Question

A spherical snow ball is melting so that its volume is decreasing at the rate of 8 c.c./sec then the rate of change of radius when the radius is 2 cm, is :

• A. The radius is increasing at the rate of \(\frac{1}{2π} \text{cm/s}\)
• B. The radius is decreasing at the rate of \(\frac{1}{2π} \text{cm/s}\)
• C. The radius is increasing at the rate of \(\frac{1}{π} \text{cm/s}\)
• D. The radius is decreasing at the rate of \(\frac{1}{π} \text{cm/s}\)

Hint:

Differentiate V = 4/3 pi r^3 with respect to time.

Answer 1

The Correct Option is B

Step 1: Surface area link
The rate of volume change equals surface area times rate of radius change: $\frac{dV}{dt} = S\frac{dr}{dt}$.

Step 2: Numbers
$S = 4\pi(2)^2 = 16\pi$. So $\frac{dr}{dt} = \frac{-8}{16\pi} = -\frac1{2\pi}$.

Step 3: Answer
Decreasing at $\frac1{2\pi}$ cm/s, option (B).

Final Answer:
Decreasing at 1/(2 pi) cm per second. \[ \boxed{\text{(B)}\ \text{decreasing at }\frac{1}{2\pi}\ \text{cm/s}} \]