Question: 4. Water is leaking from a spherical rubber container at the rate of 2 cm per minute. How fast is the surface area shrinking when the radius of the container is 9 cm? [7]
1. **State the problem:**
We have a spherical container with radius $r$ decreasing at a rate of $\frac{dr}{dt} = -2$ cm/min (since water is leaking, radius decreases).
We want to find the rate of change of the surface area $S$ when $r=9$ cm.
2. **Relevant formulas:**
The surface area of a sphere is given by:
$$S = 4\pi r^2$$
3. **Differentiate with respect to time $t$:**
$$\frac{dS}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt}$$
4. **Substitute known values:**
$$r = 9, \quad \frac{dr}{dt} = -2$$
5. **Calculate:**
$$\frac{dS}{dt} = 8\pi \times 9 \times (-2) = -144\pi$$
6. **Interpretation:**
The negative sign indicates the surface area is shrinking.
**Final answer:**
$$\frac{dS}{dt} = -144\pi \text{ cm}^2/\text{min}$$