36. draining a tank an inverted conical water tank with a height of 12 ft and a radius of 6 ft is drained…

36. draining a tank an inverted conical water tank with a height of 12 ft and a radius of 6 ft is drained through a hole in the vertex at a rate of 2 ft³/s (see figure). what is the rate of change of the water depth when the water depth is 3 ft? (hint: use similar triangles.)

36. draining a tank an inverted conical water tank with a height of 12 ft and a radius of 6 ft is drained through a hole in the vertex at a rate of 2 ft³/s (see figure). what is the rate of change of the water depth when the water depth is 3 ft? (hint: use similar triangles.)

Answer

Explanation:

Step1: Express radius in terms of height

By similar triangles, (\frac{r}{h}=\frac{6}{12}=\frac{1}{2}), so (r = \frac{h}{2}).

Step2: Volume formula of cone

The volume of a cone (V=\frac{1}{3}\pi r^{2}h). Substitute (r = \frac{h}{2}) into it, we get (V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{\pi}{12}h^{3}).

Step3: Differentiate volume with respect to time

Differentiate (V=\frac{\pi}{12}h^{3}) with respect to (t) using the chain - rule. (\frac{dV}{dt}=\frac{\pi}{12}\times3h^{2}\frac{dh}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}).

Step4: Solve for (\frac{dh}{dt})

We know that (\frac{dV}{dt}=- 2) (negative because the volume is decreasing). When (h = 3), substitute into (\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}). (-2=\frac{\pi}{4}(3)^{2}\frac{dh}{dt}), then (\frac{dh}{dt}=-\frac{8}{9\pi}).

Answer:

The rate of change of the water depth is (-\frac{8}{9\pi}\text{ ft/s}).