The surface area of a spherical ball is increasing at the rate of $4π \text{cm}^2$/second. The rate at which the radius is increasing when the surface area is $16π \text{cm}^2$ is

Question

The diagonal of a square is changing at the rate of $ 0.5 \, \text{cm/sec} $. Then the rate of change of area, when the area is 400 $ \text{cm}^2 $, is equal to:

Answer 1

Step 1: Relate
$S = 4\pi r^2$ gives $\dot S = 8\pi r \dot r$.

Step 2: Radius
$r = \sqrt{16\pi / 4\pi} = 2$ cm.

Step 3: Rate
$\dot r = 4\pi / (8\pi \cdot 2) = 1/4$ cm/s. Option (B).

Final Answer:
Option (B). \[ \boxed{0.25 \text{ cm/s}} \]

The surface area of a spherical ball is increasing at the rate of \(4π \text{cm}^2\)/second. The rate at which the radius is increasing when the surface area is \(16π \text{cm}^2\) is

Show Hint

The Correct Option is B

Solution and Explanation

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