Wrm up Recll from lst time, given polr curve r = r( ),, dx dy dx = dy d = (r( )sin( )) d (r( ) cos( )) = r0 ( )sin( )+r( ) cos( ) r 0 ( ) cos( ) r( )sin( ).. Lst time, we clculted tht for crdioid r( ) =+sin( ), dy = cos( )( + sin( )) dx =(+sin( ))( sin( )). For wht re tngent lines to this crdioid horizontl? verticl?. Clculte dy/dx for following polr curves. Wht re slopes t which ech curve crosses x nd y xes? () r( ) = () r( ) = (c) r( ) = cos( )
9. Are AREAS AND LENGTHS IN PLAR CRDIN In this section we develop formul for re of Thus Form Recll tht re of wedge of circle with is givenweyneed to use given y ngle polr eqution. from formul 80_ch09_ptg0_hr_5-5.qk_80_ch09_ptg0_hr_5-5 /7/ :9 AM Pge 55 circle r A so rn pproxi nd ( / ) = = A = r r. where, s in Figure, r is rdius nd is rdin m = Formul follows from fct tht re of sector Let r = f ( )FIGURE e positive continuous function over ngle: Afor # % "% r[, ],. AREAS (See lso Exercise 69INinP r nd SECTIN 9. AND LENGTHS It ppers from F Let e region, illustrted in Figure, ounded = nd =. let R e region ounded y r ndfigure rys nd y rys nd, where f isin positive sums re co Ri Thus from Formul we hve 0 $ ( %. We divide intervl %, & into su ' 80_ch09_ptg0_hr_5-5.qk_80_ch09_ptg0_hr_5-5 /7/ = i f( i*),,..., n nd equl width ). The rys i th "A i # f $ i*%& " regions with centrl ngle = ) i- i $ i$. If we choos %, i &, n re )Ai of ith region is pproxim = nd so n i$ pproximtion to totl re A of is =of circle with centrl ngle ) nd It rdius f i*".ppe (See F refore n Î A of Unless reg = polr orwise not A # f $ i*%& " = i My not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prty content m Copyright 0 Cengge Lerning. All Rights Reserved. Editoril review hs deemed tht ny suppressed does not mterilly ffect Figure overll lerning Cenggepproximtion Lerning reserves right to remove It content ppers from experience. tht indditionl cont im FIGURE FIGURE sums in re Riemnn sums for function t$ % # f " Thus from Form We cn pproximte re of R y slicing it into wedges 80_ch09_ptg0_hr_5-5.qk_80_ch09_ptg0_hr_5-5 /7/ :9 AM Pge 55 * = f( i ) = i- Are = Î i = lim n " n l # i # f $ i*%& Formul " y is # foften $ %& nd so n pproxi It reforeppers plusile (nd cn in fct e proved) t = A of polr region is = Let r = f ( ) e positive continuous function for over [, ], SECTIN 9.nd AREAS AND LENGTHS IN P It ppers from F = nd =. let R e region ounded y r ndfigure rys A y sums %& understn with in d re Ri # f $ FIGURE Thus from Formul we hve = f( *) r=cos = = When we pply ry t y # f $rotting i*%& " i- π = "A = i Formul is often written s nd so n pproximtion to totl reva EXAMPLE of is F = It refore ppe n r cos. Î A of polr reg A A " SLUTIN = y# f $r i*%&d " The i i i Notice from Figur It ppers from Figure tht pproximtion in from im thtrottes π FIGURE =_ FIGURE sums in re Riemnn sums for function t$ % # understnding tht r The f $ %. Note similrity f We cn pproximte re of Rwith y slicing it into wedges = i f( i*) is pproximtely When we pply Formul or it is helpful to think A of re of ech wedge n π FIGURE r=cos y rotting ry throughlim tht strts with ngle is nd en = = Formul often i- Z Z " # f $ i*%& " y # f $ %& A = ri Î =. A of polr region is = SLUTIN The curve r cos ws sketched in Exm V EXAMPLE F FIGURE n l # i A= = re enclosed f ( ).y one loop of VSoEXAMPLE rfind r= sin(nd cn in fct e proved) t It refore ppers plusile r cos. f ( i ) π =_ Notice from Figure tht region enclosed y crdio right outside f $ %& understn d Form thtrottes to π A%y 'with. #Therefore 5π from $%' = SLUTIN The = 6
Exmple For r = r( ) positive, A = r Find re enclosed y one loop of curve r = cos( ). - R r=cos(θ) A = = r (cos( )) - ne choice of nd : = /, = /. A = Z / / cos ( ) = Z / / ( + cos( )) = ( + / sin( )) = / ( /+0) ( /+0) = /6 = /8.
