91Ó°ÊÓ

The Pythagorean Theorem is a fundamental concept in geometry. It's used to relate the lengths of the sides of a right triangle. When you have a right triangle, the theorem says 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. Mathematically this is written as:\[c^2 = a^2 + b^2\] where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides.In the context of our car problem, the north and east paths of the cars form the two smaller sides of a right triangle. These are 25 miles and 60 miles respectively after 1 hour. The hypotenuse \(D\) represents the direct distance between the two cars, which can be found using the Pythagorean theorem:\[D = \sqrt{25^2 + 60^2} = 65\, \text{miles}\]This tells us that after 1 hour, the cars are 65 miles apart in a straight line.

Differentiation is a powerful tool in calculus used to find the rate at which a quantity changes. In our problem, we use differentiation to determine how quickly the distance between the two cars increases. We start by recognizing the relationship given by the Pythagorean theorem: \(D^2 = x^2 + y^2\). Differentiating both sides with respect to time \(t\) helps us find the rate of change of \(D\) with respect to time.After differentiation, the equation becomes:\[2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}\]This new equation tells us how the rates at which \(x\) and \(y\) change (the speeds of the cars) affect the rate at which \(D\) changes. Essentially, it connects all the moving parts of the problem.

The rate of change is a crucial concept in understanding how fast something is happening over time. In this context, it refers to how quickly the distance between the two cars increases. By using the differentiated formula from the previous section, we can calculate this rate accurately.Substituting the given values \(x = 60\, \text{miles}\), \(y = 25\, \text{miles}\), \(\frac{dx}{dt} = 60\, \text{mph}\), \(\frac{dy}{dt} = 25\, \text{mph}\), and \(D = 65\, \text{miles}\) into our differentiated equation:\[2 \times 65 \times \frac{dD}{dt} = 2 \times 60 \times 60 + 2 \times 25 \times 25\]Simplifying gives us:\[130 \frac{dD}{dt} = 8450\]This leads to:\[\frac{dD}{dt} = \frac{8450}{130} \approx 65\, \text{mph}\]Therefore, the rate of change in the distance between the cars is approximately 65 miles per hour after one hour. This means at that moment, the gap is widening at a speed almost identical to the speed of the faster car.