The curves sketched in Examples 6 and 8 are symmetric about the polar axis

The curves sketched in examples 6 and 8 are symmetric

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The curves sketched in Examples 6 and 8 are symmetric about the polar axis, since . The curves in Examples 7 and 8 are symmetric about because and . The four-leaved rose is also symmetric about the pole. These symmetry properties could have been used in sketching the curves. For instance, in Example 6 we need only have plotted points for and then reflected about the polar axis to obtain the complete circle. Tangents to Polar Curves To find a tangent line to a polar curve , we regard as a parameter and write its parametric equations as Then, using the method for finding slopes of parametric curves (Equation 10.2.1) and the Product Rule, we have We locate horizontal tangents by finding the points where (provided that ). Likewise, we locate vertical tangents at the points where (pro- vided that ). Notice that if we are looking for tangent lines at the pole, then and Equation 3 sim- plifies to r r 2 O (r, ¨) (_r, ¨) O (r, ¨) (r, _¨) ¨ (a) (b) (c) FIGURE 14 O (r, ¨) (r, π-¨) π-¨ ¨ 2 cos cos cos 2 cos 2 sin sin 0 2 r f y r sin f sin x r cos f cos dy dx dy d dx d dr d sin r cos dr d cos r sin 3 dy d 0 dx d 0 dx d 0 dy d 0 r 0 dr d 0 if dy dx tan
684 CHAPTER 10 For instance, in Example 8 we found that when or . This means that the lines and (or and ) are tangent lines to at the origin. (a) For the cardioid of Example 7, find the slope of the tangent line when . (b) Find the points on the cardioid where the tangent line is horizontal or vertical. SOLUTION Using Equation 3 with , we have (a) The slope of the tangent at the point where is (b) Observe that Therefore there are horizontal tangents at the points , , and vertical tangents at and . When , both and are 0, so we must be careful. Using l’Hospital’s Rule, we have By symmetry, Thus there is a vertical tangent line at the pole (see Figure 15). r cos 2 0 4 3 4 4 3 4 y x y x r cos 2 r 1 sin 3 r 1 sin dy dx dr d sin r cos dr d cos r sin cos sin 1 sin cos cos cos 1 sin sin cos 1 2 sin 1 2 sin 2 sin cos 1 2 sin 1 sin 1 2 sin 3 dy dx 3 cos 3 1 2 sin 3 1 sin 3 1 2 sin 3 1 2 ( 1 s 3 ) ( 1 s 3 2 )( 1 s 3 ) 1 s 3 ( 2 s 3 )( 1 s 3 ) 1 s 3 1 s 3 1 dy d cos 1 2 sin 0 when 2 , 3 2 , 7 6 , 11 6 dx d 1 sin 1 2 sin 0 when 3 2 , 6 , 5 6 2, 2 ( 1 2 , 7 6 ) ( 1 2 , 11 6 ) ( 3 2 , 6 ) ( 3 2 , 5 6 ) 3 2 dy d dx d lim l 3 2 dy dx lim l 3 2 1 2 sin 1 2 sin lim l 3 2 cos 1 sin 1 3 lim l 3 2 cos 1 sin 1 3 lim l 3 2 sin cos lim l 3 2 dy dx EXAMPLE 9 ”    ,       ’ ”    ,    ’ ”    ,       ’ 6 3 2 6 1 2 ”    ,        ’ 11π 6 1 2 3 2 π 6 (0, 0) m=_1 ”1+      ,     ’ π 3 œ „3 2 ”2,     ’ π 2 FIGURE 15 Tangent lines for r=1+ sin  ¨
POLAR COORDINATES 685 NOTE Instead of having to remember Equation 3, we could employ the method used to derive it. For instance, in Example 9 we could have written Then we have which is equivalent to our previous expression.

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