The radius of a spherical balloon is increasing at the rate of 0.5 cm/s. The rate at which its surface area is increasing when the radius is 8 cm is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(32\pi\) cm2/s
Let r be the radius and S the surface area at time t, so \(S = 4\pi r^2\). By the chain rule:
\( \displaystyle \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \)
With r = 8 cm and \(\frac{dr}{dt} = 0.5\) cm/s:
\( \displaystyle \frac{dS}{dt} = 8\pi \times 8 \times 0.5 \)
\( = 32\pi \) cm2/s
Why the other options are incorrect
Option (1) \(64\pi\) cm2/s — Incorrect. This is \(8\pi r\) alone, the rate with respect to r; it leaves out the factor \(\frac{dr}{dt} = 0.5\).
Option (2) \(128\pi\) cm2/s — Incorrect. This is \(4\pi r^2 \times 0.5\), the rate of increase of the volume (in cm3/s).
Option (4) \(16\pi\) cm2/s — Incorrect. This is half the correct value; the derivative of \(4\pi r^2\) is \(8\pi r\), not \(4\pi r\).