Question

A spherical balloon is inflated and its radius is increasing at 4 cm/second. At what rate would the volume be increasing when its radius is 14 cm?

• A. \(\pi(56)^2 \text{ cm}^3\text{/sec}\)
• B. \(9856 \text{ cm}^3\text{/sec}\)
• C. \(10{,}000 \text{ cm}^3\text{/sec}\)
• D. None of these

Hint:

Differentiate the sphere volume formula \(V = \frac{4}{3}\pi r^3\) with respect to time to link \(\frac{dV}{dt}\) with \(\frac{dr}{dt}\), then substitute r = 14 cm and dr/dt = 4 cm/sec.

Answer 1

The Correct Option is B

Step 1: Notice a shortcut.
For a sphere, the derivative of the volume with respect to the radius is exactly the surface area:
\[ \frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \]
This is just the curved surface area formula of the sphere at radius \(r\).

Step 2: Use the chain rule with this shortcut.
By the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$. So the rate of change of volume equals the surface area at that instant times the rate at which the radius grows.

Step 3: Find the surface area at r = 14 cm.
\[ \text{Surface area} = 4\pi r^2 = 4\pi (14)^2 = 4\pi \times 196 = 784\pi \]

Step 4: Multiply by the rate of growth of the radius.
The radius grows at 4 cm/sec, so:
\[ \frac{dV}{dt} = 784\pi \times 4 = 3136\pi \]
Taking $\pi = \frac{22}{7}$:
\[ 3136 \times \frac{22}{7} = 448 \times 22 = 9856 \]

Step 5: Match with the options.
This numeric value, 9856 cm3/sec, is exactly option (2). The symbolic form in option (1) is the same quantity before the final numeric step, so it is not the number the question is after, and 10,000 cm3/sec in option (3) does not match our working at all.
\[ \boxed{9856 \text{ cm}^3\text{/sec}} \]