Question

The radius of a circular plate is increasing at the rate of $0.01\ \text{cm/sec}$. When the radius is $12\ \text{cm}$, the rate at which its area increases is

• A. $24\pi\ \text{sq. cm/sec}$
• B. $0.24\pi\ \text{sq. cm/sec}$
• C. $1.2\pi\ \text{sq. cm/sec}$
• D. $60\pi\ \text{sq. cm/sec}$

Hint:

Think of rate of change of area as a ring unrolling: $\Delta \text{Area} \approx \text{Circumference} \times \Delta r$. Multiplying the circumference ($2\pi \cdot 12 = 24\pi$) by the tiny change ($0.01$) instantly gives $0.24\pi$ without writing down full equations!

Answer 1

The Correct Option is B

Step 1: Identify the quantities.
The radius $r$ grows with time, and we want how fast the area $A$ grows at the instant $r=12$ cm.
Step 2: Note the given rate.
The radius increases at $\dfrac{dr}{dt}=0.01$ cm/sec.
Step 3: Write the area relation.
For a circular plate, $A=\pi r^2$.
Step 4: Differentiate with respect to time.
Using the chain rule, $\dfrac{dA}{dt}=2\pi r\,\dfrac{dr}{dt}$.
Step 5: Substitute the values.
At $r=12$, $\dfrac{dA}{dt}=2\pi(12)(0.01)$.
Step 6: Simplify.
$2\times12\times0.01=0.24$, so $\dfrac{dA}{dt}=0.24\pi$ sq. cm/sec, matching option (2).
\[ \boxed{0.24\pi\ \text{sq. cm/sec}} \]