Question

The surface area of a spherical balloon is increase at the rate of 2cm2cm/sec. At what rate tbe volume of the balloon is increasing when radius of the balloon is 6cm?​

Answer 1

Step-by-step explanation:

hope this attachment will help you

Answer 2

$\huge\bf\orange{\mid{\fbox{\underline{\underline{\maltese{Answer}\maltese}}}}\mid}$

Given $\frac{dA}{dt}$ = 2 cm²/sec , r: = 6cm Surface area of sphere A = 4πr²

$\frac{dA}{dt} = 8πr \frac{dr}{dt}$

$=> 2 = 8 π × 6 × \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{1}{24π} cm/sec$

Now V = $\frac{4}{3}πr³$

=> $\frac{dV}{dt} = \frac{4}{3}π3r²\frac{dr}{dt}$

=> $\frac{dV}{dt} = \frac{4.π.(6)²}{24π}$

=> $\frac{dV}{dt} = 6\:\:cm³/sec$

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️