# 10.4 Polar coordinates: graphs  (Page 3/16)

 Page 3 / 16

## Finding zeros and maximum values for a polar equation

Using the equation in [link] , find the zeros and maximum $\text{\hspace{0.17em}}|r|\text{\hspace{0.17em}}$ and, if necessary, the polar axis intercepts of $\text{\hspace{0.17em}}r=2\mathrm{sin}\text{\hspace{0.17em}}\theta .$

To find the zeros, set $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ equal to zero and solve for $\text{\hspace{0.17em}}\theta .$

Substitute any one of the $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ values into the equation. We will use $\text{\hspace{0.17em}}0.$

$\begin{array}{l}\begin{array}{l}\\ r=2\mathrm{sin}\left(0\right)\end{array}\hfill \\ r=0\hfill \end{array}$

The points $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(0,±n\pi \right)\text{\hspace{0.17em}}$ are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.

To find the maximum value of the equation, look at the maximum value of the trigonometric function $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ which occurs when $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}±2k\pi \text{\hspace{0.17em}}$ resulting in $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{2}\right)=1.\text{\hspace{0.17em}}$ Substitute $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}\mathrm{\theta .}$

$\begin{array}{l}r=2\mathrm{sin}\left(\frac{\pi }{2}\right)\hfill \\ r=2\left(1\right)\hfill \\ r=2\hfill \end{array}$

Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of $\text{\hspace{0.17em}}|r|:\text{\hspace{0.17em}}$ $\text{\hspace{0.17em}}r=3\mathrm{cos}\text{\hspace{0.17em}}\theta .$

Tests will reveal symmetry about the polar axis. The zero is $\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right),\text{\hspace{0.17em}}$ and the maximum value is $\text{\hspace{0.17em}}\left(3,0\right).$

## Investigating circles

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves : cardioids , limaҫons, lemniscates, rose curves , and Archimedes’ spirals . We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

## Formulas for the equation of a circle

Some of the formulas that produce the graph of a circle in polar coordinates are given by $\text{\hspace{0.17em}}r=a\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a\mathrm{sin}\text{\hspace{0.17em}}\theta ,$ where $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is $\text{\hspace{0.17em}}\frac{|a|}{2},$ or one-half the diameter. For $\text{\hspace{0.17em}}r=a\mathrm{cos}\text{\hspace{0.17em}}\theta ,$ the center is $\text{\hspace{0.17em}}\left(\frac{a}{2},0\right).\text{\hspace{0.17em}}$ For $\text{\hspace{0.17em}}r=a\mathrm{sin}\text{\hspace{0.17em}}\theta ,$ the center is $\text{\hspace{0.17em}}\left(\frac{a}{2},\pi \right).\text{\hspace{0.17em}}$ [link] shows the graphs of these four circles.

## Sketching the graph of a polar equation for a circle

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

First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros    and maximum $\text{\hspace{0.17em}}|r|\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}r=4\mathrm{cos}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ First, set $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}\theta$ . Thus, a zero occurs at $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}±k\pi .\text{\hspace{0.17em}}$ A key point to plot is $\text{\hspace{0.17em}}\left(0,\text{​}\text{​}\frac{\pi }{2}\right)\text{\hspace{0.17em}}.$

To find the maximum value of $\text{\hspace{0.17em}}r,$ note that the maximum value of the cosine function is 1 when $\text{\hspace{0.17em}}\theta =0±2k\pi .\text{\hspace{0.17em}}$ Substitute $\text{\hspace{0.17em}}\theta =0\text{\hspace{0.17em}}$ into the equation:

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

The maximum value of the equation is 4. A key point to plot is $\text{\hspace{0.17em}}\left(4,\text{\hspace{0.17em}}0\right).$

As $\text{\hspace{0.17em}}r=4\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is symmetric with respect to the polar axis, we only need to calculate r -values for $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ over the interval $\text{\hspace{0.17em}}\left[0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\pi \right].\text{\hspace{0.17em}}$ Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to [link] . The graph is shown in [link] .

 $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{3\pi }{4}$ $\frac{5\pi }{6}$ $\pi$ $r$ 4 3.46 2.83 2 0 −2 −2.83 −3.46 4

## Investigating cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own.

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5