An inverted conical water tank with a height of $12 \mathrm{ft}$ and a radius of $6 \mathrm{ft}$ is drained through a hole in the vertex at a rate of $3 \mathrm{ft}^{3} / \mathrm{s}$ (see figure). What is the rate of change of the water depth when the water depth is $4 \mathrm{ft}$ ? (Hint: Use similar triangles.)

Step 1 ：Given that the volume V of a cone with radius r and height h is given by the formula: \(V = \frac{1}{3}\pi r^2 h\)

Step 2 ：Given that the height of the cone is 12 ft and the radius is 6 ft, we can use similar triangles to express r in terms of h. Since the height is twice the radius in the original cone, we have: \(r = \frac{h}{2}\)

Step 3 ：Substituting r into the volume formula gives: \(V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{12}\pi h^3\)

Step 4 ：Differentiating both sides with respect to time t gives: \(\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}\)

Step 5 ：We know that \(\frac{dV}{dt} = -3\) ft³/s (the negative sign indicates that the volume is decreasing as the water drains out), and we want to find \(\frac{dh}{dt}\) when h = 4 ft. Substituting these values into the equation gives: \(-3 = \frac{1}{4}\pi (4)^2 \frac{dh}{dt}\)

Step 6 ：Solving for \(\frac{dh}{dt}\) gives: \(\frac{dh}{dt} = \frac{-3}{\pi}\) ft/s

Step 7 ：So, the rate of change of the water depth when the water depth is 4 ft is \(\boxed{\frac{-3}{\pi}}\) ft/s