91Ó°ÊÓ

The Pythagorean theorem is a fundamental principle in geometry. It explains the relationship between the sides of a right triangle. This theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. It is expressed with the formula: \[ c^2 = a^2 + b^2 \]In our exercise, this theorem is used to relate the positions of two moving ships. Ship A and Ship B create a right triangle, where the distance between the ships is the hypotenuse \(d\), and \(x\) and \(y\) are the legs of the triangle. Specifically, at any time \(t\):

• \( x = \text{distance ship A traveled east} \)
• \( y = \text{distance ship B traveled north} \)

The Pythagorean relationship then becomes \( d = \sqrt{x^2 + y^2} \). This equation is essential for calculating the distance parameter involved in many related rates problems, where the movement of objects is considered.

Derivatives are a key component in calculus, used to determine how a quantity changes with respect to another variable. In our exercise, we're interested in how the distance \(d\) between the ships changes over time \(t\). A derivative gives us the rate of change, so by differentiating the distance equation \(d = \sqrt{x^2 + y^2}\) with respect to time, we can find the rate at which this distance is changing: \[\frac{dd}{dt} = \frac{1}{2}(x^2 + y^2)^{-1/2}(2x\frac{dx}{dt} + 2y\frac{dy}{dt})\]Upon simplification, this reduces to:\[\frac{dd}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}}\]This formula now relates the rates of change of \(x\) and \(y\) with respect to\(t\) (\(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)), to the rate of change of the distance \(d\). Calculus makes it possible to solve such dynamic problems where variables are interdependent.

Differentiation is the process of finding a derivative, which provides critical insight into the changing behavior of functions. In our problem context, it involves differentiating the distance formula with respect to time \(t\) to discover how quickly this distance changes.By applying differentiation to the formula \(d = \sqrt{x^2 + y^2}\), we follow several steps:

• Apply the chain rule, which is a crucial differentiation technique, to handle composite functions like \(\sqrt{x^2 + y^2}\).
• Use the power rule to differentiate powers of \(x\) and \(y\).
• Apply the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to account for the dynamic nature of the ships' movement.

By substituting known velocities into these differentials, such as those given for ships A and B, you can calculate the exact rate of distance change at a specific moment (e.g., at 4:00 PM). Differentiation thus enables us to handle the complexity of real-world motion in mathematical form.

The distance formula comes in handy when you need to calculate the distance between two points in a plane. This formula is an extension of the Pythagorean theorem, molded for practical applications. When considering two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is calculated as:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In the case of our exercise, the ships’ changing positions determine the coordinates you use. By setting up these coordinates for Ship A and Ship B at any given time \(t\), you can apply:

• For Ship A: \((x, 0)\), where \(x = -150 + 35t\) km
• For Ship B: \((0, y)\), where \(y = 25t\) km

This configuration simplifies into the specific problem's distance equation \(d = \sqrt{x^2 + y^2}\). Understanding the fundamental distance formula bridges the gap between static spatial problems and dynamic scenarios like our moving ships.