Problem 73 - Algebra and Trigonometry

Question

Use a graphing utility to graph the polar equation. $$r=2 \cos \left(\theta-\frac{\pi}{4}\right)$$

Step-by-Step Solution

Answer

The graph of the polar equation \(r=2 \cos \left(\theta-\frac{\pi}{4}\right)\) is a looped-heart shape, due to the characteristics of the cosine function. It encompasses the pole at its origin and starts and finishes at that point, with a maximum distance of 2 from the pole.

Step 1: Understand the Polar Equation

A polar equation is of the form \(r = f(\theta)\), where r is the distance from the origin (pole) and \(\theta\) is the angle measured from the positive x-axis. The given equation is \(r=2 \cos \left(\theta-\frac{\pi}{4}\right)\), which is a cosine function of \(\theta\). We can see it has a phase shift (shift in \(\theta\)) of \(\pi/4\) and amplitude of 2.

Step 2: Convert the Polar Equation to an Appropriate Form

Before graphing, we need to put the equation into a form that's more recognizable for a polar graph. That form is \(r = a + b \cos(\theta)\). So, we rewrite the given equation as \(r=2( 1 + \cos (\theta - \frac{\pi}{4}))\). Now it looks similar to the equations of a classical limaçon, a mathematical shape.

Step 3: Graph the Polar Equation

Now we can graph the equation. Using a graphing utility, plot \(r\) on the radial-axis, and \(\theta\) on the angular-axis. The graph will look like a 'looped' heart shape, starting then returning to the polar origin. This is due to the nature of the cosine function which oscillates between the values of -1 and 1, resulting in the shape described.