# 65 2 2 2 2 2 2 2 4 sin cos 4 cos2 4 y x r r r θ θ

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65.22222224sincos4(cos2 )4yxrrrθθθ===4cos232sec2 ,044rθθ= ±ππ= ±<<67.2216xy+=21644 (since =4 duplicates)rrrr== ±=69.y= 3sin33sin3cscrrrθθθ===71.The graph consists of wide, overlapping “petals”on a flower. 024θπproduces a completegraph, with 24 petals. For larger domains thegraph repeats.
Chapter 9 Parametric Equations and Polar Coordinates31573.A “line” that is tangent to the hole is the outsideboundary for a successful putt. A tangent line to acircle must be perpendicular to a radius. Thus, atriangle formed by the points (0, 0), (d, 0) and thepoint of tangency has a right angle at the point oftangency, a hypotenuse of length dand the legopposite the acute angle of length h. Thus anypath with acute angle Asmaller than the acuteangle of this triangle must have sin.hhAdd<<Thus11sinsin.hhAdd<<75.2()rAis given:221()1.sin(/)ArAdbh d=+1()rAwas found in exercise 74:22221()coscos()rAdAdAdh=The constants 12and AAwere found inExercise 73:1112sinand sin.hhAAdd= −=Section 9.55.cossin3cos;sinsin3 sinxryrθθθθθθ====()()()()()123332333coscos3sinsin(3 );3cos3 sincossin33 33cossincossinAt,3.33coscossinsindyddxddydxdddydxθθθθθθθθθθθθθππππ==+π+ππ=====π −π7.coscos2cos;sincos2sinxryrθθθθθθ====cos2sin2cossin 2 ;coscos22sinsin 2cos0cos02sin 0sin 01At0,,which is undefined. cos0sin 02cos0sin 00dyddxddydxdddydxθθθθθθθθθθθθθ= −=====9.cos3sincos;sin3sinsinxryrθθθθθθ====22223cos3sin;6cossin6cos0sin 00At0,0.33cos 03sin0dyddxddydxdddydxθθθθθθθθθ=======11.cossin 4cos;sinsin 4sinxryrθθθθθθ====()()()()44444coscos4sinsin 4 ;4cos4sincossin 44cossincossin22At,1.4224coscossinsindyddxddydxdddydxθθθθθθθθθθθθθππππ==+π+ππ=====π −π13.coscos3cos;sincos3 sinxryrθθθθθθ====()()()()()()()()()362622126622cos3 sin3cossin3 ;coscos33sinsin3coscos3sinsin1At ,63.cossin3cossin3 3dyddxddydxdddydxθθθθθθθθθθθθθππππππππ= −=π=====
Chapter 9 Parametric Equations and Polar Coordinates31615.ris at a maximum when 2sin3132.263nnθθθπππ= ±= ±+π= ±+sin3cos;sin3 sin3coscos3sinsin3 ;3cos3 sincossin33cos3 sincossin33coscos3sinsin331For : ( ,)sincos, sinsin,6262622dyddxdxydydxdddydxxyθθθθθθθθθθθθθθθθθθθθθθθθθ====++==πππππ===.