5. The radius \( r \) of a sphere is increasing at a constant rate. At the time when the surface area and the radius of the sphere are increasing at the same numerical rate, what is the radius of the sphere?






Answer is: option1

\(\frac{1}{8\pi}\)

Solution:

We start with the equation for the surface area of a sphere:

\[ A = 4\pi r^2 \]

Differentiating both sides with respect to \( t \):

\[ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} \]

Since it is given that:

\[ \frac{dA}{dt} = \frac{dr}{dt} \]

We substitute this into the equation:

\[ \frac{dr}{dt} = 8\pi r \frac{dr}{dt} \]

Dividing both sides by \( 8\pi r \), we get:

\[ r = \frac{1}{8\pi} \]

Thus, the correct answer is: \( \frac{1}{8\pi} \)

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