Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

The earth's position in the solar system at time \(t\) can be described approximately by \(P(93 \cos (2 \pi t), 93 \sin (2 \pi t))\), where the sun is at the origin and distances are measured in millions of miles. Suppose that an asteroid has position \(Q(60 \cos [2 \pi(1.51 t-1)], 120 \sin [2 \pi(1.51 t-1)]) .\) When, over the time period \([0,20]\) (i.e., over the next 20 years), does the asteroid come closest to the earth? How close does it come?

Step-by-Step Solution

Answer

Use differentiation and numerical methods to find the time \( t \) in \([0,20]\) where the distance \( D(t) \) is minimized.

Step 1: Understand the Problem

We have two parametric equations representing the positions of Earth and an asteroid at time \( t \). These equations are Earth: \( P(93\cos(2\pi t), 93\sin(2\pi t)) \) and Asteroid: \( Q(60\cos[2\pi(1.51t-1)], 120\sin[2\pi(1.51t-1)]) \). The task is to find when the distance between Earth and the asteroid is minimized during the period \([0,20]\).

Step 2: Set Up the Distance Formula

The distance \( D \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Applying this to our parametric equations gives: \[D(t) = \sqrt{(60\cos[2\pi(1.51t-1)] - 93\cos(2\pi t))^2 + (120\sin[2\pi(1.51t-1)] - 93\sin(2\pi t))^2}\]

Step 3: Simplify and Define Objective Function

Since the square root function is monotonically increasing, minimizing \( D^2 \) will also minimize \( D \). Therefore, let's define the objective function to be minimized as:\[D^2(t) = (60\cos[2\pi(1.51t-1)] - 93\cos(2\pi t))^2 + (120\sin[2\pi(1.51t-1)] - 93\sin(2\pi t))^2\]

Step 4: Differentiate and Find Critical Points

To find the critical points where \( D^2(t) \) might be minimized, we need to differentiate \( D^2(t) \) with respect to \( t \) and set the derivative equal to zero: \( \frac{d}{dt}D^2(t) = 0 \). This derivative can be complex, so numerical methods or computer algebra systems might be used to find the roots in the interval \([0,20]\).

Step 5: Evaluate Distance at Critical Points

Compute \( D(t) \) at each critical point found in Step 4, as well as at the boundaries \( t = 0 \) and \( t = 20 \). Compare these values to determine at which point(s) the minimum distance occurs.

Step 6: Conclusion

The critical point which results in the smallest value of \( D(t) \) during the interval \([0,20]\) is when the asteroid and Earth are closest. Calculate \( D(t) \) for that specific \( t \) to obtain the minimum distance.

Key Concepts

Parametric Equations

In calculus, parametric equations are essential when describing paths of objects where coordinates are defined as functions of a parameter, often time represented by \( t \). Here, we have equations for two celestial bodies. The earth's and the asteroid's positions over time are expressed as:

• Earth: \( P(93\cos(2\pi t), 93\sin(2\pi t)) \)
• Asteroid: \( Q(60\cos[2\pi(1.51t-1)], 120\sin[2\pi(1.51t-1)]) \)

These equations give us each body's coordinates in a circular motion, with Earth having a radius of 93 million miles and the asteroid having elliptical coordinates with different x and y radii of 60 and 120 million miles respectively. This framework allows us to track both the earth's and asteroid's path relative to time, helping us identify key interactions like proximity. Parametric equations transform the complex orbital paths into manageable mathematical expressions. Breaking movements into sine and cosine components offers insights into periodic motion, reflective of orbits, making this an effective application.

Minimization Techniques

Finding the point where the asteroid comes closest to Earth involves minimizing the distance between them over a given period. With these cyclical paths defined through parametric equations, we turn to minimization techniques that focus on finding the smallest possible value of an objective function. To simplify the problem, we utilize the fact that the square root function, which describes distance, is monotonically increasing. Therefore, minimizing the distance squared, \( D^2(t) \), is equivalent to minimizing distance, \( D(t) \). The expression to minimize is: \[ D^2(t) = (60\cos[2\pi(1.51t-1)] - 93\cos(2\pi t))^2 + (120\sin[2\pi(1.51t-1)] - 93\sin(2\pi t))^2 \] Using this squared distance, we avoid dealing with square roots, making our equations easier to handle mathematically. These techniques often require differentiation to find the least value of the target function as part of optimization.

Distance Formula

The distance formula in geometry helps us find the measure of the straight-line distance between two points in a plane. Given two points, \((x_1, y_1)\) and \((x_2, y_2)\), their distance is calculated using: \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Applying this to Earth's and the asteroid's parametric equations, we calculate: \[ D(t) = \sqrt{(60\cos[2\pi(1.51t-1)] - 93\cos(2\pi t))^2 + (120\sin[2\pi(1.51t-1)] - 93\sin(2\pi t))^2} \] The formula considers the differences in x-coordinates and y-coordinates, then squares each. Sum of these squares under a square root gives the Euclidean distance, a vital measure in physical spaces like orbits. Understanding this concept enables us to quantify how far the asteroid strays or approaches Earth, critical in orbital analysis.

Critical Points Finding

Critical points indicate where a function reaches potential maximum or minimum values, crucial in revealing where the asteroid is closest or farthest from Earth. To find these, we differentiate the squared distance function, \( D^2(t) \), with respect to \( t \), and solve for its roots where the derivative equals zero: \[ \frac{d}{dt}D^2(t) = 0 \] This derivative operation isolates points where the rate of change in distance is zero, highlighting moments of closest approach. Given the complexity of these functions due to trigonometric identities and transformations, one might use numerical methods or computational tools for practical evaluation. After finding critical points within the interval \([0,20]\), evaluate the distance function at these points and boundaries, \( t = 0 \) and \( t = 20 \), to determine minimal distance.