a spherical ballon is being inflated at a constant rate of 25 cm^3/sec. At what rate in cm/sec is the radius of the ballon changing when the radius of the ballon is 2 cm? (VOLUME SPHERE FORMULA 4/3 pi r^3)

Question

a spherical ballon is being inflated at a constant rate of 25 cm^3/sec. At what rate in cm/sec is the radius of the ballon changing when the radius of the ballon is 2 cm?   (VOLUME SPHERE FORMULA 4/3 pi r^3)

Damon · Answer

the surface area of the surface of the sphere, 4 pi r^2 times change in radius is the differential change in volume. v = (4/3) pi r^3 dv/dt = 4 pi r^2 dr/dt dr/dt = dv/dt /(4 pi r^2)