91Ó°ÊÓ

The latitude and longitude of a point \(P\) in the Northern Hemisphere are related to spherical coordinates \(\rho, \theta, \phi\) as follows. We take the origin to be the center of the earth and the positive \(z\) -axis to pass through the North Pole. The positive \(x\) -axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of \(P\) is \(\alpha=90^{\circ}-\phi^{\circ}\) and the longitude is \(\beta=360^{\circ}-\theta^{\circ} .\) Find the great-circle distance from Los Angeles (lat. \(34.06^{\circ} \mathrm{N},\) long. \(118.25^{\circ} \mathrm{W}\) ) to Montreal (lat. \(45.50^{\circ} \mathrm{N},\) long. \(73.60^{\circ} \mathrm{W} ) .\) Take the radius of the earth to be 3960 \(\mathrm{mi} .\) (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.)

Step by step solution

01

Convert coordinates to spherical coordinates

Start by converting the latitude and longitude of both locations into spherical coordinates. For Los Angeles and Montreal, calculate \( \phi \) and \( \theta \) using \( \phi = 90^{\circ} - \text{latitude} \) and \( \theta = 360^{\circ} - (\text{longitude} + 180^{\circ}) \) for western longitudes.\- Los Angeles: \( \phi_{LA} = 90^{\circ} - 34.06^{\circ} = 55.94^{\circ}, \theta_{LA} = 360^{\circ} - (118.25^{\circ} + 180^{\circ}) = 61.75^{\circ} \)\- Montreal: \( \phi_{M} = 90^{\circ} - 45.50^{\circ} = 44.50^{\circ}, \theta_{M} = 360^{\circ} - (73.60^{\circ} + 180^{\circ}) = 106.4^{\circ} \)

02

Convert spherical coordinates to Cartesian coordinates

Next, convert the spherical coordinates into Cartesian coordinates using the formulas: \[ x = \rho \sin(\phi) \cos(\theta), \quad y = \rho \sin(\phi) \sin(\theta), \quad z = \rho \cos(\phi) \]. Taking \( \rho = 3960 \text{ mi}\) for the Earth's radius:\( \begin{align*} x_{LA} &= 3960 \cdot \sin(55.94^{\circ}) \cdot \cos(61.75^{\circ}), \ y_{LA} &= 3960 \cdot \sin(55.94^{\circ}) \cdot \sin(61.75^{\circ}), \ z_{LA} &= 3960 \cdot \cos(55.94^{\circ}) \end{align*} \)\( \begin{align*} x_{M} &= 3960 \cdot \sin(44.50^{\circ}) \cdot \cos(106.4^{\circ}), \ y_{M} &= 3960 \cdot \sin(44.50^{\circ}) \cdot \sin(106.4^{\circ}), \ z_{M} &= 3960 \cdot \cos(44.50^{\circ}) \end{align*} \)

03

Calculate the dot product of the position vectors

Calculate the dot product of the position vectors \( \mathbf{v_{LA}} \) and \( \mathbf{v_{M}} \). The dot product \( \mathbf{v_{LA}} \cdot \mathbf{v_{M}} \) is given by: \[ \mathbf{v_{LA}} \cdot \mathbf{v_{M}} = x_{LA} \cdot x_{M} + y_{LA} \cdot y_{M} + z_{LA} \cdot z_{M} \]. Compute this value using the Cartesian coordinates of both locations.

04

Calculate the angle between the vectors

The dot product formula can also be expressed as: \[ \mathbf{v_{LA}} \cdot \mathbf{v_{M}} = \|\mathbf{v_{LA}}\| \|\mathbf{v_{M}}\| \cos(\gamma) \] where \( \gamma \) is the central angle between the two points. Since both \( \|\mathbf{v_{LA}}\| \) and \( \|\mathbf{v_{M}}\| \) equal the Earth's radius (3960 mi), solve for \( \cos(\gamma) \) and hence \( \gamma \): \[ \cos(\gamma) = \frac{\mathbf{v_{LA}} \cdot \mathbf{v_{M}}}{3960^2} \]. Use the inverse cosine to find \( \gamma \).

