Question

For relatively small heights above Earth, a simple radical function can be used to approximate the distance to the horizon. a) If Earth's radius is assumed to be \(6378 \mathrm{km},\) determine the equation for the distance, \(d,\) in kilometres, to the horizon for an object that is at a height of \(h\) kilometres above Earth's surface. b) Identify the domain and range of the function. c) How can you use a graph of the function to find the distance to the horizon for a satellite that is \(800 \mathrm{km}\) above Earth's surface? d) If the function from part a) were just an arbitrary mathematical function rather than in this context, would the domain or range be any different? Explain.

Short answer

The equation is \(d \approx \sqrt{2 \times 6378 \times h}\). The domain is \(h \geq 0\) and the range is \(d \geq 0\). For \(h = 800\), calculate \(d \approx 31883 \text{ km}\).

Step-by-step solution

Determine the formula using the Pythagorean theorem

To find the distance to the horizon, use the Pythagorean theorem. The Earth’s radius is given as 6378 km. Let \(h\) be the height above the Earth's surface and \(r = 6378\) km be the radius of the Earth. The distance \(d\) to the horizon forms a right triangle with the Earth’s radius and the Earth’s radius plus the height. The formula is: \[ (r + h)^2 = r^2 + d^2 \]

Solve the equation for \(d\)

Substitute \(r = 6378\) and solve for \(d\). Rearrange the equation to \[ d^2 = (r + h)^2 - r^2 \]Then, \[ d = \sqrt{(r + h)^2 - r^2} \].

Simplify the radical expression

Simplify the expression within the square root: \[ d = \sqrt{6378^2 + 2 \times 6378 \times h + h^2 - 6378^2} \]This simplifies to \[ d = \sqrt{2 \times 6378 \times h + h^2} \]Further simplifying, we can approximate as \[ d \approx \sqrt{2 \times 6378 \times h} \] when \(h\) is relatively small.

Identify the domain and range

In this context, the height \(h\) must be non-negative, resulting in \(h \geq 0\). Since \(d = \sqrt{2 \times 6378 \times h}\), and the square root function is defined for non-negative values, the domain of the function is \[\text{Domain} : h \geq 0 \]For range, since \(d\) starts from 0 and increases as \(h\) increases, \[\text{Range} : d \geq 0\].

Use a graph to find the distance for a satellite

Use the equation \(d \approx \sqrt{2 \times 6378 \times h}\) and substitute \(h = 800\). Plot the function and look for the corresponding \(d\)-value when \(h = 800\). Solving it directly gives: \[ d \approx \sqrt{2 \times 6378 \times 800}\]. Calculate to find \(d\).

Discuss changes in domain and range for an arbitrary function

For an arbitrary function \(d = \sqrt{2 \times k \times h}\) where \(k\) is a constant, the domain would still be \(h \geq 0\) because the square root function requires non-negative input. The range would be \(d \geq 0\) as well, since the output of a square root is non-negative.

Key concepts

Radical Functions

A radical function involves a variable under a radical sign, most commonly a square root. For instance, the distance to the horizon formula, \(d = \sqrt{2 \times 6378 \times h}\), is a radical function because it contains the square root of an expression involving the variable \(h\). These functions often model relationships where the rate of change decreases as the input increases.
In our formula, as the height \(h\) above the Earth's surface increases, the distance \(d\) to the horizon increases, but at a decreasing rate. Other examples of radical functions include those used in physics and engineering, such as escaping velocity or the intensity of light as a function of distance.
Understanding how to work with radical functions is crucial. Always check whether the input (under the radical) is non-negative, as square roots of negative numbers are not real numbers.

Pythagorean Theorem

The Pythagorean Theorem is fundamental in geometry. It 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.
Mathematically, it is written as: \[a^2 + b^2 = c^2\]
In the context of our problem, we used this theorem to find the distance to the horizon. By treating the radius of the Earth, the height above the Earth, and the distance to the horizon as sides of a right triangle, we could form the equation:
\( (r + h)^2 = r^2 + d^2 \).
Solving this resulted in our radical function for \(d\). Understanding how to apply the Pythagorean theorem is vital in many areas of math and physics.

Domain and Range

The domain of a function is the set of all possible input values that the function can accept. The range is the set of all possible output values the function can produce.
For our horizon distance function: \(d = \sqrt{2 \times 6378 \times h}\), the domain is determined by the variable \(h\). Since \(h\) represents the height above the Earth's surface and must be non-negative, the domain is \[h \geq 0\].
The range of our function is influenced by the domain and the nature of the square root. Since the smallest value of \(h\) is zero, and the square root function returns only non-negative results, the smallest value for \(d\) is also zero. Thus, the range is \[d \geq 0\].
Understanding both domain and range is crucial in graphing and in interpreting the behavior and limits of functions.

Simplifying Expressions

Simplifying expressions helps to make calculations easier and more comprehensible. Consider the expression from our horizon distance formula: \(d = \sqrt{(6378 + h)^2 - 6378^2}\).
By simplifying inside the square root, we have:
\[d = \sqrt{6378^2 + 2 \times 6378 \times h + h^2 - 6378^2}\]
Further simplifying this gives:
\[d = \sqrt{2 \times 6378 \times h + h^2}\]
When \(h\) is small relative to \(6378\), the term \(h^2\) becomes negligible, allowing us to approximate:
\[d \approx \sqrt{2 \times 6378 \times h}\]
Learning to simplify expressions, especially those involving radicals, is a key skill in algebra and helps streamline problem-solving in mathematics.

Graphing Functions

Graphing functions provides a visual way to understand relationships between variables. For our distance to the horizon function, \[d = \sqrt{2 \times 6378 \times h}\], graphing helps students see how \(d\) changes as \(h\) increases.
To graph this, we can create a table of values where we calculate \(d\) for different values of \(h\). For example, plot \(h = 0, 100, 200, 300\), and so on.
Graphically, this function will produce a curve that starts at the origin (0,0) and rises more slowly as \(h\) increases because of the square root operation.
If given a specific height, like \(h = 800\), you can directly find the corresponding value of \(d\) on the graph. For \(h = 800\):
\[d \approx \sqrt{2 \times 6378 \times 800}\]
Using a graph provides a clearer and faster way to interpret changes and relationships between the variables.