Problem 64 - Calculus - AP Edition

Question

The mean orbital radius \(r\) (in units of \(10^{5} \mathrm{km}\) ) of a moon of Saturn can be modeled by the equation \(r=1.93 t^{2 / 3}\) where \(t\) is the time in (Earth) days for the moon to complete one orbit about the planet. Use this model to estimate the instantaneous rate of change of \(r\) with respect to \(t\) when \(t=0.602\) day (the orbital period of Saturn's moon Atlas).

Step-by-Step Solution

Answer

The instantaneous rate of change is approximately 1.806.

Step 1: Identify the Given Function

The given function for the mean orbital radius of the moon is \( r = 1.93 t^{2/3} \). This function relates the radius \( r \) with the time \( t \), taken in Earth days, to complete one orbit.

Step 2: Differentiate the Function

To find the instantaneous rate of change of \( r \) with respect to \( t \), we differentiate \( r = 1.93 t^{2/3} \) with respect to \( t \). The derivative \( \frac{dr}{dt} \) gives the rate of change. Using the power rule, we have:\[ \frac{dr}{dt} = 1.93 \cdot \frac{2}{3} \cdot t^{-1/3} = \frac{3.86}{3} t^{-1/3} = 1.29 t^{-1/3}. \]

Step 3: Evaluate the Derivative at Given \(t\)

We evaluate \( \frac{dr}{dt} \) at \( t = 0.602 \) to find the instantaneous rate of change. Substitute \( t = 0.602 \) into the derivative:\[ \frac{dr}{dt} \bigg|_{t=0.602} = 1.29 (0.602)^{-1/3}. \]

Step 4: Calculate the Value

Compute \( (0.602)^{-1/3} \) and multiply by 1.29 to find the rate of change. First, calculate \( (0.602)^{-1/3} \). Using a calculator gives approximately 1.40, so \( \frac{dr}{dt} \approx 1.29 \times 1.40 \). The final result is approximately 1.806.

Key Concepts

Instantaneous Rate of Change

The concept of instantaneous rate of change is about understanding how a quantity is changing at a very specific moment in time. In this context, it isn't about how the quantity changes over an interval, but rather at a singular point. Imagine you are driving a car, and you want to know exactly how fast you're going at a specific instant, not over a period of, let's say, the entire trip.

To find this instantaneous rate, we use the derivative of a function. This derivative gives us the rate at which one quantity changes relative to another. In our exercise, the function we're working with is the orbital radius as a function of time. By calculating the derivative, we get a formula that tells us how fast the radius is changing with time. This is crucial in fields like physics and engineering, where precise measurements can make a big difference.

• Provides a 'snapshot' of the rate of a change at a specific moment.
• Mathematically determined using derivatives.
• Essential for real-time analysis and predictions.

Differentiation

Differentiation is a natural extension of the concept of instantaneous rate of change. It’s the mathematical process we use to compute derivatives. In simple terms, differentiation is like finding the slope of a curve at any point.

In the context of the given exercise, to determine how the orbital radius changes with respect to time, we differentiated the function \( r = 1.93 t^{2/3} \). By applying the power rule, we transformed this function into the derivative \( \frac{dr}{dt} = 1.29 t^{-1/3} \). This derivative formula now allows us to calculate the rate of change at any potential orbit time \( t \).

• Involves applying mathematical rules to find the derivative.
• Power rule was used in the exercise for functions in the form \( at^{n} \).
• Useful for predicting and understanding dynamic changes in systems.

Orbital Mechanics

Orbital mechanics is the study of the motion of objects in space. It is governed by gravity and other forces. This field is crucial for understanding how celestial bodies like moons and planets move.

In this exercise, we look at the orbital mechanics of Saturn's moon Atlas. Specifically, we are interested in how changes in time affect its mean orbital radius. Modeling this relationship is significant because it helps scientists predict orbital paths and their changes over time. The equation provided in the exercise, \( r = 1.93 t^{2/3} \), approximates this behavior, demonstrating how advanced mathematics is used in practical applications like space exploration.

• Explores how celestial bodies behave under gravitational forces.
• Helps create models for predicting the motion of moons and planets.
• Key in planning space missions and studying planetary systems.