To calculate: The sections represented by the polar equation r=186−6cos⁡θ and graph it by the use of graphing utility.

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Elberte · Accepted Answer

Step 1  The equation r=186−6cos⁡θ is not in standard form as the constant tern in the denominator is not 1.  To obtain 1 as the constant term in the denominator divide the numerator and denominator by 4.  Thus,  r=186−6cos⁡θ=31−cos⁡θ  The equation r=31−cos⁡θ   is in the standard form of r=ep1−ecos⁡θ.  On comparing the provided equation to the standard form it can be obtained that e=1.  Since e=1, thus the given polar equation represents a parabola.  Step 2  Now, use Ti-83 to plot the graph of the function:  a) Press the [MODE] key then select the polar function and the radian mode.  b) Press the [Y=] key, then there will appear the equations for y.  c) Enter the equations in r1.   Here, r1=186−6cos⁡θ.  d) Press [WINDOW] and then edit the values as:  Xmin=−7,Xmax=7,XScale=1,Ymin=−7andYScale=1  e) press the [GRAPH] key to plot the graph.  The graph:

To calculate: The sections represented by the polar equation displaystyle{r}=frac{18}{{{6}-{6} cos{theta}}} and graph it by the use of graphing utility.

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