Problem 49 - Calculus Early Transcendentals

Question

The planets orbit the Sun in elliptical orbits with the Sun at one focus (see Section 10.4 for more on ellipses). The equation of an ellipse whose dimensions are \(2 a\) in the \(x\) -direction and \(2 b\) in the \(y\) -direction is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\). a. Let \(d^{2}\) denote the square of the distance from a planet to the center of the ellipse at \((0,0) .\) Integrate over the interval \([-a, a]\) to show that the average value of \(d^{2}\) is \(\left(a^{2}+2 b^{2}\right) / 3\). b. Show that in the case of a circle \((a=b=R)\), the average value in part (a) is \(R^{2}\). c. Assuming \(0

Step-by-Step Solution

Answer

Answer: The average value of d^2 is \(\frac{a^2 + 2b^2}{3}\).

Step 1: Find expression for d^2

The square of the distance (\(d^2\)) can be represented as \(d^2 = x^2 + y^2\).

Step 2: Express y^2 in terms of x using the ellipse equation

To do this, we write the ellipse equation and solve for y^2: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) \(y^2 = b^2(1 - \frac{x^2}{a^2})\) Now, we can rewrite the expression for \(d^2\) in terms of x: \(d^2 = x^2 + b^2(1 - \frac{x^2}{a^2})\)

Step 3: Integrate d^2 over the interval [-a, a] and find the average

To find the average value of \(d^2\), we need to integrate it over the interval [-a, a] and then divide by the range (2a): \(\frac{1}{2a}\int_{-a}^a (x^2 + b^2(1 - \frac{x^2}{a^2})) dx\) Now integrate d^2: \(\frac{1}{2a} [(\frac{1}{3}x^3 + b^2x - \frac{b^2}{3a^2}x^3) ]_{-a}^a\) After evaluating the integral and simplifying, we get: \(\frac{1}{2a} [(2a^2 + 2b^2) - 0]\) So, the average value of \(d^2 = \frac{a^2 + 2b^2}{3}\)

Step 4: Show that for a circle (a = b = R), the average value is R^2

To show this, we need to plug a = b = R into the expression for the average value of \(d^2\): \(\frac{R^2 + 2R^2}{3} = \frac{3R^2}{3} = R^2\)

Step 5: Find expression for D^2 and express y^2 in terms of x given Sun's coordinates

The square of the distance between the planet and the Sun (\(D^2\)) can be represented as \(D^2 = (x - \sqrt{a^2 - b^2})^2 + y^2\). Using the equation of the ellipse, we found that \(y^2 = b^2(1 - \frac{x^2}{a^2})\) in Step 2.

Step 6: Integrate D^2 over the interval [-a, a] and find the average

To find the average value of \(D^2\), we need to integrate it over the interval [-a, a] and then divide by the range (2a): \(\frac{1}{2a}\int_{-a}^a ((x - \sqrt{a^2 - b^2})^2 + b^2(1 - \frac{x^2}{a^2})) dx\) Now integrate D^2: \(\frac{1}{2a} [(\frac{1}{3}x^3 - (2\sqrt{a^2 - b^2})x^2 + 4(a^2 - b^2)x + b^2x - \frac{b^2}{3a^2}x^3) ]_{-a}^a\) After evaluating the integral and simplifying, we get: \(\frac{1}{2a}[8a^3 - 8a^3 + 4a^3 + 2a^3b^2]\) So, the average value of \(D^2 = \frac{4a^2 - b^2}{3}\).

Key Concepts

Average Value Theorem

The Average Value Theorem is a handy tool in calculus, useful for finding the average value of a function over a certain interval. Here's a simple explanation to guide you.

Suppose you have a continuous function \( f(x) \) on an interval \([a, b]\). The average value of this function over that interval is calculated by:

• Integrating the function over the interval \([a, b]\).
• Dividing the result by the interval's length, \( b-a \).

So, the formula looks like this:\[\text{Average value of } f(x) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]

In the provided exercise, we apply the Average Value Theorem to calculate the average square of the distance from the planet to the center of the ellipse, \(d^2\), by integrating \(d^2\) over \([-a, a]\). Similarly, we apply the same process to find the average square of the distance from the planet to the Sun, \(D^2\).

By using this theorem, we help describe how these average distances are derived over the span of a planet's orbit, showcasing the elegance and usefulness of calculus in astronomy.

Ellipses

Ellipses are a fascinating geometric shape and essential in understanding planetary orbits. An ellipse is a set of points where the sum of the distances to two fixed foci is constant. In the context of celestial mechanics, planets move in elliptical orbits around the Sun, which is located at one of the focal points of such an ellipse.

The standard equation describing an ellipse in coordinate geometry is given by:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
where:

• \(a\) is the semi-major axis, detailing how far the ellipse stretches along the x-axis.
• \(b\) is the semi-minor axis, describing the stretch along the y-axis.

For planetary orbits, the semi-major axis is crucial. It defines the size of the orbit and is a major factor in determining the orbital period.

When both axes are equal \(a = b\), the ellipse becomes a circle. That's why part (b) of the exercise simplifies to a circle, and naturally, the calculations reflect the symmetry where \(a = b = R\). Understanding ellipses in the context of coordinate geometry allows us to model and predict planetary motion with remarkable accuracy.

Coordinate Geometry

Coordinate Geometry, also known as Analytic Geometry, helps us represent geometric shapes numerically and analyze their properties using algebra. It's a crucial foundation for understanding the trajectories of celestial bodies.

By setting up coordinate systems on a plane, we can use equations to describe shapes like ellipses, circles, and more. In the original exercise:

• The ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) allows us to analyze the planet's orbit.
• The center of the ellipse is at the origin, \((0, 0)\), simplifying calculations for distances.
• The Sun's position \((\sqrt{a^2 - b^2}, 0)\) as one of the foci, helps in determining the distance between celestial objects.

Using coordinate geometry, we decompose complex planetary motions into manageable equations and calculations. This branch of mathematics is indispensable in fields such as astronomy and physics, as it forms the basis of many fundamental principles related to motion and distance.