QuestionAn inverted cylindrical cone, 40 ft deep and 20 ft across at the top, is being filled with water at a rate of $19 \mathrm{ft}^{3} / \mathrm{min}$. At what rate is the water rising in the tank when the depth of the water is:
1 foot? Answer= $\square$ 10 feet? Answer= $\square$ 39 feet? Answer= $\square$

Studdy Solution
For each depth, plug in the value of $h$:
1. **When the depth is 1 foot:**
$\frac{dh}{dt} = \frac{19 \cdot 16}{\pi \cdot 1^2} = \frac{304}{\pi} \approx 96.78 \text{ ft/min}$
2. **When the depth is 10 feet:**
$\frac{dh}{dt} = \frac{19 \cdot 16}{\pi \cdot 10^2} = \frac{304}{100\pi} \approx 0.97 \text{ ft/min}$
3. **When the depth is 39 feet:**
$\frac{dh}{dt} = \frac{19 \cdot 16}{\pi \cdot 39^2} = \frac{304}{1521\pi} \approx 0.0636 \text{ ft/min}$