8.4 Polar coordinates: graphs  (Page 5/16)

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Sketch the graph of $\text{\hspace{0.17em}}r=3-2\mathrm{cos}\text{\hspace{0.17em}}\theta .$

Another type of limaçon, the inner-loop limaçon , is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer (1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing . A century later, the father of mathematician Blaise Pascal , Étienne Pascal(1588-1651), rediscovered it.

Formulas for inner-loop limaçons

The formulas that generate the inner-loop limaçons are given by $\text{\hspace{0.17em}}r=a±b\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a±b\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b>0,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{\hspace{0.17em}}a The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See [link] for the graphs.

Sketching the graph of an inner-loop limaçon

Sketch the graph of $\text{\hspace{0.17em}}r=2+5\text{cos}\text{\hspace{0.17em}}\theta .$

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =1.98.\text{\hspace{0.17em}}$ The maximum $\text{\hspace{0.17em}}|r|\text{\hspace{0.17em}}$ is found when $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =1\text{\hspace{0.17em}}$ or when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ Thus, the maximum is found at the point (7, 0).

Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.

 $\theta$ $0$ $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7

As expected, the values begin to repeat after $\text{\hspace{0.17em}}\theta =\pi .\text{\hspace{0.17em}}$ The graph is shown in [link] .

Investigating lemniscates

The lemniscate is a polar curve resembling the infinity symbol $\text{\hspace{0.17em}}\infty \text{\hspace{0.17em}}$ or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.

Formulas for lemniscates

The formulas that generate the graph of a lemniscate    are given by $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{cos}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{sin}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\ne 0.\text{\hspace{0.17em}}$ The formula $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{sin}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is symmetric with respect to the pole. The formula $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{cos}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is symmetric with respect to the pole, the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ and the polar axis. See [link] for the graphs.

Sketching the graph of a lemniscate

Sketch the graph of $\text{\hspace{0.17em}}{r}^{2}=4\mathrm{cos}\text{\hspace{0.17em}}2\theta .$

The equation exhibits symmetry with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution $\text{\hspace{0.17em}}u=2\theta .$

So, the point $\left(0,\frac{\pi }{4}\right)$ is a zero of the equation.

Now let’s find the maximum value. Since the maximum of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}u=1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}u=0,\text{\hspace{0.17em}}$ the maximum $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}2\theta =1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}2\theta =0.\text{\hspace{0.17em}}$ Thus,

$\begin{array}{c}\text{\hspace{0.17em}}{r}^{2}=4\mathrm{cos}\left(0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=4\left(1\right)=4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=±\sqrt{4}\text{\hspace{0.17em}}=2\end{array}$

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},$ and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to [link] .

 $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $r$ 2 $\sqrt{2}$ 0 $\sqrt{2}$ 0

Plot the points on the graph, such as the one shown in [link] .

Investigating rose curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

Rose curves

The formulas that generate the graph of a rose curve    are given by $\text{\hspace{0.17em}}r=a\mathrm{cos}\text{\hspace{0.17em}}n\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a\mathrm{sin}\text{\hspace{0.17em}}n\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is even, the curve has $\text{\hspace{0.17em}}2n\text{\hspace{0.17em}}$ petals. If $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is odd, the curve has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ petals. See [link] .

The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
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Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
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lamoussa
14
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more than 6000
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SLIMANE
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Marco
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how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations