Problem 83 - Precalculus

Question

Determine the values of \(\theta\) at which \(r=4 \cos \theta\) and \(r \cos \theta=1\) intersect. Graph both equations.

Step-by-Step Solution

Answer

The curves intersect at \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \).

Step 1: Convert equations to rectangular form

For the first equation, we have \( r = 4 \cos \theta \). We convert this polar equation to rectangular form: \( r = x \cos \theta + y \sin \theta \). Therefore, \( x^2 + y^2 = 4 \frac{x}{r} \). Rearranging, we get \( (x^2 + y^2)^2 = 16x^2 \).For the second equation, \( r \cos \theta = 1 \), we have \( x = 1 \). Thus, the second equation converts to a vertical line at \( x = 1 \) in rectangular form.

Step 2: Solve equations simultaneously

Substituting the equation \( x = 1 \) into \( (x^2 + y^2)^2 = 16x^2 \), we get \( (1^2 + y^2)^2 = 16 \times 1^2 \). Simplifying, \( (1 + y^2)^2 = 16 \). Solving for \( y \), we get \( 1 + y^2 = 4 \) or \( 1 + y^2 = -4 \). The latter is not possible, so \( y^2 = 3 \). Therefore, \( y = \pm \sqrt{3} \).

Step 3: Find corresponding values of \(\theta\)

From \( x = 1 \) and \( y = \pm \sqrt{3} \), we find \( \theta \) using \( \tan \theta = \frac{y}{x} \). For \( y = \sqrt{3} \), \( \tan \theta = \sqrt{3} \) which gives \( \theta = \frac{\pi}{3} \). For \( y = -\sqrt{3} \), \( \tan \theta = -\sqrt{3} \), which gives \( \theta = \frac{5\pi}{3} \).

Step 4: Verify the solution

Substitute \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \) back into both equations to ensure they satisfy \( r = 4 \cos \theta \) and \( r \cos \theta = 1 \). Both values of \( \theta \) satisfy the original equations, confirming that the points intersect at these angles.

Step 5: Graph the equations

To graph, draw the circle from \( r = 4 \cos \theta \) as a circle with radius 4 centered at (2,0) in Cartesian coordinates, and the vertical line at \( x = 1 \). The points of intersection at \( \left(1,\sqrt{3}\right) \) and \( \left(1,-\sqrt{3}\right) \) appear where the line crosses the circle.

Key Concepts

Rectangular Coordinates

To understand the concept of rectangular coordinates, let's take a closer look at the idea of converting polar equations to rectangular form. Rectangular coordinates use the familiar

• **x-coordinate** (horizontal position)
• **y-coordinate** (vertical position)

For the equation of the circle, we start with the polar form: \( r = 4 \cos \theta \). In converting this to rectangular coordinates, we make use of trigonometric identities: \( x = r \cos \theta \) and \( y = r \sin \theta \). Rearrange it to get the familiar equation of a circle: \( (x^2 + y^2)^2 = 16x^2 \).
This process changes the perspective from polar coordinates, where distance and angle are used, to the familiar x-y plane used in graphing equations.

Intersection of Curves

Understanding the intersection of curves involves finding the points where two different equations have matching solutions. In this case, we have two equations:

• \( r = 4 \cos \theta \) (a circle in polar form)
• \( r \cos \theta = 1 \) (a line in polar form, representing \( x = 1 \) in rectangular coordinates)

We need to find the points \((x, y)\) that satisfy both equations, representing the locations where the circle and line intersect.
By substituting \( x = 1 \) in the equation of the circle, the calculations give us \( y = \pm \sqrt{3} \), indicating that the two curves meet at \((1, \sqrt{3})\) and \((1, -\sqrt{3})\). These intersection points are crucial for visualizing how two different geometric shapes interact in a shared space.

Converting Equations

To convert equations between different coordinate systems, we utilize trigonometric relationships. This involves changing from polar coordinates (\(r, \theta\)) to rectangular coordinates (\(x, y\)). For conversion, remember:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( r^2 = x^2 + y^2 \)

These conversions allow equations initially formulated for circular sweeps (polar) to be expressed in straight line graphs (rectangular). By working through these transformations, what once seemed like a circle drawn through angles and radii is clearly shown as familiar shapes on a Cartesian plane.
This skill is not only useful when solving problems involving intersections, but it's also crucial for higher-level mathematics involving calculus and analysis.

Solving Trigonometric Equations

Solving trigonometric equations is often necessary in finding specific values within problems involving angles and distances. Here, in determining the values of \( \theta \) at which \( r = 4 \cos \theta \) and \( r \cos \theta = 1 \) intersect, we set up the trigonometric relationship for \( \tan \theta \) as:

• \( \tan \theta = \frac{y}{x} \)

This translates to solving for \( \theta \) in:

• For \( y = \sqrt{3} \), \( \tan \theta = \sqrt{3} \), giving \( \theta = \frac{\pi}{3} \)
• For \( y = -\sqrt{3} \), \( \tan \theta = -\sqrt{3} \), giving \( \theta = \frac{5\pi}{3} \)

By verifying and substituting these \( \theta \) values back into the original equations, we confirm these angles indeed represent the intersections.
Having a good grasp of solving trigonometric equations solidifies understanding of both circular and coordinate grid-based problems.