Lecture Note
University
California State UniversityCourse
MATH 150B | Calculus IIPages
7
Academic year
2023
yung dump
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0
11.4 Graphing in Po lar Coordinates 4 Symmetry Tests for Polar Graphs: II Symmetry about x-axis: If the point (r, o) lies on the graph, then (r, - Q) or (-r, II- o) lies on the graph. y (r,0) X 2 Symmetry about y-axis: If the point (r,e) lies on the graph then the point (r,-G) (-r,11-0) or (r, II- Q) or (-r, - -0) lies on the graph y (r,TT-0) or (-r,-e) (r,0) x (3) Symme try about the origin: If the point (r,o) lies on the graph / then the point (- r, o) or (r, O+II) lies on the y (r,e) graph. x (-r,e) or (Y, O+TT) slope Let r=f(0) . Recall the parametric equations: x=rccs / y= rsinG r==((0) =f(o) coso = (f(o) sinG slope of the curve r=f(0) at (r,o) is Proof dy = dy/do Tx Ix/de dy / r sin O + r Cos D = = fsine +f cos o r cos O - sing fcos C - -f sing (r,0) Note that when the curve v= ((O) passes through the origin at Oo do = tan o. dx (0,001
Exp Find the slope of r = cos 20 at 0=0,= 5 when 0=0 r=1 (r, o)=(1,0) r = -25in20 slope is dx dy / = sino + rcoso = -2sin(0)sin(o) +(1) cos(0) r'cose - rsing -asin(0) cos(o) (1) sinco (1,0) (1,0) undefined when O = II r=-1 (r,0)=(-1, E) (1,II) the slope is dy -2sin(TT) sin (12) +(-1) COSTIE) x = o (1,0) dx = sin(II) COS (IF) - (-1) sin(F) (-1, II) (-VII) Four-leaved rose Exp sketch the grave of the following curves, Identify the symmethy [I] r=1-cose (r,0) on the graph r=1-cost r=1-cos(-)) (r,-G) on the graph the curve is symmetric about x-axis. 1- - cos (-0)=1-c056# - r the curve is not symmetric 1- - Cos (T-O)=1+cose # r about y-axis 1- COS o #-r the curve is not symmetic 1- cos(0+11) = It COS O Fr about the origin. COSE (2/20) O r=1-cose (1,0) " III 211 O O 0 II 1/2 (1/45 3 - COS O (2,TT) x II O I (-2,0) 11 2 II 2IT 2 2II 12 3 1) r r=1-Cose TI 2 O 1 12/3/ = (11-24) - cardioid 0 TT 2IT
1 r= 1+ cos O is symmetric about x-axis since 1+ cos (-0) = It cos O = r not symmetric about y-axis since I + cost-0)= It cos 6 #-r and I + Cos (II- 0)= It cos II (05(-)) sin II sin (-e) = 1- cost 6 # r TherPore, it is not symmetric about origin. y r=l+cose O r 0 I 2 1.5 2 I 2 x II 3 1/2 1 IT 0 cardioid
11.4 Lecture Problems I 4 r = It sino is not symmetric about X- axis since 1 + sin(-) = 1- sinG #r and I + sin (II- o) = It sin II cos(-)) + sin(-) COSTT =1+0-sin@ (-1) = \ + sin #- r symmetric about y-axis since I + sin (IT- 0) = 1+ sinTT cos(-0) + sin(-G)COSTT = 1 + o - sin@ (-1) = I + sincerely O = r Therefore, no symmetry about origin y r=l+sine -III II II O II 2 2 6 6 2 cardioid r 0 -1 2 I 3/2 2 x
y [4] V= I -sin o (e) symmetic about y-axis (16-1) x "(-1,0) it (i.o)
* r= COS 20 II 14/ 2 5 FL 2 6 o y 12 2 is enough symmetriy about x- axis since () (t,I) COS20 = cosz(-) to, F x (1,0) (r,e) (1,TT) (1/2/25) (r,- 0) or (-r,TT-e) (1) symmetry about y- axis since r=c0520 = Cos2(11-0) (-r,o) r,e) (05(2IT-26) (r, II-G) = COS 2TT cos(-2G) - sin 217 sin(-2c) = (1) cos 20 - 0 = COS20 = r about origin too. That is r= cos20 = cos2 symmetry (IT+O) Hence, = cos(211'+26) (v,0) = COS 21T COS 20 -sinziisin20 = (1) (052G o = COS20 (-r,6) = r or (r, IIto)
7 r=sin (o) is 2 symmetric about y-axis since sin (- 2/2 sin(Q) r symmetric about x-axis since sin(II- 0)= 2 sinTT cost 1 + sin (- cos(II) = 0 sin (Q) (-1) = sin/ o)=r Hence, it is symmetric about origin II II 2II II O II 3 2 3 3 r 1 0 1/2 2 V2 1/2 I r: sin 1/5 -1 x ,Tay
Graphing in Polar Coordinates
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