Question

Suppose is a closed curve in the plane and that -Y dr + 2? + y2 2 dy = 671 z? + y2 How many self-intersection points must have, at least? By "self-intersection point", I mean a point where the curve intersects itself other than its endpoints. For example, a simple closed curve has zero self-intersection points, and a figure 8 has one self-intersection point. Hint: If a curve has self-intersection points, then it can be divided up into a union of multiple simple closed curves, and the above integral is the sum of the same integral over each of the simple closed curves. So in the following example the a curve with two self-intersection points is divided into three simple closed curves (C is the piecewise smooth curve forming the outermost part of the original curve, and Cy and C3 are the loops formed in the middle). C C C See Example 5 on page 1100 for the value of the above integral if Cis positively oriented simple closed curve enclosing the origin. You can use Green's Theorem to determine what its value is if C is a positively oriented simple closed curve not enclosing the origin. 1 O2 More than 2 There is not enough information to decide

Answer 1

# We know that the line integral of along any closed positively oriented curve c ip aro. where 'n' & caueel ronding number around origin. - Note :-) winding number around sorgen if the number of times 'C' revolves around origin. # tere given that ulty? dy So here windinglion number ip 3 so by definition of winding number must have at least a self intersection point . i. Number of self intersection point ig at least two, pay da no t - GA = 2x (3). e n=3 -> Now consider the Cerve'c' below GO x Now J - 2 diz + hry J Fidre la here F= dr = f.dre tf Fd dy Katega ningu), <dn,dyy F. dret » [ Fidere = { if origen lies inside a then winding number o ff.dre 27 (1)=27 If origin lies inside G or Cz then winding number =2. :: ff.de - 24(2) 247, so in each case Sif.dresa, F.dk + Į C2

10,09 20,9) fourth option && correct. ข/ flerce Ciş a triangle with vertices at 10,0), 1.0) (014). "(3, 0) Let chi line segment from (0,0) to (310) Cai line segment from (310) to 1014) Ez: line segment from (0,4) to (0,0) Hence C = 9 V CZ UCZ and foxy? ds = f nyids + f ny ds + / ry³ds 4 (° )

Now Now and lostsil -0 on G::aVector equatibn of line from (0,0) to (310) i$ 10,0) ++1310) forest) <37,0) 1.1=8t'i y=0 Now a't) = 3; ylt 1=0 jo ds = Verkt ) )+ry'et) j24/7/3?lagdt ay3 = 35.0 =0 i J ny3ds = S'onadt Again ong! Vector equation of g {e (310)+1(0,4) =(3,4t) si M53, y = ut [osta] WH) = 0, y't )=4 sods= Vot16 at ydt mys = 3 (4+)3 = 19243 {nyeds = 19984 544 - 19284 Finally ong:- Vectoro equations of cz 4 (0,4) + +10,0), = 0,47 3) 2=0,4 =4 2) ny3=0 . ny3ds putting above values u ve get 0 + 192 70 = 192 Cell = 192 Jayas: (Aus)