Question

A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15cm/min. At what rate must air be removed when the radius is 9cm?

Answer 1

consider the formula for the volume of the sphere
volume=(4/3)pi(radius)^3
now differentiate the volum with respect to time
dV/dt=4pi(r^2)(dr/dt)
dV/dt=3pi(9^2)(15)
dV/dt=3645pi cm^3/min
so the air must must be removed at 3645pi cm^3/min

Answer 2

Yep, good job @Trai7 .

Answer 3

Thanks

Answer 4

how to defferentiate the volume with respect to time

Answer 5

\[\large \cfrac{d}{dr}(V) = \cfrac{d}{dr}(\cfrac43 \pi r^3)\]\[\large V' = \cfrac{4}{3}\pi \cfrac{d}{dr}(r^3)\]\[\large V' = \cfrac{4}{3} \cdot 3r^2 \]\[\large V' = 4\pi r^2\]

Answer 6

Understand? :)