Question

A spherical balloon is being inflated at the rate of 35cc/min. The rate of increase in the surface area (in \(\mathrm{cm^2}\)/min.) of the balloon when its diameter is 14 cm, is :

• A. 10
• B. \(\sqrt{10}\)
• C. 100
• D. \(10\sqrt{10}\)

Hint:

Relate \(dV/dt\) and \(dS/dt\) through \(dr/dt\) using \(V=\frac{4}{3}\pi r^3\) and \(S=4\pi r^2\).

Answer 1

The Correct Option is A

Diameter 14 cm, so \(r = 7\) cm.
\(V=\frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Given \(\frac{dV}{dt}=35\): \(4\pi(49)\frac{dr}{dt}=35 \Rightarrow \frac{dr}{dt}=\frac{5}{28\pi}\).
\(S=4\pi r^2 \Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(7)\cdot\frac{5}{28\pi}=10\).
Answer: (1) 10