there is an inverted right pyramid which is held vertically. its base is a sqaure with sides 5 cm and its height is 10cm. the container is initially full of water. water is leaking from a small hole at the tip of the pyramid at a constant rate of 2cm^3/s. when the depth of water is 6cm, find the rate of change of the area of the water surface

To solve this problem, we'll use related rates to find the rate of change of the area of the water surface when the depth of water is 6 cm. 

### Step 1: Understanding the Geometry and Variables

1. The inverted pyramid has a square base with side length \( 5 \, \text{cm} \) and a height of \( 10 \, \text{cm} \).
2. Water is leaking at a rate of \( \frac{dV}{dt} = -2 \, \text{cm}^3/\text{s} \) (negative because the water is decreasing).
3. We need to find the rate of change of the surface area \( A \) of the water when the depth of the water is \( h = 6 \, \text{cm} \).

### Step 2: Relate the Variables

The pyramid's cross-sectional area at a certain height \( h \) forms a smaller, similar square pyramid.

1. Let the side length of the square cross-section at height \( h \) be \( s \).
2. By similarity of triangles, the ratio of corresponding dimensions of the pyramids is:
   \[
   \frac{s}{5} = \frac{h}{10} \implies s = \frac{5h}{10} = \frac{h}{2}.
   \]

### Step 3: Volume of Water as a Function of Height

The volume of the water \( V \) at any height \( h \) can be expressed as:

\[
V = \frac{1}{3} s^2 h.
\]

Substituting \( s = \frac{h}{2} \):

\[
V = \frac{1}{3} \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \cdot \frac{h^2}{4} \cdot h = \frac{h^3}{12}.
\]

### Step 4: Differentiate Volume with Respect to Time

Differentiating \( V = \frac{h^3}{12} \) with respect to time \( t \):

\[
\frac{dV}{dt} = \frac{1}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{h^2}{4} \frac{dh}{dt}.
\]

Given \( \frac{dV}{dt} = -2 \):

\[
-2 = \frac{h^2}{4} \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-8}{h^2}.
\]

### Step 5: Rate of Change of the Surface Area

The area of the surface of the water \( A \) at depth \( h \) is:

\[
A = s^2 = \left(\frac{h}{2}\right)^2 = \frac{h^2}{4}.
\]

Differentiating \( A \) with respect to \( t \):

\[
\frac{dA}{dt} = \frac{1}{4} \cdot 2h \frac{dh}{dt} = \frac{h}{2} \frac{dh}{dt}.
\]

Substituting \( \frac{dh}{dt} = \frac{-8}{h^2} \):

\[
\frac{dA}{dt} = \frac{h}{2} \cdot \frac{-8}{h^2} = \frac{-8}{2h} = \frac{-4}{h}.
\]

### Step 6: Evaluate at \( h = 6 \, \text{cm} \)

Substituting \( h = 6 \):

\[
\frac{dA}{dt} = \frac{-4}{6} = -\frac{2}{3} \, \text{cm}^2/\text{s}.
\]

### Final Answer

The rate of change of the area of the water surface when the depth of water is 6 cm is:

\[
\frac{dA}{dt} = -\frac{2}{3} \, \text{cm}^2/\text{s}.
\]

This negative value indicates that the area of the water surface is decreasing at this rate.

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Would you like more details on any step, or do you have any further questions?

### Related Questions

1. How does the rate of change of surface area vary if the depth changes?
2. What would happen to the rate of surface area change if the leaking rate changes?
3. How would the problem change if the pyramid were upright instead of inverted?
4. Can you find the rate of change of the depth of the water when it is 8 cm instead of 6 cm?
5. How would the rate of change of surface area be affected if the base size of the pyramid were different?

**Tip:** Understanding geometric similarity is key in related rates problems, as it often simplifies complex relationships between changing dimensions!

Learn how to find the rate of change of the area of the water surface in an inverted pyramid. The problem involves key geometric concepts, related rates, and calculus. Suitable for students in Grades 11-12.

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