Problem 49 - Thomas Calculus

Question

(Continuation of Example \(5 . )\) The simultaneous solution of the equations $$ \begin{aligned} r^{2} &=4 \cos \theta \\ r &=1-\cos \theta \end{aligned} $$ in the text did not reveal the points \((0,0)\) and \((2, \pi)\) in which their graphs intersected. a. We could have found the point \((2, \pi),\) however, by replacing the \((r, \theta)\) in Equation \((1)\) by the equivalent \((-r, \theta+\pi)\) to obtain $$ \begin{aligned} r^{2} &=4 \cos \theta \\\\(-r)^{2} &=4 \cos (\theta+\pi) \\\ r^{2} &=-4 \cos \theta \end{aligned} $$ Solve Equations \((2)\) and \((3)\) simultaneously to show that \((2, \pi)\) is a common solution. (This will still not reveal that the graphs intersect at \((0,0) . )\) b. The origin is still a special case. (It often is.) Here is one way to handle it: Set \(r=0\) in Equations \((1)\) and \((2)\) and solve each equation for a corresponding value of \(\theta .\) since \((0, \theta)\) is the origin for any \(\theta\) , this will show that both curves pass through the origin even if they do so for different \(\theta\) -values.

Step-by-Step Solution

Answer

\((2, \pi)\) is a common solution for modified equations; both curves pass through the origin at different \(\theta\).

Step 1: Define the Equations

Equation (1) is given as \( r^2 = 4 \cos \theta \) and Equation (2) is \( r = 1 - \cos \theta \). We are also using a modified version where \( (-r)^2 = 4 \cos(\theta + \pi) \).

Step 2: Simplify the Cosine of the Angle Sum

To solve for \( (-r)^2 = 4 \cos(\theta+\pi) \), we use the identity \( \cos(\theta+\pi) = -\cos\theta \). This simplifies the equation to \( (-r)^2 = -4 \cos \theta \), or \( r^2 = 4 \cos \theta \).

Step 3: Solve the Modified Equations

Now, solve \( r = 1 - \cos \theta \) and \( r^2 = -4 \cos \theta \). Since both should be satisfied, plug \( r = 1 - \cos \theta \) into \( r^2 = -4 \cos \theta \).

Step 4: Substitute and Solve for \( \theta \)

By substituting, we have: \((1 - \cos \theta)^2 = -4 \cos \theta\). Expanding gives \(1 - 2 \cos \theta + \cos^2 \theta = -4 \cos \theta\). Reorganize to \(\cos^2 \theta + 2\cos \theta + 1 = 0\).

Step 5: Solve the Quadratic Equation

The quadratic equation \(\cos^2 \theta + 2\cos \theta + 1 = 0\) is a perfect square trinomial and can be written as \((\cos \theta + 1)^2 = 0\). Thus, \(\cos \theta = -1\). This implies \(\theta = \pi\).

Step 6: Verify the Solution

Check that this value of \(\theta = \pi\) in \(r = 1 - \cos \theta\) gives \(r = 2\). Therefore, \((r, \theta) = (2, \pi)\) is a common solution.

Step 7: Handle the Origin Case

For the origin, set \(r = 0\) in both equations. For Equation (1), \(0^2 = 4 \cos \theta\), leading to \(\cos \theta = 0\), showing \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\). For Equation (2), \(0 = 1 - \cos \theta\) gives \(\cos \theta = 1\), indicating any \(\theta\) that makes the point the origin. Both curves pass through the origin at these \(\theta\)-values.

Key Concepts

Simultaneous Equations

Simultaneous equations are sets of equations that are solved together to find a common solution. In this exercise, we are dealing with two polar equations:

• Equation (1): \( r^2 = 4 \cos \theta \)
• Equation (2): \( r = 1 - \cos \theta \)

We want to find points \((r, \theta)\) that satisfy both equations simultaneously. Solving these equations requires substituting one equation into the other. For example, replacing \( r = 1 - \cos \theta \) from Equation (2) into Equation (1) allows us to find a solution for \(\theta\) and verify it matches both equations. This is a common technique used to discover points of intersection or common solutions in mathematical problems.

Trigonometric Identities

Trigonometric identities are useful tools in simplifying equations involving trigonometric functions like sine, cosine, and tangent. In this problem, the identity used is the cosine angle sum formula: \( \cos(\theta + \pi) = -\cos \theta \). This simplifies the modified equation \((-r)^2 = 4 \cos(\theta + \pi)\) to \(r^2 = -4 \cos \theta \). Thus, through the identity, the modified equation aligns with the original Equation (1). By simplifying the expression using this identity, we can handle transformations in polar coordinates, revealing essential relationships between the equations that were not immediately obvious before the modification.

Graph Intersections

Graph intersections occur where two graphs meet at the same points in the coordinate plane. In this exercise, the question revolves around polar coordinates, where the points of intersection can have unique implications.

• The point \((2, \pi)\) is one such intersection found through a modified approach using trigonometric identities.
• The intersection at \((0,0)\) requires setting the radius \(r\) to zero, highlighting the unique nature of intersections at the origin in polar coordinates.

These intersections help us understand where and how different curves overlap, particularly when dealing with circular and angular components in polar graphs, providing insight beyond Cartesian graphs.

Quadratic Equations

Quadratic equations frequently arise in various mathematical contexts, including geometry and trigonometry. In the exercise, by substituting the terms, we ended with a quadratic equation of the form \(\cos^2 \theta + 2\cos \theta + 1 = 0\). This is a perfect square trinomial that further simplifies to \((\cos \theta + 1)^2 = 0\). Solving this quadratic gives the solution \(\cos \theta = -1\), which corresponds to the angle \(\theta = \pi\).

• Quadratic equations often form when solving systems involving squares or similar transformations.
• The solutions provide critical points where the polar graphs intersect, validating the points of intersection determined through simultaneous equations.

Understanding how to handle and solve quadratic equations allows us to break down complex expressions into more manageable forms, leading to easier interpretation of graphical results.