Problem 75 - Calculus Early Transcendentals

Question

Show that the polar equation $$r^{2}-2 r r_{0} \cos \left(\theta-\theta_{0}\right)=R^{2}-r_{0}^{2}$$ describes a circle of radius \(R\) whose center has polar coordinates \(\left(r_{0}, \theta_{0}\right)\).

Step-by-Step Solution

Answer

Question: Show that the given polar equation \(r^{2}-2 r r_{0} \cos(\theta-\theta_{0})=R^{2}-r_{0}^{2}\) represents a circle of radius \(R\) centered at \(\left(r_{0}, \theta_{0}\right)\). Answer: Yes, the given polar equation represents a circle of radius \(R\) centered at \(\left(r_{0}, \theta_{0}\right)\).

Step 1: Write the circle equation in Cartesian coordinates.

A circle with radius \(R\) and center \((a, b)\) can be represented by the equation \((x-a)^{2} + (y-b)^{2} = R^{2}\). We are given the circle's center in polar coordinates \((r_0, \theta_0)\). We need to convert this to Cartesian coordinates to obtain \(a = r_0 \cos\theta_0\) and \(b = r_0 \sin\theta_0\).

Step 2: Convert the Cartesian coordinates equation to polar coordinates.

Given \(x = r\cos\theta\) and \(y = r\sin\theta\), replace \(x\) and \(y\) in the circle equation with their respective polar coordinate expressions: \([(r\cos\theta) - (r_0\cos\theta_0)]^2 + [(r\sin\theta) - (r_0\sin\theta_0)]^2 = R^2\).

Step 3: Expand and simplify the equation.

Expanding the equation, we have: \(r^2\cos^2\theta - 2rr_0\cos\theta\cos\theta_0 + r_0^2\cos^2\theta_0 + r^2\sin^2\theta - 2rr_0\sin\theta\sin\theta_0 + r_0^2\sin^2\theta_0 = R^2\). Now combine the terms with the same trigonometric functions: \(r^2(\cos^2\theta + \sin^2\theta) + r_0^2(\cos^2\theta_0 + \sin^2\theta_0) - 2rr_0(\cos\theta\cos\theta_0 + \sin\theta\sin\theta_0) = R^2\). Since \(\cos^2\theta + \sin^2\theta = 1\) and \(\cos^2\theta_0 + \sin^2\theta_0 = 1\), we can further simplify the equation to: \(r^2 + r_0^2 - 2rr_0(\cos\theta\cos\theta_0 + \sin\theta\sin\theta_0) = R^2\). Now use the identity \(\cos(\theta - \theta_0) = \cos\theta\cos\theta_0 + \sin\theta\sin\theta_0\) to simplify the equation: \(r^2 - 2rr_0\cos(\theta - \theta_0) = R^2 - r_0^2\).

Step 4: Compare the simplified equation with the given equation.

The simplified equation we obtained is: \(r^2 - 2rr_0\cos(\theta - \theta_0) = R^2 - r_0^2\). The given polar equation is: \(r^{2}-2 r r_{0} \cos\left(\theta-\theta_{0}\right)=R^{2}-r_{0}^{2}\). Both equations are the same. Thus, the given polar equation describes a circle of radius \(R\) centered at the polar coordinates \((r_0, \theta_0)\).

Key Concepts

Converting Polar to Cartesian Coordinates

Understanding how to convert polar coordinates to Cartesian coordinates is essential in visualizing equations in a more familiar xy-coordinate system. The conversion leverages the relationship between the coordinates in the two systems. For a point with polar coordinates \(r, \theta\), the corresponding Cartesian coordinates \(x, y\) can be calculated using the equations \(x = r \cos\theta\) and \(y = r \sin\theta\).

Let's apply this to determine the center of our circle. Given polar coordinates \(r_0, \theta_0\), the Cartesian conversion implies \(a = r_0 \cos\theta_0\) and \(b = r_0 \sin\theta_0\), where \(a\) and \(b\) represent the Cartesian coordinates of the circle’s center. Thus, picturing the polar equation as a circle in Cartesian space becomes straightforward, where the circle’s center is placed at \(a, b\) in the xy-plane.

Trigonometric Identities

Trigonometric identities are fundamental in transforming and simplifying mathematical expressions involving angles. Among the most pivotal is the Pythagorean identity: \(\cos^2\theta + \sin^2\theta = 1\). This identity allows us to combine and reduce terms when expanding equations that have both sine and cosine functions of the same angle.

In the context of our exercise, we apply this identity to the discussion of polar coordinates. By acknowledging that \(\cos^2\theta + \sin^2\theta = 1\) and similarly, \(\cos^2\theta_0 + \sin^2\theta_0 = 1\), we can drastically simplify the circular equation. Moreover, the identity \(\cos(\theta - \theta_0) = \cos\theta\cos\theta_0 + \sin\theta\sin\theta_0\) becomes incredibly useful to further simplify the equations involving the cosine difference, eventually leading us to uncover the original polar equation of the circle.

Circle Equation Expansion and Simplification

When dealing with equations of geometric shapes such as the circle, it is often necessary to manipulate and simplify them for ease of understanding and comparison. The circle equation in its expanded Cartesian form, \( (x - a)^{2} + (y - b)^{2} = R^{2} \), becomes more complex when transformed into polar coordinates.

During the simplification process, we expand the terms \( (r\cos\theta - r_0\cos\theta_0)^{2} \) and \( (r\sin\theta - r_0\sin\theta_0)^{2} \) by squaring and combining like terms. The use of trigonometric identities is crucial here; it helps us to convert sums of trigonometric functions into compact forms. Specifically, as we saw in the exercise, the compound angle identity helped collapse an expanded expression into the much neater \( -2rr_0\cos(\theta - \theta_0) \). This step was key to matching the transformed Cartesian equation back to its original polar form, satisfying the exercise’s objective to validate that the equation indeed represents a circle with radius \(R\) and center \(r_0, \theta_0\).