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The problem in the image involves a boat being pulled toward a dock by a rope. The given information indicates:

- The rope is being pulled in at a rate of 4 feet per second.
- The length of the rope is the hypotenuse of a right triangle formed by the boat, the dock, and the length of the rope.
- The vertical distance (height) from the dock to where the rope is tied is constant at 5 feet.

We are asked to find how fast the distance between the boat and the dock is decreasing when the boat is 3 feet away from the dock.

### Approach:

We will use related rates to solve this problem by applying Pythagoras' Theorem. Let:
- \( x \) be the horizontal distance between the boat and the dock.
- \( y \) be the length of the rope.
- \( \frac{dy}{dt} \) is given as the rate at which the rope is being pulled in, which is \( -4 \, \text{ft/sec} \) (negative since the length of the rope is decreasing).

By the Pythagorean Theorem:
\[
x^2 + 5^2 = y^2
\]
\[
x^2 + 25 = y^2
\]

Differentiate both sides with respect to \( t \) (time):
\[
2x \frac{dx}{dt} = 2y \frac{dy}{dt}
\]
Simplify:
\[
x \frac{dx}{dt} = y \frac{dy}{dt}
\]
Now, solve for \( \frac{dx}{dt} \) (the rate at which the distance between the boat and the dock is decreasing):
\[
\frac{dx}{dt} = \frac{y \frac{dy}{dt}}{x}
\]

### Step 1: Find \( y \) when \( x = 3 \)
When the boat is 3 feet from the dock:
\[
x^2 + 25 = y^2
\]
\[
3^2 + 25 = y^2
\]
\[
9 + 25 = y^2
\]
\[
y^2 = 34 \implies y = \sqrt{34}
\]

### Step 2: Substitute values into the equation for \( \frac{dx}{dt} \)
\[
\frac{dx}{dt} = \frac{\sqrt{34} \times (-4)}{3}
\]
Now, let's calculate the value. I'll do that next.The rate at which the distance between the boat and the dock is decreasing when the boat is 3 feet from the dock is approximately **7.775 feet per second** (rounded to the nearest thousandth).

Would you like more details or further clarification?

Here are five related questions you might find useful:
1. How would the rate change if the rope were pulled in faster, say at 6 feet per second?
2. How can we determine the rate of change at different distances from the dock?
3. What happens to the rate of change as the boat gets closer to the dock?
4. How would the problem change if the height from the dock to the boat were different?
5. Can this problem be solved using any other mathematical method?

**Tip:** In related rates problems, always ensure to differentiate both sides of the equation with respect to time carefully.

A boat is being pulled toward a dock by a rope at a rate of 4 feet per second. How fast is the distance between the dock and the boat decreasing when the boat is 3 feet from the dock?

Learn how to solve a related rates problem involving a boat being pulled toward a dock. The solution includes applying the Pythagorean Theorem and differentiating with respect to time to find the rate of change when the boat is 3 feet from the dock.

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