An inverted conical water tank with a height of

8 ft8 ft

and a radius of

4 ft4 ft

is drained through a hole in the vertex at a rate of

6 ft cubed divided by s6 ft3/s

​(see figure). What is the rate of change of the water depth when the water depth is

4 ft4 ft​?

​(Hint: Use similar​ triangles.)

8 ft8 ft

4 ft4 ft

Outflow 6 ft cubed divided by sOutflow 6 ft3/s

...

Question content area right

Part 1

When the water depth is

4 ft4 ft​,

the rate of change of the water depth is about

To solve this problem, we need to determine the rate of change of the water depth when the water depth is \( h = 4 \, \text{ft} \). The water is draining from an inverted conical tank, and we are given that the volume of water is decreasing at a rate of \( \frac{dV}{dt} = -6 \, \text{ft}^3/\text{s} \).

### Step 1: Relate the volume of water in the tank to its depth.
The volume \( V \) of a cone is given by:

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

where \( r \) is the radius of the surface of the water and \( h \) is the height (or depth) of the water.

### Step 2: Express \( r \) in terms of \( h \) using similar triangles.
Since the tank is an inverted cone, the ratio of the radius to the height remains constant. The original tank has a height of 8 feet and a radius of 4 feet. Using similar triangles, we have:

\[
\frac{r}{h} = \frac{4}{8} = \frac{1}{2}
\]

Thus, the radius \( r \) of the water at any height \( h \) is:

\[
r = \frac{h}{2}
\]

### Step 3: Substitute \( r = \frac{h}{2} \) into the volume formula.
Now, substitute \( r = \frac{h}{2} \) into the volume formula:

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

So the volume of water as a function of the height \( h \) is:

\[
V = \frac{\pi}{12} h^3
\]

### Step 4: Differentiate the volume with respect to time.
We want to find the rate of change of the height of the water, \( \frac{dh}{dt} \). To do this, we differentiate the volume \( V \) with respect to time \( t \):

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

Simplifying:

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

### Step 5: Solve for \( \frac{dh}{dt} \).
We are given that \( \frac{dV}{dt} = -6 \, \text{ft}^3/\text{s} \) and we want to find \( \frac{dh}{dt} \) when \( h = 4 \, \text{ft} \). Substituting these values into the equation:

\[
-6 = \frac{\pi}{4} (4)^2 \frac{dh}{dt}
\]

Simplifying:

\[
-6 = \frac{\pi}{4} \cdot 16 \cdot \frac{dh}{dt}
\]

\[
-6 = 4\pi \cdot \frac{dh}{dt}
\]

\[
\frac{dh}{dt} = \frac{-6}{4\pi} = \frac{-3}{2\pi} \, \text{ft/s}
\]

### Final Answer:
The rate of change of the water depth when the water depth is 4 feet is approximately:

\[
\frac{dh}{dt} \approx \frac{-3}{2\pi} \approx -0.477 \, \text{ft/s}
\]

This means that the water depth is decreasing at a rate of approximately \( 0.477 \, \text{ft/s} \).

---

Would you like to dive into more details, or do you have any further questions?

Here are 5 related questions to deepen your understanding:
1. How does the relationship between height and radius change if the cone’s dimensions are different?
2. What happens to the rate of change \( \frac{dh}{dt} \) if the draining rate \( \frac{dV}{dt} \) changes?
3. Can you derive a similar formula for a cylindrical tank instead of a conical one?
4. What is the physical meaning of a negative \( \frac{dh}{dt} \)?
5. How would the problem change if the tank were draining faster, say at \( 10 \, \text{ft}^3/\text{s} \)?

**Tip:** When working with similar triangles, always look for proportional relationships between the dimensions; they often help simplify the variables involved.

An inverted conical water tank with a height of 8 ft and a radius of 4 ft is drained through a hole in the vertex at a rate of 6 ft^3/s. What is the rate of change of the water depth when the water depth is 4 ft? (Hint: Use similar triangles.)

Learn how to calculate the rate of change of the water depth in an inverted conical water tank using related rates and similar triangles. The problem involves a cone with a height of 8 ft and a radius of 4 ft, and water draining at a rate of 6 ft^3/s. Detailed step-by-step solution provided. Suitable for high school calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation