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When we're dealing with right triangles, the Pythagorean theorem is an essential tool to know about. This theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. The formula is written as:

• \( a^2 + b^2 = c^2 \)

In the context of our boat problem, the hypotenuse \( s \) is the length of the rope, while \( x \), the distance from the boat to the dock, and the 5-foot height, make up the other two sides.
By applying the Pythagorean theorem, we were able to relate all these sides into a manageable equation. With \( s = 13 \) and the fixed height of 5 ft, we derived \( x^2 + 5^2 = 13^2 \).
This formula provides the backbone for finding how the distances change as the rope is pulled and helps us set the stage for solving related rates problems.

Differentiation is a core concept in calculus, used to find how a quantity changes concerning another. When we differentiate an equation, we essentially find the rate of change, which is precisely what we need for solving related rates problems.

• In this problem, we have the equation: \( x^2 + 25 = s^2 \).
• To examine how the relationship between the rope's length \( s \) and the distance \( x \) changes with time, we differentiate both sides of this equation concerning time \( t \).

The differentiation process involves applying the chain rule, resulting in the equation:

• \( 2x \frac{dx}{dt} = 2s \frac{ds}{dt} \).

Here, \( \frac{dx}{dt} \) and \( \frac{ds}{dt} \) represent how fast the boat approaches the dock and how fast the rope is pulled, respectively. This step allows us to express the change in horizontal distance \( x \) using the known change in rope length \( \frac{ds}{dt} \). Hence, differentiation provides the key operating principle in such calculus-related problems.

Calculus problem solving involves a systematic approach that often uses differentiation and the Pythagorean theorem to relate different quantities. In a related rates problem, like the boat and rope scenario, these tools help us quantify how one rate affects another.

• Firstly, we set up a clear visual representation of the problem. For our problem, we imagined a right triangle formed by the boat, dock, and rope.
• Secondly, we established a relevant equation: the Pythagorean theorem. This equation described how the different parts of the triangle related to each other.
• The third step was to differentiate this equation with respect to time, already knowing one rate of change: the rate at which the rope length \( s \) decreases.

After substitution with known values (\( s = 13 \), \( \frac{ds}{dt} = -0.6 \), and \( x = 12 \)), we solved for \( \frac{dx}{dt} \).
Finally, interpreting the result, we recognized that the boat approached the dock at 0.65 ft/sec.
Thus, problem-solving in calculus doesn’t just solve mathematical puzzles but offers profound insights into how systems behave over time, making it incredibly useful in the real world.