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A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in [link] .
Given a polar equation, test for symmetry.
Test the equation $\text{\hspace{0.17em}}r=2\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for symmetry.
Test for each of the three types of symmetry.
1) Replacing $\text{\hspace{0.17em}}(r,\theta )\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}(-r,-\theta )\text{\hspace{0.17em}}$ yields the same result. Thus, the graph is symmetric with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi}{2}.$ | $\begin{array}{ll}-r=2\mathrm{sin}(-\theta )\hfill & \hfill \\ -r=\mathrm{-2}\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill & \text{Even-oddidentity}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=2\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill & \text{Multiply}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\mathrm{-1}\hfill \\ \text{Passed}\hfill & \hfill \end{array}$ |
2) Replacing $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}-\theta \text{\hspace{0.17em}}$ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. | $\begin{array}{ll}r=2\mathrm{sin}(-\theta )\hfill & \hfill \\ r=\mathrm{-2}\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill & \text{Even-oddidentity}\hfill \\ r=\mathrm{-2}\mathrm{sin}\text{\hspace{0.17em}}\theta \ne 2\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill & \hfill \\ \text{Failed}\hfill & \hfill \end{array}$ |
3) Replacing $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ with $\u2013r\text{\hspace{0.17em}}$ changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. | $\begin{array}{l}-r=2\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ \text{}r=\mathrm{-2}\mathrm{sin}\text{\hspace{0.17em}}\theta \ne 2\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ \text{Failed}\hfill \end{array}$ |
Test the equation for symmetry: $\text{\hspace{0.17em}}r=-2\mathrm{cos}\text{\hspace{0.17em}}\theta .$
The equation fails the symmetry test with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi}{2}\text{\hspace{0.17em}}$ and with respect to the pole. It passes the polar axis symmetry test.
To graph in the rectangular coordinate system we construct a table of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ values. To graph in the polar coordinate system we construct a table of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ values. We enter values of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ into a polar equation and calculate $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ However, using the properties of symmetry and finding key values of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ means fewer calculations will be needed.
To find the zeros of a polar equation, we solve for the values of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that result in $\text{\hspace{0.17em}}r=0.\text{\hspace{0.17em}}$ Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ We use the same process for polar equations. Set $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}\theta .$
For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ into the equation that result in the maximum value of the trigonometric functions. Consider $\text{\hspace{0.17em}}r=5\mathrm{cos}\text{\hspace{0.17em}}\theta ;\text{\hspace{0.17em}}$ the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when $\text{\hspace{0.17em}}\theta =0,\text{\hspace{0.17em}}$ so our polar equation is $\text{\hspace{0.17em}}5\mathrm{cos}\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ and the value $\text{\hspace{0.17em}}\theta =0\text{\hspace{0.17em}}$ will yield the maximum $\text{\hspace{0.17em}}\left|r\right|.$
Similarly, the maximum value of the sine function is 1 when $\text{\hspace{0.17em}}\theta =\frac{\pi}{2},\text{\hspace{0.17em}}$ and if our polar equation is $\text{\hspace{0.17em}}r=5\mathrm{sin}\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ the value $\text{\hspace{0.17em}}\theta =\frac{\pi}{2}\text{\hspace{0.17em}}$ will yield the maximum $\text{\hspace{0.17em}}\left|r\right|.\text{\hspace{0.17em}}$ We may find additional information by calculating values of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.
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