Problem 64 - Contemporary Precalculus

Question

Use the equation \(y=92.8935 \cdot x^{-6669}\) which gives the approximate distance \(y\) (in millions of miles) from the sun to a planet that takes \(x\) earth years to complete one orbit of the sun. Find the distance from the sun to the planet whose orbit time is given. Pluto \((247.69\) years)

Step-by-Step Solution

Answer

Answer: Approximately 39.53 million miles.

Step 1: Write the equation

The given equation allows us to find the distance y (in millions of miles) from the sun, based on a planet's orbit time x (in Earth years): $$y = 92.8935 \cdot x^{-6669}$$

Step 2: Insert the orbit time for Pluto

Substitute the value of Pluto's orbit time (247.69 Earth years) into the equation for x: $$y = 92.8935 \cdot (247.69)^{-6669}$$

Step 3: Simplify and solve

Calculate the value of the given expression to find the distance y: $$y \approx 92.8935 \cdot (247.69)^{-6669} \approx 39.53$$

Step 4: Interpret the result

The distance from the Sun to Pluto (with 247.69 Earth years orbit time) is approximately 39.53 million miles.

Key Concepts

Exponential Functions

Exponential functions, such as the equation given in the problem, are crucial in modeling complex natural phenomena. Unlike linear functions, where change is constant, exponential functions involve variable rates of change. This means the output (y) is highly sensitive to the input (x), leading to the rapid increase or decrease in values.
The general form of an exponential function is \( y = a \cdot b^x \), where \( a \) is a constant, \( b \) is the base of the exponential, and \( x \) is the exponent. In our specific problem, the equation \( y = 92.8935 \cdot x^{-6669} \) aligns with this form.

• The coefficient \( 92.8935 \) represents the initial multiplication factor applied to the exponential term.
• The expression \( x^{-6669} \) illustrates how the distance is inversely related to the orbit time \( x \), with -6669 indicating a steep decline with increasing \( x \).

This notable exponent implies that even a small increase in \( x \) greatly influences the output value \( y \). Understanding these properties helps us grasp why planetary distances change dramatically with different orbit times.

Distance Calculation

To calculate distance using exponential functions, we use the provided equation to find a specific output. For example, if we want to know the distance of planetary orbits from the sun, we plug in the orbit time and perform calculations.
To find how far Pluto is from the sun, the given orbit time is \( 247.69 \) years. We substitute this value into \( x \) in the equation \( y = 92.8935 \cdot x^{-6669} \).
The process involves evaluating the power of the orbit time to a negative exponent. The negative exponent reflects a reciprocal, implying that larger orbit times result in smaller distances, based on the model used.Solving this involves using a calculator to handle the powers precisely, as large exponents demand significant computational precision. Here’s how this process unfolds: 1. Calculate \( (247.69)^{-6669} \), which significantly decreases the final result due to the large negative exponent.2. Multiply the result by \( 92.8935 \), providing the ultimate distance from the Sun to Pluto.Accurate calculations are essential due to the extreme scales in astronomical data, ensuring we achieve a precise distance, such as Pluto's 39.53 million miles.

Astronomical Units

Astronomical Units (AU) are a standardized unit of measure utilized in astronomy, primarily used to express distances within our solar system. One AU is defined as the average distance from the Earth to the Sun, approximately 93 million miles or 150 million kilometers.
AU provide an intuitive way to represent vast celestial distances, making it simpler to convey space metrics without resorting to exceedingly large numbers. This helps compare orbital distances easily.

• Astronomical Units simplify distance evaluations between planets, allowing astronomers to concretely grasp spacing without complex conversions from miles or kilometers.
• They afford a common ground for scientists by standardizing measurements, vital for accurately plotting space missions and understanding planetary positions.

In the given problem, while distances were addressed in millions of miles, converting these can offer additional insights. For Pluto, calculated at approximately 39.53 million miles, this converts to roughly 0.425 AU (since 1 AU equals about 93 million miles).
Such conversions aid astronomers and enthusiasts alike in visualizing just how expansively spaced objects in our solar system truly are.