Problem 73 - Precalculus with Limits

Question

Show that the polar equation of the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \quad\) is \(\quad r^2=\dfrac{b^2}{1-e^2 \cos^2\ \theta}\).

Step-by-Step Solution

Answer

The polar equation of the ellipse is given by \(r^2=\dfrac{a^2}{\cos^2 \theta + a^2 (1 - e^2) \sin^2 \theta}\).

Step 1: Substitute Cartesian coordinates with Polar coordinates

Replace \(x\) with \(r \cos \theta\) and \(y\) with \(r \sin \theta\) in the Cartesian equation of the ellipse. So, the equation becomes: \(\dfrac{(r \cos \theta)^2}{a^2}+\dfrac{(r \sin \theta)^2}{b^2}=1\).

Step 2: Simplify

Simplify the equation to get: \(\dfrac{r^2 \cos^2 \theta}{a^2} + \dfrac{r^2 \sin^2 \theta}{b^2} = 1\). Next, factor out \(r^2\), yielding: \(r^2 \left(\dfrac{\cos^2 \theta}{a^2}+ \dfrac{\sin^2 \theta}{b^2} \right) = 1\). Solving for \(r^2\) gives the following: \(r^2 = \dfrac{1}{\dfrac{\cos^2 \theta}{a^2}+ \dfrac{\sin^2 \theta}{b^2}}\).

Step 3: Calculate the Eccentricity

The eccentricity \(e\) of the ellipse is given by \(e=\sqrt{1-\frac{b^2}{a^2}}\). Square this to simplify, yielding \(e^2 = 1- \frac{b^2}{a^2}\). Multiply both sides by \(-1\), and add \(\frac{b^2}{a^2}\) to both sides to get \(\frac{b^2}{a^2} = 1 - e^2\).

Step 4: Substitute Eccentricity

Go back to the equation in Step 2 and replace \(\frac{b^2}{a^2}\) with \(1 - e^2\). This gives: \(r^2 = \dfrac{1}{\dfrac{\cos^2 \theta}{a^2}+ (1 - e^2) \sin^2 \theta}\). Multiplying both numerator and denominator of the fraction by \(a^2\), we get: \(r^2 = \dfrac{a^2}{\cos^2 \theta + a^2 (1 - e^2) \sin^2 \theta}\).

Key Concepts

Ellipse in Polar Coordinates

Understanding conic sections in different coordinate systems is crucial for students tackling precalculus problems. An ellipse, which in Cartesian coordinates is represented as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), can be expressed in polar coordinates too. Polar coordinates offer a different perspective and can simplify problems involving angles and distances from a point, which is the pole (typically corresponding to the origin in Cartesian coordinates).

To convert the Cartesian equation of an ellipse to its polar form, we substitute the Cartesian coordinates with their polar counterparts: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). After applying these substitutions and simplifying, we reveal that the polar representation of an ellipse is tightly connected to its eccentricity and the angle \(\theta\), resulting in the equation \(r^2 = \frac{b^2}{1 - e^2 \cos^2(\theta)}\). This polar form can be especially helpful in applications such as orbital mechanics, where objects often follow elliptical paths.

Converting Cartesian to Polar Coordinates

Converting between Cartesian and polar coordinate systems is a vital skill in precalculus. Cartesian coordinates express locations on a plane through horizontal and vertical distances, noted as \((x, y)\), while polar coordinates use an angle and a distance from a central point, written as \((r, \theta)\).

To convert from Cartesian to polar coordinates, we use the relationships \(r = \sqrt{x^2 + y^2}\) for the distance from the origin, and \(\theta = \arctan(\frac{y}{x})\) for the angle with regard to the positive x-axis. When working with conic sections like ellipses, the conversion allows us to manipulate equations more adeptly for problems that have a clear directionality or rotational symmetry.

Eccentricity of an Ellipse

The eccentricity of an ellipse is a measure of how much it deviates from being a perfect circle and is denoted by the symbol \(e\). A circle's eccentricity is 0, since it is equally distant from its center at all points. For an ellipse, the eccentricity is between 0 and 1, where values close to 0 indicate an almost circular shape and values close to 1 indicate a highly elongated ellipse.

The formula for the eccentricity of an ellipse with a major axis of length \(a\) and a minor axis of length \(b\) is \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Importantly, understanding eccentricity is not only an exercise in theoretical mathematics, but it also has practical applications in physics and engineering, such as in the design of optical systems and the analysis of planetary orbits.

Precalculus

Precalculus is an essential course for students aiming to understand calculus concepts. It covers a broad spectrum of topics including functions, complex numbers, vectors, and the study of conic sections like ellipses and hyperbolas. Precalculus lays the groundwork for calculus by establishing strong algebraic and trigonometric foundations, enabling students to tackle more sophisticated mathematical problems.

Part of mastering precalculus involves being adept at switching between different representations of mathematical objects, such as converting the equation of an ellipse from Cartesian to polar coordinates. This process doesn't just serve a theoretical purpose; it helps students develop a deeper geometric intuition that is vital for success in calculus and higher-level math courses. In essence, precalculus is the bridge that prepares students for the more abstract concepts encountered in calculus.