A tetrahedral tank with each side equal to 4 meters has an opening at the top through which the water is getting stored @0.25cubic meter/minute. Find the rate at which the level of the water is rising after 20 minutes.

To find the rate at which the water level is rising after 20 minutes in a tetrahedral tank, we first need to establish the relationship between the height of the water level and the volume of water in the tank.

### Step 1: Volume of the Tetrahedron
Given that the tetrahedral tank has sides of length \( a = 4 \) meters, the volume \( V \) of a regular tetrahedron can be calculated using the formula:

\[
V = \frac{a^3 \sqrt{2}}{12}
\]

Substituting \( a = 4 \) meters:

\[
V = \frac{4^3 \sqrt{2}}{12} = \frac{64 \sqrt{2}}{12} = \frac{16 \sqrt{2}}{3} \text{ cubic meters}
\]

### Step 2: Volume of Water as a Function of Height
For a regular tetrahedron, the volume of water \( V(t) \) as a function of height \( h \) (from the vertex to the water surface) can be related by the proportional relationship:

\[
V(t) = \frac{h^3 \sqrt{2}}{12}
\]

We differentiate this equation with respect to time \( t \) to find the rate at which the water level is rising.

### Step 3: Differentiate with Respect to Time
Given \( \frac{dV}{dt} = 0.25 \, \text{cubic meters per minute} \), we differentiate the volume function with respect to time \( t \):

\[
\frac{dV}{dt} = \frac{3h^2 \sqrt{2}}{12} \cdot \frac{dh}{dt}
\]

Simplifying:

\[
\frac{dV}{dt} = \frac{h^2 \sqrt{2}}{4} \cdot \frac{dh}{dt}
\]

Now, solve for \( \frac{dh}{dt} \):

\[
\frac{dh}{dt} = \frac{4 \cdot \frac{dV}{dt}}{h^2 \sqrt{2}} = \frac{4 \times 0.25}{h^2 \sqrt{2}} = \frac{1}{h^2 \sqrt{2}} \text{ meters per minute}
\]

### Step 4: Calculate Height After 20 Minutes
To find the height \( h \) after 20 minutes, we need to first calculate the volume of water after 20 minutes:

\[
V(t) = 0.25 \times 20 = 5 \text{ cubic meters}
\]

Using the volume formula:

\[
5 = \frac{h^3 \sqrt{2}}{12}
\]

Solving for \( h \):

\[
h^3 = \frac{5 \times 12}{\sqrt{2}} = \frac{60}{\sqrt{2}} = 30 \sqrt{2}
\]

\[
h = \sqrt[3]{30 \sqrt{2}} \text{ meters}
\]

### Step 5: Plugging \( h \) into the Rate Equation
Finally, substitute \( h \) into the equation for \( \frac{dh}{dt} \):

\[
\frac{dh}{dt} = \frac{1}{(\sqrt[3]{30 \sqrt{2}})^2 \sqrt{2}} \text{ meters per minute}
\]

Simplifying further gives you the rate at which the water level is rising after 20 minutes.

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Learn how to calculate the rate at which the water level rises in a tetrahedral tank after 20 minutes. This problem involves applying geometry to find the volume of a tetrahedron and differentiation to determine the rate of change. Suitable for advanced high school students studying geometry and calculus.

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