A spherical balloon is being inflated at the rate 35cc/sec. Find the r

1.	A spherical balloon is being inflated at the rate 35cc/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.
Answer» Given \(\frac{dr}{dt}\) = 35cc/sec 2r = 14 ⇒ r = 7, \(\frac{ds}{dt}\) = ?, \(\frac{dr}{dt}\) = ? v = \(\frac{4}{3}\)πr³ s = 4πr² \(\frac{dv}{dt}\) = \(\frac{4}{3}\)π³r²\(\frac{dr}{dt}\) \(\frac{ds}{dt}\) = 4π²r\(\frac{dr}{dt}\) 35 = 4π . 7²\(\frac{dr}{dt}\) = 4π × 2 × 7 × \(\frac{5}{28\pi}\) \(\frac{dr}{dt}\) = \(\frac{35}{196\pi}\) = \(\frac{5}{28\pi}\) = 10cm²/sec

A spherical balloon is being inflated at the rate 35cc/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.

Answer»

Given \(\frac{dr}{dt}\) = 35cc/sec 2r = 14

⇒ r = 7, \(\frac{ds}{dt}\) = ?, \(\frac{dr}{dt}\) = ?

v = \(\frac{4}{3}\)πr³ s = 4πr²

\(\frac{dv}{dt}\) = \(\frac{4}{3}\)π³r²\(\frac{dr}{dt}\)

\(\frac{ds}{dt}\) = 4π²r\(\frac{dr}{dt}\)

35 = 4π . 7²\(\frac{dr}{dt}\) = 4π × 2 × 7 × \(\frac{5}{28\pi}\)

\(\frac{dr}{dt}\) = \(\frac{35}{196\pi}\) = \(\frac{5}{28\pi}\) = 10cm²/sec