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An angler hooks a trout and reels in his line at 4 in divided by s4 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​ (forming a right angled triangle). Find the horizontal speed of the fish when it is 21 ft from the angler.

This problem is related to related rates in calculus, where the distances and rates are changing over time.

### Setup:
- We are given a right triangle where the vertical leg is the height from the tip of the fishing rod to the water, which is constant at 13 feet.
- The horizontal distance between the fish and the angler, denoted as \( x(t) \), is changing over time.
- The hypotenuse, the length of the fishing line, denoted as \( L(t) \), is also changing as the angler reels in the line.
- The rate at which the length of the line is decreasing is given as \( \frac{dL}{dt} = -4 \, \text{in/s} \), which needs to be converted to feet per second (since other measurements are in feet).

Given:
- \( \frac{dL}{dt} = -4 \, \text{in/s} = -\frac{1}{3} \, \text{ft/s} \).
- The height (vertical leg) is constant at \( y = 13 \, \text{ft} \).
- We are asked to find the horizontal speed of the fish (rate of change of \( x(t) \), \( \frac{dx}{dt} \)) when the fish is 21 feet horizontally away from the angler.

### Applying Pythagoras:
From the Pythagorean theorem, the relation between the sides of the right triangle is:
\[
L(t)^2 = x(t)^2 + y^2
\]
Since \( y = 13 \, \text{ft} \), we have:
\[
L(t)^2 = x(t)^2 + 13^2
\]
\[
L(t)^2 = x(t)^2 + 169
\]

### Differentiating with respect to time \( t \):
Now, differentiate both sides with respect to \( t \):
\[
2L(t) \frac{dL}{dt} = 2x(t) \frac{dx}{dt}
\]
Simplifying:
\[
L(t) \frac{dL}{dt} = x(t) \frac{dx}{dt}
\]

### Finding \( L(t) \) when \( x(t) = 21 \, \text{ft} \):
Using the Pythagorean relation again:
\[
L(t)^2 = 21^2 + 13^2 = 441 + 169 = 610
\]
\[
L(t) = \sqrt{610} \, \text{ft} \approx 24.7 \, \text{ft}
\]

### Substituting values:
Now substitute \( L(t) = 24.7 \, \text{ft} \), \( \frac{dL}{dt} = -\frac{1}{3} \, \text{ft/s} \), and \( x(t) = 21 \, \text{ft} \) into the differentiated equation:
\[
24.7 \times \left( -\frac{1}{3} \right) = 21 \times \frac{dx}{dt}
\]
\[
\frac{-24.7}{3} = 21 \times \frac{dx}{dt}
\]
\[
-8.23 = 21 \times \frac{dx}{dt}
\]
\[
\frac{dx}{dt} = \frac{-8.23}{21} \approx -0.392 \, \text{ft/s}
\]

### Conclusion:
The horizontal speed of the fish when it is 21 feet from the angler is approximately \( 0.392 \, \text{ft/s} \) towards the angler (the negative sign indicates that the distance is decreasing).

Would you like more details or further clarifications?

Here are 5 related questions to explore this topic further:
1. How would the solution change if the angler reeled in the line at a faster rate?
2. What would happen if the fish were moving away from the angler rather than towards them?
3. How would you calculate the rate at which the hypotenuse is changing when the fish is a different distance from the angler?
4. Could you find the rate at which the angle between the fishing line and the horizontal is changing?
5. How would the horizontal speed change if the height of the rod above the water were different?

**Tip:** Always ensure that units are consistent when solving related rates problems to avoid calculation errors!

An angler hooks a trout and reels in his line at 4 inches per second. Assume the tip of the fishing rod is 13 feet above the water, and the fish is pulled horizontally toward the angler (forming a right-angled triangle). Find the horizontal speed of the fish when it is 21 feet away from the angler.

Solve a related rates problem involving an angler reeling in a fish. Given the height of the fishing rod and the horizontal distance, learn how to calculate the horizontal speed of the fish using the Pythagorean theorem and differentiation. Suitable for students in grades 11-12.

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