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: Understanding the Question:
The problem presents a rate-of-change scenario where the radius of a circular plate expands over time. Given the instantaneous radius and its rate of growth, we need to compute the corresponding rate of increase of the circle's area.

Step 2: Key Formula or Approach:
The area $A$ of a circle as a function of its radius $r$ is given by:
$$A = \pi r^2$$ Differentiating this geometric equation with respect to time $t$ using the chain rule yields:
$$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}$$

Step 3: Detailed Explanation:
Let's collect the given values from the text:
Current radius, $r = 12\ \text{cm}$
Rate of change of radius, $\frac{dr}{dt} = 0.01\ \text{cm/sec}$
Substitute these values into our derivative rate expression:
$$\frac{dA}{dt} = 2\pi \cdot (12) \cdot (0.01)$$ $$\frac{dA}{dt} = 24\pi \cdot 0.01 = 0.24\pi\ \text{sq. cm/sec}$$ This matches option (B).

Step 4: Final Answer:
The rate at which the area increases is $0.24\pi\ \text{sq. cm/sec}$, which corresponds to option (B).