(1 point) A conical water tank with vertex down has a radius of 11 feet at the top and is 25 feet high. If water flows into the tank at a rate of 10 ft3/min
, how fast is the depth of the water increasing when the water is 14 feet deep?

To solve this problem, we will use **related rates**, where we relate the rate at which the volume of water is changing to the rate at which the depth of the water is increasing.

### Step 1: Understand the relationship
- Given:
  - Radius of the tank at the top \( R = 11 \) ft.
  - Height of the tank \( H = 25 \) ft.
  - Water is flowing into the tank at a rate of \( \frac{dV}{dt} = 10 \) ft³/min.
  - We want to find how fast the depth \( h \) of the water is increasing when \( h = 14 \) ft, i.e., \( \frac{dh}{dt} \) when \( h = 14 \) ft.

### Step 2: Set up the formula for the volume of a cone
The formula for the volume \( V \) of a cone is:

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

Where:
- \( r \): the radius of the water's surface.
- \( h \): the height (or depth) of the water.

### Step 3: Relate \( r \) and \( h \)
Since the tank is conical, \( r \) and \( h \) are related. Using similar triangles, we can express \( r \) in terms of \( h \):

\[
\frac{r}{h} = \frac{11}{25} \implies r = \frac{11}{25} h
\]

### Step 4: Substitute \( r \) into the volume formula
Substitute \( r = \frac{11}{25} h \) into the volume formula:

\[
V = \frac{1}{3} \pi \left(\frac{11}{25} h\right)^2 h
\]

\[
V = \frac{1}{3} \pi \frac{121}{625} h^3
\]

\[
V = \frac{121 \pi}{1875} h^3
\]

### Step 5: Differentiate with respect to \( t \)
To find \( \frac{dh}{dt} \), we differentiate both sides with respect to \( t \):

\[
\frac{dV}{dt} = \frac{121 \pi}{1875} \cdot 3h^2 \cdot \frac{dh}{dt}
\]

Simplifying this gives:

\[
\frac{dV}{dt} = \frac{363 \pi}{1875} h^2 \cdot \frac{dh}{dt}
\]

### Step 6: Substitute the known values
We know \( \frac{dV}{dt} = 10 \) ft³/min and \( h = 14 \) ft:

\[
10 = \frac{363 \pi}{1875} \cdot 14^2 \cdot \frac{dh}{dt}
\]

### Step 7: Solve for \( \frac{dh}{dt} \)
Calculate \( 14^2 = 196 \):

\[
10 = \frac{363 \pi}{1875} \cdot 196 \cdot \frac{dh}{dt}
\]

Simplifying further:

\[
10 = \frac{71268 \pi}{1875} \cdot \frac{dh}{dt}
\]

\[
\frac{dh}{dt} = \frac{10 \cdot 1875}{71268 \pi}
\]

### Step 8: Simplify the expression
Let's calculate this:

\[
\frac{dh}{dt} = \frac{18750}{71268 \pi} \approx 0.0836 \, \text{ft/min}
\]

### Final Answer
The depth of the water is increasing at a rate of approximately **0.0836 ft/min** when the water is 14 feet deep.

---

Would you like a deeper explanation or have any questions?

### Related Questions
1. How do you derive the formula for the volume of a cone?
2. What is the significance of using similar triangles in related rates problems?
3. How does changing the rate of water flow affect \( \frac{dh}{dt} \)?
4. Can you solve this problem using a different method, such as implicit differentiation?
5. How does the shape of the tank (cylindrical vs. conical) change the solution process?

### Tip
When solving related rates problems, always draw a diagram and write out known quantities and relationships before starting calculations. This helps to visualize the problem and organize the necessary formulas.

A conical water tank with vertex down has a radius of 11 feet at the top and is 25 feet high. If water flows into the tank at a rate of 10 ft³/min, how fast is the depth of the water increasing when the water is 14 feet deep?

Solve a related rates problem involving a conical water tank with a 25-foot height and 11-foot radius. Learn how fast the water depth is increasing when water flows in at 10 ft³/min and the depth reaches 14 feet. This solution uses calculus, chain rule, and geometry of cones.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation