Related Rates Spherical Baloon

Question

The surface area of a spherical balloon is increasing at 3 cm^2/sec. At what rate is its volume changing when its radius is 10cm?

Answer 1

V= 4pi(r^3)/3

dV/dt = 4pi(r^2)r'

SA =4pi(r^2)

dSA/dt = 8pirr'

SA'(10) = 80pi(r') = 3

r' = 3/80pi

V'(10) = 4pi(10^2)r' = 400pi(r') = 400pi(3/80pi) = 400(3)/80

= 3(5) = 15 cm^3/sec

rate of change of the sphere's volume when radius =10 cm is 15 cm^3 per second

Answer 2

Volume V = 4/3πr³; surface area S = 4πr²; dV/dt = 4πr²dr/dt; dS/dt = 8πrdr/dt.

From here (dV/dt)/(dS/dt) = r/2; dV/dt = dS/dt·r/2 = 3cm/s²·10cm/2 = 15cm³/s