Problem 106 - A Graphical Approach to College Algebra

Question

Solve each problem. Two ships leave the same harbor at the same time. The first ship heads north at 20 mph and the second ship heads west at 15 mph. (a) Draw a sketch depicting their positions after t hours. (b) Write an expression that gives the distance between the ships after t hours.

Step-by-Step Solution

Answer

The distance between the ships is \( 25t \) miles after \( t \) hours.

Step 1: Sketch the Situation

Visualize the scenario with a simple coordinate system, with the harbor as the origin point. After \( t \) hours, the first ship traveling north at 20 mph would be \( 20t \) miles north of the harbor. The second ship traveling west at 15 mph would be \( 15t \) miles west of the harbor. The paths of the ships form a right triangle with sides \( 20t \) and \( 15t \).

Step 2: Identify the Right Triangle

Recognize the paths of the ships as the legs of a right triangle. The distance between the ships is the hypotenuse of this triangle. The legs of the triangle are \( 20t \) (northward distance) and \( 15t \) (westward distance).

Step 3: Apply the Pythagorean Theorem

Use the Pythagorean theorem to find the hypotenuse, which represents the distance between the ships, described by the equation \( c = \sqrt{a^2 + b^2} \). Here, \( a = 20t \) and \( b = 15t \).

Step 4: Calculate the Expression for Distance

Plug the values into the distance formula: \[ c = \sqrt{(20t)^2 + (15t)^2} \]Simplify the expression: \[ c = \sqrt{400t^2 + 225t^2} = \sqrt{625t^2} = 25t \] This expression \( 25t \) gives the distance between the ships after \( t \) hours.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry. It is used to calculate the length of the hypotenuse in a right triangle. 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:

• \( c = \sqrt{a^2 + b^2} \)

Where \( a \) and \( b \) are the triangle's legs, and \( c \) is the hypotenuse. In the exercise above, after \( t \) hours, the ships' paths create a right triangle with the legs being \( 20t \) and \( 15t \) miles. Consequently, the Pythagorean Theorem helps us determine the distance between the ships, which turns out to be \( 25t \) miles.
This showcases how the Pythagorean Theorem can be applied to real-world scenarios involving right triangles.

Coordinate System

A coordinate system helps us visualize and describe the positions of objects. It’s a key tool in geometry and navigation. Here, we utilize a simple two-dimensional coordinate system to illustrate the paths of the two ships.
The harbor is set at the origin, point \((0, 0)\).
The first ship's journey to the north is represented along the vertical y-axis, moving in a positive direction. Conversely, the second ship's westward path is shown along the negative x-axis.

• Northward ship: Position \((0, 20t)\)
• Westward ship: Position \((-15t, 0)\)

Visualizing the problem in this manner allows us to clearly see the horizontal and vertical distances from the harbor, making it easier to apply the Pythagorean Theorem.

Distance Formula

The distance formula, derived from the Pythagorean Theorem, calculates the distance between two points in a coordinate plane. It is particularly useful in determining the straight-line distance between two locations. The standard distance formula is:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

In this exercise, since the ships travel at right angles to each other, the formula simplifies due to

• \( x_1 = 0 \), \( y_1 = 0 \) (origin, harbor)
• \( x_2 = -15t \), \( y_2 = 20t \) (positions of the ships after \( t \) hours)

Therefore, the expression becomes \( d = \sqrt{(0 - (-15t))^2 + (20t - 0)^2} \).
This reduces to \( d = \sqrt{(15t)^2 + (20t)^2} \), which is \( d = 25t \) using simplification. Hence, the distance formula confirms the distance derived using the Pythagorean Theorem.