Problem 66 - Algebra and Trigonometry with Analytic Geometry

Question

An airplane flying at a speed of \(400 \mathrm{mi} / \mathrm{hr}\) flies from a point \(A\) in the direction \(153^{\circ}\) for 1 hour and then flies in the direction \(63^{\circ}\) for 1 hour. (a) In what direction does the plane need to fly in order to get back to point \(A\) ? (b) How long will it take to get back to point \(A\) ?

Step-by-Step Solution

Answer

Fly in the exact opposite direction determined in Step 3; calculate time using displacement magnitude over speed from Step 5.

Step 1: Determine the displacement vectors

For the first hour of flight, the direction is \(153^\circ\), and the plane flies at \(400\) miles per hour. The first displacement vector can be represented in terms of its components:\[\vec{D_1} = 400 (\cos 153^\circ \hat{i} + \sin 153^\circ \hat{j})\]For the second hour of flight, the direction is \(63^\circ\), so the second displacement vector is:\[\vec{D_2} = 400 (\cos 63^\circ \hat{i} + \sin 63^\circ \hat{j})\]

Step 2: Calculate the total displacement

Add the two displacement vectors to obtain the total displacement vector \(\vec{D}\):\[\vec{D} = \vec{D_1} + \vec{D_2}\]Calculate each component:- \(\hat{i}\) component: \(400\cos 153^\circ + 400\cos 63^\circ\)- \(\hat{j}\) component: \(400\sin 153^\circ + 400\sin 63^\circ\)

Step 3: Calculate the reverse direction

The opposite direction the plane must fly back to the origin \(A\) is given by the angle \(\theta\) calculated from:\[\tan \theta = \frac{\text{OPPOSITE axis component}}{\text{ADJACENT axis component}}\]With the components known, compute \(\theta\) using the inverse tangent function and apply adjustments to get the bearing back.

Step 4: Determine the return distance

The total displacement distance is given by the magnitude of the displacement vector \(\vec{D}\):\[|\vec{D}| = \sqrt{(\hat{i}\text{-component})^2 + (\hat{j}\text{-component})^2}\]

Step 5: Calculate time to return

With the distance known, use speed to find the time:\[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\]Since the speed remains \(400\) mi/hr, calculate the time it takes to fly back to point \(A\).

Key Concepts

Displacement Vectors

When understanding problems involving movement and direction, displacement vectors are a critical concept. Essentially, a displacement vector represents a change in position. It is a vector quantity, meaning it has both magnitude and direction. In the context of our airplane problem, two displacement vectors are used to track the plane's journey.

• The first displacement occurs when the plane travels in the direction of 153° for 400 miles, represented by the vector: \( \vec{D_1} = 400 (\cos 153^\circ \hat{i} + \sin 153^\circ \hat{j}) \).
• The second displacement takes place in the direction of 63° for the same distance, given by: \( \vec{D_2} = 400 (\cos 63^\circ \hat{i} + \sin 63^\circ \hat{j}) \).

To find out where the plane ends up and how it needs to return to point A, we need to sum these vectors. This involves adding each of their components together. The result is the total displacement vector \( \vec{D} \), which tells us the final location relative to our starting point.

Trigonometry

Trigonometry is a mathematical method used to relate the angles and sides of triangles, which proves extremely useful for solving problems involving vector analysis.
For instance, to determine the components of a displacement vector, we use trigonometric functions like sine and cosine. Consider these trigonometric relationships:

• Cosine (cos) of an angle determines the adjacent side component (along the x-axis, represented by \(\hat{i} \)).
• Sine ( sin) of an angle determines the opposite side component (along the y-axis, represented by \(\hat{j} \)).

Thus, the components of a vector are derived using: \[ x = V \cos \theta \] \[ y = V \sin \theta \] where \( \theta \) is the direction of the vector. This is exactly how we determine the individual components of the plane's displacement vectors along the x and y axes.
Trigonometry ensures that we accurately decompose vectors into parts that are manageable and mathematically tractable.

Inverse Tangent Function

The inverse tangent function, also known as arctangent, plays a key role in solving problems that require calculating angles from given side lengths. When considering vector addition and determining directions, finding the resulting angle from vector components becomes essential.
Once we have the total displacement vector \(\vec{D}\), which has components along both the x-axis (\(\hat{i}\)) and y-axis (\(\hat{j}\)), we can find the angle the resultant vector makes with a reference axis using:

• \( \tan \theta = \frac{\text{OPPOSITE axis component}}{\text{ADJACENT axis component}} \)

To extract the angle \(\theta\), we use the inverse tangent function: \[ \theta = \tan^{-1}\left(\frac{\hat{j}\text{-component}}{\hat{i}\text{-component}}\right) \]
Once computed, we may need to adjust this angle to get the correct bearing (direction relative to north), ensuring our calculations align with standard navigational directions. This function provides a clear and straightforward approach to finding precise directions, which are crucial for navigation and orientation tasks.