Problem 46 - Contemporary Precalculus

Question

A plane flies in a direction of \(105^{\circ}\) from airport \(A\). After a time, it turns and proceeds in a direction of \(267^{\circ} .\) Finally, it lands at airport \(B, 120\) miles directly south of airport \(A\) How far has the plane traveled? [ Note: Aerial navigation directions are explained in Exercise \(41 \text { of Section } 8.2 .]\)

Step-by-Step Solution

Answer

Answer: To find the total distance traveled by the plane, follow the steps outlined in the solution, solving the system of equations for \(t\) and \(s\). Then, calculate the sum of the distances traveled during both legs of the plane's journey to find the total distance.

Step 1: Represent the initial and final positions as points

Let airport A be represented by point A and airport B by point B. Since the plane lands directly south of airport A, then airport B can be represented as the point B(0, -120).

Step 2: Find the vectors representing the plane's travel

The first leg of the plane's journey has a heading of \(105^{\circ}\) and the second leg has a heading of \(267^{\circ}\). We can represent these headings as vectors: First leg: \(u = (\cos105^\circ, \sin105^\circ)\) Second leg: \(v = (\cos267^\circ, \sin267^\circ)\)

Step 3: Define scalar multiples of the vectors

Let \(t\) be the distance traveled in the first leg and \(s\) be the distance traveled in the second leg. Then the scalar multiples of the vectors \(u\) and \(v\) can be represented as follows: First leg: \(t \cdot u = (t \cos105^\circ, t \sin105^\circ)\) Second leg: \(s \cdot v = (s \cos267^\circ, s \sin267^\circ)\)

Step 4: Write the equation for the plane's final position

The plane's final position can be represented as the sum of its initial position and the scalar multiples of the vectors \(u\) and \(v\). That is: \(A + t \cdot u + s \cdot v = B\) Substitute the coordinates of points A and B: \((0, 0) + (t \cos105^\circ, t \sin105^\circ) + (s \cos267^\circ, s \sin267^\circ) = (0, -120)\) Simplify the equation: \((t \cos105^\circ + s \cos267^\circ, t \sin105^\circ + s \sin267^\circ) = (0, -120)\)

Step 5: Solve for \(t\) and \(s\) using system of equations

Now we have a system of two equations, which can be solved for \(t\) and \(s\): \(t \cos105^\circ + s \cos267^\circ = 0\) \(t \sin105^\circ + s \sin267^\circ = -120\) Solve the system of equations for \(t\) and \(s\) to find the distances traveled during both legs of the plane's journey.

Step 6: Find the total distance traveled

Once you have the values for \(t\) and \(s\), add them to find the total distance traveled by the plane: Total distance traveled = \(t + s\)

Key Concepts

Vectors and Their Role

Vectors are essential in both mathematics and physics as they represent quantities that have both magnitude and direction. When dealing with problems that involve navigation, such as the journey of a plane, vectors help in defining the path in terms of horizontal and vertical components.
In this exercise, two vectors represent the two legs of the plane's flight. The first vector, calculated using the direction of \(105^\circ\), and the second vector, with a direction of \(267^\circ\), allow us to describe how far and in which direction the plane travels during each segment.

• Vectors are denoted in component form, like \((a, b)\), where \(a\) and \(b\) are the direction's cosine and sine respective components.
• They provide a way to decompose complex movements into manageable calculations.
• These vectors are combined to track and predict the movement of objects, such as planes in navigation scenarios.

Solving Systems of Equations

Systems of equations are sets of equations with multiple variables that can be solved simultaneously. They are a critical tool in determining values that satisfy multiple conditions.
In this exercise, the system of equations arises from setting up the vector equations for the plane's journey:

• Equation 1: \( t \cos 105^{\circ} + s \cos 267^{\circ} = 0 \)
• Equation 2: \( t \sin 105^{\circ} + s \sin 267^{\circ} = -120 \)

The purpose of these equations is to find the exact distances \(t\) and \(s\) that meet the plane's path requirements. This system reflects the conditions required to transition from the initial point \(A\) to the final destination point \(B\).
Once these equations are solved, you can determine how far the plane flew during each leg of the flight.

Understanding Navigation

Navigation, especially in an aerial context, heavily relies on trigonometry and vector analysis to predict and define paths. When a plane flies, it must often travel at set angles relative to fixed directions (e.g., North) to reach its destination efficiently.
In this problem,

• The plane initially heads in a direction of \(105^\circ\) from the North, and later turns to head \(267^\circ\).
• These angles help determine the precise vectors needed to calculate the plane's trajectory and ultimate landing point.

Navigation calculations are pivotal in aviation as they ensure safety and accuracy. By breaking down flight paths via vectors and trigonometry, calculations assure that planes can travel over long distances accurately!

Geometry in Navigation

Geometry forms the basis of all navigational calculations, as it allows us to visually represent and solve practical problems relating to spatial relationships. In navigation exercises like this, the challenge is often to visualize paths and bearings in connection with coordinates on a plane.

• The direct south path between airports \(A\) and \(B\) offers a simple yet effective way to understand direction and distance.
• The problem uses the right triangle formed by the coordinates of point \(B\) \((0, -120)\) and the origin \((0, 0)\).

Geometry also melds with algebra, as equations originate from geometric interpretations of the problem. Tools from geometry, such as angles and distances, form the blueprint for understanding how navigation operates within real-life scenarios such as aviation.