If the volume of a spherical balloon is changing at a rate of 10 cm3

Question

/s, how fast is the surface area
changing when the radius is 6.5 cm?

Answer 1

since v = 4/3 pi r^3
and a = 4 pi r^2,

v = 1/3 ar

so,

dv/dt = 4pi r^2 dr/dt = a dr/dt
so, dr/dt = (dv/dt)/a

dv/dt = 1/3 (r da/dt + a dr/dt)
= 1/3 (r da/dt + a*(dv/dt)/a)
= 1/3 (r da/dt + dv/dt)
= 1/2 r da/dt
So,
10 = 1/2 (6.5) da/dt
da/dt = 3.077 cm^2/s