt})$

$\frac{dr}{dt}$ = $\frac{a}{(4πr²)}$ — (1)

Surface area = $4πr²$

$\frac{d (surface area)}{dt} = 4π (2R) (\frac{dr}{dt})$

$= 8πr (\frac{a}{4πr²})= \frac{2a}{r}$

$\frac{d (surface\:\: area)}{dt} = \frac{2a}{ b}$

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