05

Calculate the great-circle distance

The great-circle distance \( d \) is given by \[ d = 3960 \cdot \gamma \] where \( \gamma \) is in radians. Convert the angle from degrees to radians using \( 1^{\circ} = \frac{\pi}{180} \) radians and solve for \( d \). This gives us the distance between Los Angeles and Montreal along the Earth's surface.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Great-circle Distance

In geography, the great-circle distance is one of the most fundamental concepts related to spherical geometry. A great circle is the largest possible circle that can be drawn on a sphere. This is because it is defined as the intersection of the sphere with a plane that passes through the center of the sphere.
For the Earth, this means the great circle is the shortest path between two points on the surface. Calculating the great-circle distance involves determining the central angle between those two points and converting that to a physical distance across the Earth's surface.

• To find the great-circle distance between two cities, like Los Angeles and Montreal, we first need to determine the angular distance between them.
• This requires calculating the central angle between their spherical coordinates.
• The formula used is: \[ d = ho \cdot \gamma \] where - \( d \) is the great-circle distance, - \( \rho \) is the Earth's radius (3960 miles in this case), and - \( \gamma \) is the central angle in radians.

By understanding the concept of a great circle, we can efficiently navigate the shortest distance between two points on our planet, saving both time and resources.

Latitude and Longitude Conversion

Latitude and longitude are the coordinates used to define any location on Earth. However, for computations involving distances on a sphere, like finding the great-circle distance, these coordinates must be converted into angles used in spherical coordinates.
The equations for conversion are derived from spherical coordinate geometry. More specifically:

• The latitude (\( \alpha \)) is converted to an angle (\( \phi \)) using the formula: \[ \phi = 90^\circ - \alpha \]
• For the longitude (\( \beta \)), \[ \theta = 360^\circ - \beta \] needs adjustments for western longitudes by adding 180° to account for the full 360° rotation.

With these conversions, any location on Earth's surface can be related to spherical coordinates, thereby allowing more complex calculations like those needed for great-circle distances.

Cartesian Coordinates

After expressing locations in spherical coordinates, the next step is to convert these into Cartesian coordinates to facilitate easier calculation of distances and vectors. This involves using the radius of the sphere (Earth in this context) and the two angles previously derived. The three-dimensional Cartesian coordinate system uses x, y, and z to define positions in space.
The conversion formulas are:

• \( x = \rho \sin(\phi) \cos(\theta) \)
• \( y = \rho \sin(\phi) \sin(\theta) \)
• \( z = \rho \cos(\phi) \)

Here, \( \rho \) is the radius of the Earth, \( \phi \) is the angle from the North Pole (converted from latitude), and \( \theta \) is the angle around the equator (adjusted longitude). This transformation allows for the calculation of vector properties like the dot product.

Dot Product Calculation

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and calculates a single number. It's used to derive the cosine of the angle between two vectors, which is essential in finding the great-circle distance.
For vectors \( \mathbf{v_{LA}} \) for Los Angeles and \( \mathbf{v_{M}} \) for Montreal, the dot product is calculated as:

• \( \mathbf{v_{LA}} \cdot \mathbf{v_{M}} = x_{LA} \cdot x_{M} + y_{LA} \cdot y_{M} + z_{LA} \cdot z_{M} \)

This value is then used to find the cosine of the angle \( \gamma \) between the vectors. The relationship is given by: \[ \cos(\gamma) = \frac{\mathbf{v_{LA}} \cdot \mathbf{v_{M}}}{\rho^2} \] where \( \rho \) is the radius of the Earth. Solving for \( \gamma \) using the inverse cosine enables the determination of the great-circle distance when multiplied by \( \rho \), completing the calculation.