An angler hooks a trout and reels in his line at
3 in divided by s3 in/s.
Assume the tip of the fishing rod is
13 ft13 ft
above the water and directly above the​ angler, and the fish is pulled horizontally directly toward the angler​ (see figure). Find the horizontal speed of the fish when it is
15 ft15 ft
from the angler. answer in in/s

Let's define the key variables and solve the problem step by step.

### Problem Setup:

- The height of the fishing rod tip above the water is \( h = 13 \, \text{ft} \).
- The horizontal distance of the fish from the angler is \( x = 15 \, \text{ft} \) at the point of interest.
- The angler reels in the line at a speed of \( \frac{3}{\sqrt{3}} \, \text{in/s} \).

### Step 1: Identify the relationship between variables

The situation involves a right triangle where:
- \( h = 13 \, \text{ft} \) is the vertical distance (constant).
- \( x \) is the horizontal distance (variable).
- The length of the fishing line \( L \) is the hypotenuse.

Using the Pythagorean theorem:
\[
L^2 = h^2 + x^2
\]
Substitute \( h = 13 \, \text{ft} \):
\[
L^2 = 13^2 + x^2 = 169 + x^2
\]

### Step 2: Differentiate with respect to time

To find the rate of change of \( x \) (horizontal speed of the fish), we differentiate the equation \( L^2 = 169 + x^2 \) with respect to time \( t \):

\[
2L \frac{dL}{dt} = 2x \frac{dx}{dt}
\]

Where:
- \( \frac{dL}{dt} \) is the rate at which the length of the fishing line changes, which is the speed at which the angler reels in the line (given as \( \frac{3}{\sqrt{3}} \, \text{in/s} \)).
- \( \frac{dx}{dt} \) is the horizontal speed of the fish that we need to find.

Simplifying:
\[
L \frac{dL}{dt} = x \frac{dx}{dt}
\]
Solve for \( \frac{dx}{dt} \):
\[
\frac{dx}{dt} = \frac{L \frac{dL}{dt}}{x}
\]

### Step 3: Convert units

First, we need to convert the given quantities to consistent units. Since \( \frac{dL}{dt} \) is given in inches per second, convert the distances from feet to inches:
- \( h = 13 \, \text{ft} = 156 \, \text{in} \)
- \( x = 15 \, \text{ft} = 180 \, \text{in} \)

### Step 4: Find \( L \) when \( x = 180 \, \text{in} \)

Using the Pythagorean theorem to find \( L \) when \( x = 180 \, \text{in} \):

\[
L = \sqrt{156^2 + 180^2} = \sqrt{24336 + 32400} = \sqrt{56736} \approx 238.16 \, \text{in}
\]

### Step 5: Calculate the horizontal speed \( \frac{dx}{dt} \)

Substitute \( L = 238.16 \, \text{in} \), \( x = 180 \, \text{in} \), and \( \frac{dL}{dt} = \frac{3}{\sqrt{3}} \, \text{in/s} \) into the equation:

\[
\frac{dx}{dt} = \frac{238.16 \times \frac{3}{\sqrt{3}}}{180} 
\]

Now, calculate this expression:

\[
\frac{dx}{dt} = \frac{238.16 \times 1.732}{180} \approx \frac{412.78}{180} \approx 2.29 \, \text{in/s}
\]

### Final Answer:
The horizontal speed of the fish is approximately \( 2.29 \, \text{in/s} \).

Do you want any further details or clarifications on the steps? 

Here are 5 related questions to think about:
1. What happens to the speed as the fish gets closer to the angler?
2. How would the problem change if the rod were higher above the water?
3. What if the fish moved diagonally rather than horizontally?
4. How would the speed change if the angler reels in the line faster?
5. What if the fish swims away instead of towards the angler?

**Tip:** Always keep track of units, especially when working with different measurement systems (like feet and inches).

An angler hooks a trout and reels in his line at 3 in/s. Assume the tip of the fishing rod is 13 ft above the water and directly above the angler, and the fish is pulled horizontally directly toward the angler. Find the horizontal speed of the fish when it is 15 ft from the angler. Answer in in/s.

Learn how to determine the horizontal speed of a fish when an angler reels in the line at 3 in/s, with the fishing rod 13 ft above the water. This problem involves related rates and the Pythagorean theorem, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation