Problem 89 - Algebra and Trigonometry

Question

At 10: 15 A.M., a radar station detects an aircraft at a point 80 miles away and 25 degrees north of due east. At 10: 25 A.M., the aircraft is 110 miles away and 5 degrees south of due east. (a) Using the radar station as the pole and due east as the polar axis, write the two locations of the aircraft in polar coordinates. (b) Write the two locations of the aircraft in rectangular coordinates. Round answers to two decimal places. (c) What is the speed of the aircraft in miles per hour? Round the answer to one decimal place.

Step-by-Step Solution

Answer

The speed of the aircraft is 342.5 mph.

Step 1: Interpretation of Polar Coordinates for First Instance

Given that the aircraft is 80 miles away and 25 degrees north of due east at 10:15 A.M., the polar coordinates of this point will be \( (r,\theta) = (80, 25^\text{°}) \).

Step 2: Interpretation of Polar Coordinates for Second Instance

Given that the aircraft is 110 miles away and 5 degrees south of due east at 10:25 A.M., the polar coordinates of this point will be \( (r,\theta) = (110, -5^\text{°}) \).

Step 3: Converting Polar Coordinates to Rectangular Coordinates (First Position)

To convert polar coordinates \( (r,\theta) = (80, 25^\text{°}) \) to rectangular coordinates, use the formulas \( x = r \times \text{cos}(\theta) \) and \( y = r \times \text{sin}(\theta) \).\( x_1 = 80 \times \text{cos}(25^\text{°}) \)\( y_1 = 80 \times \text{sin}(25^\text{°}) \). After calculation, we get \( x_1 = 80 \times 0.9063 \) and \( y_1 = 80 \times 0.4226 \). Therefore, \( (x_1, y_1) = (72.50, 33.81) \)

Step 4: Converting Polar Coordinates to Rectangular Coordinates (Second Position)

To convert polar coordinates \( (r,\theta) = (110, -5^\text{°}) \) to rectangular coordinates, use the formulas \( x = r \times \text{cos}(\theta) \) and \( y = r \times \text{sin}(\theta) \).\( x_2 = 110 \times \text{cos}(-5^\text{°}) \)\( y_2 = 110 \times \text{sin}(-5^\text{°}) \). After calculation, we get \( x_2 = 110 \times 0.9962 \) and \( y_2 = 110 \times -0.0872 \). Therefore, \( (x_2, y_2) = (109.58, -9.59) \)

Step 5: Calculate Distance Traveled

Calculate the distance traveled using the Euclidean distance formula \(\text{distance} = \text{√}((x_2 - x_1)^2 + (y_2 - y_1)^2) \).Substitute the coordinates: \(\text{distance} = \text{√}((109.58 - 72.50)^2 + (-9.59 - 33.81)^2) \).This gives \(\text{distance} = \text{√}((37.08)^2 + (-43.40)^2) \).After calculation, \(\text{distance} = \text{√}(1374.26 + 1882.76) = \text{√}(3257.02) = 57.08 \text{ miles}\).

Step 6: Calculate Speed of Aircraft

Speed is distance divided by time. The time elapsed is 10 minutes, which is \( \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours}\). Therefore, the speed of the aircraft is \(\text{Speed} = \frac{57.08 \text{ miles}}{\frac{1}{6} \text{ hours}} = 57.08 \times 6 = 342.5 \text{ miles per hour}\).

Key Concepts

Polar Coordinates

Polar coordinates represent a point in the plane using two values: the radius, or distance from a central point (the pole), and the angle from a reference direction (often due east). The polar coordinate system can simplify the understanding of systems with rotational symmetry.

For example, in the given exercise, the aircraft's position is first noted 80 miles from the radar station and 25 degrees north of due east. Its coordinate in the polar system is written as (80, 25°). The second position is given as 110 miles away and 5 degrees south of due east, so its polar coordinate is (110, -5°).

Rectangular Coordinates

Rectangular or Cartesian coordinates designate a point using horizontal (x) and vertical (y) distances from a fixed reference point (origin). Converting from polar to rectangular coordinates is key to solving many geometric problems.

To convert polar coordinates (r, θ) to rectangular, we use the formulas:
\[ x = r \times \text{cos}(θ) \]
\[ y = r \times \text{sin}(θ) \]

In the problem's first instance, the aircraft's position is converted as follows:
\[ x_1 = 80 \times \text{cos}(25°) \approx 72.50 \]
\[ y_1 = 80 \times \text{sin}(25°) \approx 33.81 \]
So, the rectangular coordinates are (72.50, 33.81).

For the second position:
\[ x_2 = 110 \times \text{cos}(-5°) \approx 109.58 \]
\[ y_2 = 110 \times \text{sin}(-5°) \approx -9.59 \]
Thus, the rectangular coordinates are (109.58, -9.59).

Euclidean Distance

Euclidean distance measures the straight-line distance between two points in a plane. It's derived from the Pythagorean theorem and is crucial for determining the movement between positions.

With the aircraft's rectangular coordinates calculated, we can find the distance traveled using:
\[ \text{distance} = \text{√}((x_2 - x_1)^2 + (y_2 - y_1)^2) \]

Substituting in our values:
\[ \text{distance} = \text{√}((109.58 - 72.50)^2 + (-9.59 - 33.81)^2) \approx \text {√}(3257.02) \approx 57.08 \text{ miles} \]

This formula helps validate the distance accurately.

Coordinate Conversion

Converting between polar and rectangular coordinates is an essential skill in trigonometry and coordinate geometry. The conversion aids in translating between two common systems of describing position.

For polar to rectangular conversion, we use:
\[ x = r \times \text{cos}(θ) \]
\[ y = r \times \text{sin}(θ) \]

In contrast, converting rectangular coordinates back to polar coordinates involves:
\[ r = \text{√}(x^2 + y^2) \]
\[ θ = \text{tan}^{-1}(y/x) \]

Understanding these transformations ensures easier manipulation of points across different coordinate systems.

Speed Calculation

Speed quantifies how quickly an object moves from one place to another. It's calculated as distance divided by time. In the exercise, the aircraft's speed is calculated over a 10-minute interval.

First, we find the time in hours:
\[ \text{time} = \frac{10 \text{ minutes}}{60 \text{ minutes/hour}} = \frac{1}{6} \text{ hours} \]

Next, we use the distance traveled (57.08 miles) and the time to find speed:
\[ \text{Speed} = \frac{57.08 \text{ miles}}{\frac{1}{6} \text{ hours}} = 57.08 \times 6 = 342.5 \text{ miles per hour} \]

This shows the aircraft's speed in MI/h, rounding to one decimal place.