You try For r = r( ) positive, A = r. Clculte re enclosed y crdioid r =+sin(x).. Grph curve r =sin( ) nd clculte re enclosed y this curve. (Creful The function sin( ) isn t lwys positive, nd we wnt re Look t picture nd decide how to rek up prolem.)
Are eween curves Let R e region ounded etween curves r = f( ) nd r = g( ) (where 0 pple g( ) pple f( )) ndrys = nd =. = r=g( ) The re of R is given y = A = r out Z = r in = (f ( ) g ( )) (r out r in) Z Are eween curves For 0 pple r in pple r out, A = ((r out r in)) Exmple: Find re of region tht lies inside circle r =sin( ) nd outside crdioid r =+sin( ). Answer: First, grph functions. r= sin(θ) R Next, identify region nd ounds. Intersection points: sin( ) =+sin( ), sosin( ) =, so = /6, 5 /6. r=+sin(θ) - So A = Z 5 /6 /6 Setting up integrl: r out = sin( ), r in = + sin( ) (Check Are oth r s positive over intervl? If not, do we hve to fix nything?) X (( sin( )) ( + sin( )) )
You try: For r = r( ) positive, A = For 0 pple r in pple r out, A = r ((r out r in)). We showed tht re of region tht lies inside circle r =sin( ) nd outside crdioid r =+sin( ) is given y integrl A = Z 5 /6 /6 (( sin( )) ( + sin( )) ). Check tht this integrl evlutes to.. For following regions, (i) grph functions nd identify region, (ii) decide how to rek your integrl up using ove re formuls, (ii) clculte re of region. () Let R e region inside r =(/) sin( ) nd outside r =/. () Let R e region enclosed y oth r =+sin( ) nd r = sin( ).
Arc length (gin) Agin We strt with Z ` = d`, where d` = p dx + dy. When we hd y = f(x), wemultipliedydx/dx to get d` = p +(dy/dx) dx. When we hd x = f(y), wemultipliedydy/dy to get d` = p (dx/dy) +dy. When we hd x(t) nd y(t), wemultipliedydt/dt to get Now we hve q d` = dx dt + dy dt dt. x( ) =r( ) cos( ) nd y( ) =r( ) sin( ), so multiply y / to get s dx d` = + (exctly sme s t cse ove) dy Arc length (gin) Agin We strt with Z ` = d`, where d` = p dx + dy. Now we hve x( ) =r( ) cos( ) nd y( ) =r( ) sin( ), so multiply y / to get s dx d` = + dy (exctly sme s t cse ove), now where dx = d r( ) cos( ) = r0 ( ) cos( ) r( )sin( ), nd dy = d r( )sin( ) = r0 ( )sin( )+r( ) cos( ).
Arc length of prmetric curves ` = = s dx + dy, where x( ) =r( ) cos( ), y( ) =r( )sin( ). Exmple: Agin Let s clculte rc length of n rc of ngle A of circle of rdius R. The circle of rdius R is given y polr curve r( ) =R. So x( ) =R cos( ) y( ) =R sin( ), nd n rc of ngle A is curve trced from =0to = A: R A d` = p ( dx/ = R sin( ) dy/ = R cos( ) R sin( )) +(Rcos( )) = p R dt So ` = Z A =0 d` = Z A 0 R = R A =0 = RA 0 = RA. You try ` = = s dx + dy, where x( ) =r( ) cos( ), y( ) =r( )sin( ).. Set up nd simplify n integrl tht gives length of curve r = from = / to =.. Clculte length of curve r = cos( ) from =0to =. (Hint: fter doing derivtives, efore setting up, recll doule ngle formuls.). Clculte length of curve r = e, from =to =.