Question

A spherical balloon is being inflated at the rate of $35\, cc$ per minute, When its radius is $7\, cm$, its surface area increases at the rate of

• A. 10 sq. cm/min
• B. 15 sq. cm/min
• C. 20 sq. cm/min
• D. 25 sq. cm/min

Answer 1

Answer: (A)

Given , $\frac{dV}{dt} = 35 ,$
where $V$ is volume of spherical balloon.
Also, $V = \frac{4}{3} \pi r^3$
$\Rightarrow \frac{d}{dt} \left(\frac{4}{3} \pi r^{3}\right) = 35$
$ \Rightarrow \frac{4}{3} \pi \times3r^{2} \frac{dr}{dt} = 35 $
$\Rightarrow \frac{dr}{dt} = \frac{35 \times3}{4\pi \times3r^{2}} $
Let $S$ be surface area of sphere then $S = 4\pi r^2$
Taking derivatives w.r.t. $'t'$
$\Rightarrow \frac{dS}{dt} = 8\pi \times r \frac{dr}{dt} = 8\pi \times r\times \frac{35 \times3}{4\pi\times3r^{2}}$
Substituter r = 7
$ \frac{dS}{dt} = \frac{2\times 35 \times 3}{3\times 7} = 10 cm^{2}/ min $
$= 2\log_{e}a$