it is given that a container is in the from of an inverted right circular cone iwht base radius 4cm and height 16cm. the contianer is held vertically. initially, the container is full filled with water. then the water is pumped out from the container at a constant rate of 60cm^3 /s. let v cm^3 and h cm be the volume and the depth of water in the container respectively after t seconds. find the depth of water when the rate of increase of the depth of water is 12 cm/s

We are given an inverted right circular cone with a base radius of \(4\) cm and height \(16\) cm. The water is being pumped out at a constant rate of \(60 \, \text{cm}^3/\text{s}\), and we are asked to find the depth of water \(h\) when the rate of increase of the depth of water is \(12 \, \text{cm}/\text{s}\).

### Step 1: Formula for the volume of water
The volume of water in the cone, at any depth \(h\), is given by the formula for the volume of a cone:

\[
V = \frac{1}{3} \pi r^2 h
\]

where \(r\) is the radius of the water level at depth \(h\), and \(V\) is the volume of water. However, the radius \(r\) of the water surface depends on the depth \(h\). From the geometry of the cone, the ratio of \(r\) to \(h\) remains constant:

\[
\frac{r}{h} = \frac{4}{16} = \frac{1}{4}
\]

So, \(r = \frac{h}{4}\).

### Step 2: Substituting for \(r\) in the volume formula
Substituting \(r = \frac{h}{4}\) into the volume formula:

\[
V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h
\]
\[
V = \frac{1}{3} \pi \frac{h^2}{16} h = \frac{\pi}{48} h^3
\]

This is the volume \(V\) of water as a function of the depth \(h\).

### Step 3: Differentiating the volume with respect to time
We are given that water is being pumped out at a rate of \(60 \, \text{cm}^3/\text{s}\), so the rate of change of volume with respect to time is:

\[
\frac{dV}{dt} = -60 \, \text{cm}^3/\text{s}
\]

Now, differentiate the volume formula with respect to time \(t\):

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{48} h^3 \right) = \frac{\pi}{48} \cdot 3h^2 \cdot \frac{dh}{dt}
\]
\[
\frac{dV}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt}
\]

### Step 4: Solving for \(h\)
We know \(\frac{dV}{dt} = -60\) and are given that the rate of increase of the depth of water is \(\frac{dh}{dt} = 12 \, \text{cm}/\text{s}\). Substituting these values into the equation:

\[
-60 = \frac{\pi}{16} h^2 \cdot 12
\]
\[
-60 = \frac{3\pi}{4} h^2
\]
\[
h^2 = \frac{-60 \times 4}{3\pi}
\]
\[
h^2 = \frac{-240}{3\pi} = \frac{-80}{\pi}
\]

Since the square of a quantity cannot be negative, it appears there's an inconsistency here, likely due to a misinterpretation of the "rate of increase" of the depth. Could you clarify if the question meant "rate of decrease"? 

Let me know if you'd like further details, or feel free to clarify! Here are a few related questions for further practice:

1. What is the relationship between the volume and depth in a conical container?
2. How do you derive the formula for the volume of a cone with varying water levels?
3. What role does related rates play in geometry problems involving changing quantities?
4. How would the formula change if the cone were a different shape, such as a cylinder?
5. What other types of problems can be solved using similar related rates techniques?

**Tip:** Always check the units and consistency of the problem when interpreting rates of change!

It is given that a container is in the form of an inverted right circular cone with base radius 4 cm and height 16 cm. The container is held vertically. Initially, the container is fully filled with water. Then the water is pumped out from the container at a constant rate of 60 cm^3/s. Let v cm^3 and h cm be the volume and the depth of water in the container respectively after t seconds. Find the depth of water when the rate of increase of the depth of water is 12 cm/s.

Solve a related rates problem involving an inverted right circular cone. The water is pumped out at 60 cm^3/s, and we calculate the depth when the rate of change of water depth is 12 cm/s. Explore key geometric concepts, volume formulas, and differentiation techniques in this step-by-step guide for Grades 10-12.

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