Question

the volume of a spherical balloon is increasing @ 20 CM cube per second find the rate of change of its surface area at the instant when its radius is 8 cm

Answer 1

Yeah sure for your help

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Hard Question at my level

Here is the answer which you are searching for :-

Let V be volume of sphere of r radius

We know that volume of sphere is 4/3 pie r^3

$\frac{dv}{dt}$ = 20 cm^3/s

$\frac{d}{dt} =$[tex] \frac4/3\pi r^{3 [/tex] = 20

$\frac{4}{3} \pi .3r.\frac{dr}{dt} =20$

( $\frac{dr}{dt} = \frac{20}{4\pi r^{2}}$ ) ------> Equation 1

Let S be surface area of ballon

S = $4\pi r^{2}$

Rate of increase of surface area = $\frac{dS}{dt} =\frac{d}{dt} ( 4\pi r^{2}$

==> $8\pi r \frac{dr}{dt} = 8\pi r . \frac{20}{4\pi r^{2}}$

==> $\frac{50}{r}$

r = 8 rate of increase of surface area = $\frac{50}8}$ = 6.25 cm ^3/ second

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@ Brainlestuser