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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 becauseand . The four-leaved rose is also symmetricabout 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 thenreflected about the polar axis to obtain the complete circle.Tangents to Polar CurvesTo find a tangent line to a polar curve, we regard as a parameter and write itsparametric equations asThen, using the method for finding slopes of parametric curves (Equation 10.2.1) and theProduct Rule, we haveWe 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 torr2O(r, ¨)(_r, ¨)O(r, ¨)(r, _¨)_¨¨(a)(b)(c)FIGURE 14O(r, ¨)(r, π-¨)π-¨¨2coscos cos 2cos 2sinsin 02rfyrsin fsin xrcos fcos dydxdyddxddrdsin rcos drdcos rsin 3dy d0dx d0dx d0dy d0r0drd0ifdydxtan
684CHAPTER 10For instance, in Example 8 we found that when or . Thismeans that the lines and (or and ) are tangent lines toat 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.SOLUTIONUsing Equation 3 with , we have(a) The slope of the tangent at the point where is(b) Observe thatTherefore there are horizontal tangents at the points , , andvertical tangents at and . When , both and are0, so we must be careful. Using l’Hospital’s Rule, we haveBy symmetry,Thus there is a vertical tangent line at the pole (see Figure 15).rcos 20434434yxyxrcos 2r1sin 3r1sin dydxdrdsin rcos drdcos rsin cos sin 1sin cos cos cos 1sin sin cos 12 sin 12 sin2sin cos 12 sin 1sin 12 sin 3dydx3cos312 sin31sin312 sin312(1s3)(1s32)(1s3)1s3(2s3)(1s3)1s31s31dydcos 12 sin 0when 2, 32, 76, 116dxd1sin 12 sin 0when 32, 6, 562, 2(12, 76) (12, 116)(32, 6)(32, 56)32dy ddx dliml32dydxliml3212 sin 12 sin liml32cos 1sin 13liml32cos 1sin 13liml32sin cos liml32dydxEXAMPLE 9” , ’” , ’” , ’5π6327π612” , ’11π61232π6(0, 0)m=_1”1+ , ’π3œ„32”2, ’π2FIGURE 15Tangent lines for r=1+sin ¨
POLAR COORDINATES685NOTEInstead of having to remember Equation 3, we could employ the method used toderive it. For instance, in Example 9 we could have writtenThen we havewhich is equivalent to our previous expression.