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: Write the volume formula.
The balloon is a sphere, so at any moment its volume is \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius at that moment.

Step 2: Relate the rate of change of volume to the rate of change of the radius.
Both \(V\) and \(r\) change with time \(t\), so we differentiate both sides with respect to \(t\) using the chain rule:
\[ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \]
This says the rate at which the volume grows depends on the current radius and on how fast the radius itself is growing.

Step 3: Plug in the numbers.
We are told the radius grows at \(\frac{dr}{dt} = 4\) cm/sec, and we want the rate at the moment \(r = 14\) cm.
\[ \frac{dV}{dt} = 4\pi (14)^2 (4) = 4 \times 196 \times 4 \times \pi = 3136\pi \]

Step 4: Evaluate the number.
Take \(\pi = \frac{22}{7}\), since 3136 divides cleanly by 7:
\[ 3136 \times \frac{22}{7} = 448 \times 22 = 9856 \]
So the volume is increasing at 9856 cm3/sec at that instant.

Step 5: Rule out the other options.
Option (1) leaves \(\pi\) unevaluated, so it is not the clean numeric rate the question is asking for. Option (3), 10,000 cm3/sec, does not match our working, and option (4) does not apply since a matching value is present.

Final Answer:
The volume is increasing at 9856 cm3/sec. \[ \boxed{9856 \text{ cm}^3\text{/sec}} \]