# Vertical tangent lines occur when r q cos q r q sin q

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Vertical tangent lines occur when r′(q) cos q– r(q) sin q= 0Tangent lines through (0, 0) (r= 0) occur when ( )sinsintan( )coscosryrqqqqqqq￲===, r′(q) ≠ 0.Thus, tangent lines through the origin are given by the rays q= q0where r(q0) = 0, r′(q0) ≠ 0.77
62/678. r= cos q+ sin qFind the slope of the tangent line when q= ¼π.22222222224444(sincos)sin(cossin)cos(sincos)cos(cossin)sinsincossincoscossincossincoscossinsincos2cossinsincos2cossinsincos2cossinsincdydxppppqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq-+++=-+-+-+++=-+--+-=--+-=22444411os2cossinsin1pppp---==-When are the tangent lines horizontal? vertical?22371511444437151188882259131444cos2sincossin0cos2sin 20sin 2cos2sin 21cos2tan 212, , , , , , cos2sincossin0cos2sin 20sin 2cos2sin 21cos2tan 212, , , qqqqqqqqqqqqppppqppppqqqqqqqqqqqqppp+-=+==-=-=-==--=-=====4591318888, , , pqpppp=When is r= 0?3744cossin0sincostan1, qqqqqqpp+==-=-=78
It’s a circle:2222222111222cossincossin0()()rrrrxyxyxxyyxyqqqq=+=++=+-+-=-+-=Circles generally have the polar formr= acos qr= asin qr= acos q+ bsin q79
10.4 Areas and Lengths in10.4 Areas and Lengths inPolar CoördinatesPolar Coördinates17 February 2006Base Case:Assume 0 < ba≤ 2π, f(q) ≥ 0 aqb, and f(q) is continuous. Find the areabounded by the polar graph r= f(q) between q= aandq= b.The area of a circle of radius f(q1*) isπf(q1*)²; the area of the given sector (i.e., r=f(q) between q0and q1) is()()()2*1012*111022Affqqpqpqqq-==-Therefore, the area bounded by the polar graph r= f(q) between q= aand q= bis given by()()()()()()2**212122**21111122222212222( )baAfffffffdllllqqqpqpqpqpppqqpqqqqqq++++++=LL, 1iiiqqq-=-.Example: Area bounded by r= 2 for 0 ≤ q ≤ ½π.212202020(2)22202Addpppqqqpp====-=80q0q= ar = f(q)q= bq0q1q1f(q1*)f(q1*)
10/683. Sketch r= 3(1 + cos q) and find the area it encloses.r= 3(1 + cos q) has the form r= a+ bcos q; a= b, so r= f(q) is a cardioid.()()()22120221202292022920029122012[3(1cos)](33cos)(12coscos)1cos22sin2sin 2200092910Addddppppppqqqqqqqqqqqqpqppppp=+=+=+++=++=+ --++4=+=+=18/683. Find the area enclosed by one loop of r= 4 sin 3q(rose petals).Note:One loop of r= 4 sin 3qdoes not go from 0 to 2π; rather, in this case, aand bare two consecutive tangential qvalues.()213202303031204sin38sin 31cos682sin 6828068643Adddppppqqqqqqqqppp==-==-=-==81
Arclength:Consider the length of the curve r= f(q), traversed only once, for aqb. Rewrite f(q) as aparametric equation with parameter qin terms of xand y.