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Chapter 7: Problem 73

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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 lima莽on, a mathematical shape.
03

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.

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Most popular questions from this chapter

Find the smallest interval for \(\theta\) starting with \(\theta \min =0\) so that your graphing utility graphs the given polar equation exactly once without retracing any portion of it. $$r=4 \sin \theta$$

Exercises \(116-118\) will help you prepare for the material covered in the next section. Use the distance formula to determine if the line segment with endpoints \((-3,-3)\) and \((0,3)\) has the same length as the line segment with endpoints \((0,0)\) and \((3,6)\)

In Exercises \(81-86,\) solve equation in the complex number system. Express solutions in polar and rectangular form. $$ x^{4}+16 i=0 $$

In Exercises \(77-80,\) convert to polar form and then perform the indicated operations. Express answers in polar and rectangular form. $$ \frac{(1+i \sqrt{3})(1-i)}{2 \sqrt{3}-2 i} $$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. A complex number \(a+b i\) can be interpreted geometrically as the point \((a, b)\) in the \(x y\) -plane.
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