Appendi A: Area worked-ou s o Odd-Numbered Eercises Do no read hese worked-ou s before aemping o do he eercises ourself. Oherwise ou ma mimic he echniques shown here wihou undersanding he ideas. Bes wa o learn: Carefull read he secion of he ebook, hen do all he odd-numbered eercises and check our answers here. If ou ge suck on an eercise, hen look a he worked-ou here. Find he radius of a circle ha has circumference inches. Le r denoe he radius of his circle in inches. Thus πr =, which implies ha r = π inches. Find he radius of a circle ha has circumference 8 more han is diameer. Le r denoe he radius of his circle. Thus he circle has circumference πr and has diameer r. Because he circumference is 8 more han diameer, we have πr = r + 8. Thus (π )r = 8, which implies ha r = 4 π. Suppose ou wan o design a 400-meer rack consising of wo half-circles conneced b parallel line segmens. Suppose also ha ou wan he oal lengh of he curved par of he rack o equal half he oal lengh of he sraigh par of he rack. Wha dimensions should he rack have? Le equal he oal lengh of he sraigh par of he rack in meers. We wan he curved par of he rack o have oal lengh. Thus we wan + = 400. Solving his equaion for, we ge = 800 meers. Thus each of he wo sraigh pieces mus be 400 meers long. Hence we ake c = 400. meers in he figure in he below. We also wan he oal lengh of he wo half-circles o be 400 meers. Thus we wan πd = 400. Hence we ake d = 400 4.44 meers in he figure above. π Suppose a rope is jus long enough o cover he equaor of he Earh. Abou how much longer would he rope need o be so ha i could be suspended seven fee above he enire equaor? Assume he equaor of he Earh is a circle. This assumpion is close enough o being correc o answer a quesion ha requires onl an approimaion. Assume he radius of he Earh is r, measured in fee (noe ha we do no need o know he value of r for his eercise). For a rope o cover he equaor, i needs o have lengh πr fee. For a rope o be suspended seven fee above he equaor, i would need o have lengh π(r + ) fee, which equals (πr + 4π) fee. In oher words, o be suspended seven fee above he equaor, he rope would need o be onl 4π fee longer han a rope covering he equaor. Because 4π 4 = 44, he rope would need o be abou 44 fee longer han a rope covering he equaor. 9 Find he area of a riangle ha has wo sides of lengh and one side of lengh 0. B he Phagorean Theorem (see figure below), he heigh of his riangle equals, which equals. d c
Appendi A: Area A riangle ha has wo sides of lengh and one side of lengh 0. Thus he area of his riangle equals. (a) Find he disance from he poin (, ) o he line conaining he poins (, ) and (, 4). Use he informaion from par (a) o find he area of he riangle whose verices are (, ), (, ), and (, 4). Thus he disance from he poin (, ) o he line conaining he poins (, ) and (, 4) is he disance from he poin (, ) o he poin ( 94, 8 ). This disance equals ( 94 ) + ( 8 ), which equals, which equals 4. We will consider he line segmen connecing he poins (, ), and (, 4) o be he base of his riangle. In par (a), we found ha he heigh of his riangle equals 4. 4 (a) To find he disance from he poin (, ) o he line conaining he poins (, ) and (, 4), we firs find he equaion of he line conaining he poins (, ) and (, 4). The slope of his line equals 4 ( ) ( ), which equals. Thus he equaion of he line conaining he poins (, ) and (, 4) is which can be rewrien as 4 =, = +. To find he disance from he poin (, ) o he line conaining he poins (, ) and (, 4), we wan o find he equaion of he line conaining he poin (, ) ha is perpendicular o he line conaining he poins (, ) and (, 4). The equaion of his line is =, which can be rewrien as = + 9. To find where his line inersecs he line conaining he poins (, ) and (, 4), we need o solve he equaion + = + 9. Simple algebra shows ha he o his equaion is = 94. Plugging his value of ino he equaion of eiher line shows ha = 8. Thus he wo lines inersec a he poin ( 94, 8 ). 4 The riangle wih verices (, ), (, ), and (, 4), wih a line segmen showing is heigh. The base of he riangle is he disance beween he poins (, ) and (, 4). This disance equals 4. Thus he area of he riangle (one-half he base imes he heigh) equals ( 4 4 which equals 4. ), [There are easier was o find he area of his riangle, bu he echnique used here gives ou pracice wih several imporan conceps.] Find he area of he riangle whose verices are (, 0), (9, 0), and (4, ). Choose he side connecing (, 0) and (9, 0) as he base of his riangle. Thus he riangle below has base 9, which equals. 4 9 The heigh of his riangle is he lengh of he red line shown here; his heigh equals he second coordinae
of he vere (4, ). In oher words, his riangle has heigh. Thus his riangle has area, which equals. Suppose (, ), (, ), and (, ) are hree verices of a parallelogram, wo of whose sides are shown here. (a) (a) Find he fourh vere of his parallelogram. Find he area of his parallelogram. Consider he horizonal side of he parallelogram connecing he poins (, ) and (, ). This side has lengh. Thus he opposie side, which connecs he poin (, ) and he fourh vere, mus also be horizonal and have lengh. Thus he second coordinae of he fourh vere is he same as he second coordinae of (, ), and he firs coordinae of he fourh vere is obained b adding o he firs coordinae of (, ). Hence he fourh vere equals (8, ). The base of his parallelogram is he lengh of he side connecing he poins (, ) and (, ), which equals. The heigh of his parallelogram is he lengh of a verical line segmen connecing he wo horizonal sides. Because one of he horizonal sides lies on he line = and he oher horizonal side lies on he line =, a verical line segmen connecing hese wo sides will have lengh. Thus he parallelogram has heigh. Because his parallelogram has base and heigh, i has area. Find he area of his rapezoid, whose verices are (, ), (, ), (, ), and (, ). =, a verical line segmen connecing hese wo sides will have lengh. Thus he rapezoid has heigh. Because his rapezoid has bases and and has heigh, i has area ( + ), which equals 9. 9 Find he area of he region in he -plane under he line =, above he -ais, and beween he lines = and =. The line = inersecs he line = a he poin (, ). The line = inersecs he line = a he poin (, ). Thus he region in quesion is he rapezoid shown above. The parallel sides of his rapezoid (he wo verical sides) have lenghs and, and hus his rapezoid has bases and. As can be seen from he figure above, his rapezoid has heigh 4. Thus he area of his rapezoid is ( + ) 4, which equals 8. Le f () =. Find he area of he region in he -plane under he graph of f, above he -ais, and beween he lines = and =. The region under consideraion is he union of wo riangles, as shown here. One base of his rapezoid is he lengh of he side connecing he poins (, ) and (, ), which equals. The oher base of his rapezoid is he lengh of he side connecing he poins (, ) and (, ), which equals. The heigh of his rapezoid is he lengh of a verical line segmen connecing he wo horizonal sides. Because one of he horizonal sides lies on he line = and he oher horizonal side lies on he line One of he riangles has base and heigh and hus has area. The oher riangle has base and heigh and hus has area. Thus he area of he region under consideraion equals +, which equals 9. Find he area inside a circle wih diameer. A circle wih diameer has radius. Thus he area inside his circle is π( ), which equals 49π 4. Find he area inside a circle wih circumference fee. Le r denoe he radius of his circle in fee. Thus πr =, which implies ha r = π. Thus
4 Appendi A: Area he area inside his circle is π( π ) square fee, which equals square fee. 4π Find he area inside he circle whose equaion is + + 0 =. To find he radius of he circle given b he equaion above, we complee he square, as follows: = + + 0 = ( ) 9 + ( + ) = ( ) + ( + ) 4. Adding 4 o boh sides of his equaion gives ( ) + ( + ) =. Thus we see ha his circle is cenered a (, ) (which is irrelevan for his eercise) and ha i has radius. Thus he area inside his circle equals π, which equals π. 9 Find a number such ha he area inside he circle is 8. + = Rewriing he equaion above as + = ( ), we see ha his circle has radius. Thus he area inside his circle is π ( ), which equals π. We wan his area o equal 8, which means we need o solve he equaion π = 8. Thus = 4 π. Use he following informaion for Eercises : A sandard DVD disk has a -cm diameer (cm is he abbreviaion for cenimeers). The hole in he cener of he disk has a 0.-cm radius. Abou 0. megabes of daa can be sored on each square cm of usable surface of he DVD disk. Wha is he area of a DVD disk, no couning he hole? The DVD disk has radius cm (because he diameer is cm). The area inside a circle wih radius cm is π square cm. The area of he hole is 0. π square cm, which is 0.π square cm. Subracing 0.π from π, we see ha he DVD disk has area.4π square cm, which is approimael. square cm. A movie compan is manufacuring DVD disks conaining one of is movies. Par of he surface of each DVD disk will be made usable b coaing wih laser-sensiive maerial a circular ring whose inner radius is. cm from he cener of he disk. Wha is he minimum ouer radius of his circular ring if he movie requires 00 megabes of daa sorage? Because 0. megabes of daa can be sored in each square cm of usable surface, he usable surface mus have area a leas 00 square cm, or 0. abou. square cm. If he ouer radius of he circular ring of usable area is r, hen he usable area will be πr. π. Thus we solve he equaion πr. π =. for r, geing r 4.9 cm. Suppose a movie compan wans o sore daa for era feaures in a circular ring on a DVD disk. If he circular ring has ouer radius.9 cm and 00 megabes of daa sorage is needed, wha is he maimum inner radius of he circular ring for he era feaures? Because 0. megabes of daa can be sored in each square cm of usable surface, he usable surface mus have area a leas 00 square cm, 0. or abou.984 square cm. If he inner radius of he circular ring of usable area is r, hen he usable area will be.9 π πr. Thus we solve he equaion.9 π πr =.984 for r, geing r.9 cm. Find he area of he region in he -plane under he curve = 4 (wih ) and above he -ais.
Square boh sides of he equaion = 4 and hen add o boh sides. This gives he equaion + = 4, which is he equaion of a circle of radius cenered a he origin. However, he equaion = 4 forces o be nonnegaive, and hus we have onl he op half of he circle. Thus he region in quesion, which is shown above, has half he area inside a circle of radius. Hence he area of his region is π, which equals π. 9 Using he answer from Eercise, find he area of he region in he -plane under he curve = 4 (wih ) and above he -ais. The region in his eercise is obained from he region in Eercise b sreching vericall b a facor of. Thus b he Area Srech Theorem, he area of his region is imes he area of he region in Eercise. Thus his region has area π. 4 Using he answer from Eercise, find he area of he region in he -plane under he curve = 4 (wih ) and above he - 9 ais. Define a funcion f wih domain he inerval [, ] b f () = 4. Define a funcion h wih domain he inerval [, ] b h() = f ( ). Thus h() = f ( ) = 4 ( ) = 4 9. Hence he graph of h is obained b horizonall sreching he graph of f b a facor of (see Secion.). Thus he region in his eercise is obained from he region in Eercise b sreching horizonall b a facor of. Thus b he Area Srech Theorem, his region has area π. 4 Find he area of he region in he -plane under he curve = + 4, above he -ais, and beween he lines = and =. The curve = + 4 is obained b shifing he curve = 4 up uni. Thus we have he region above, which should be compared o he region shown in he o Eercise. To find he area of his region, we break i ino wo pars. One par consiss of he recangle shown above ha has base 4 and heigh (and hus has area 4); he oher par is obained b shifing he region in Eercise up uni (and hus has area π, which is he area of he region in Eercise ). Adding ogeher he areas of hese wo pars, we conclude ha he region shown above has area 4 + π. In Eercises 4 0, find he area inside he ellipse in he -plane deermined b he given equaion. 4 + = Rewrie he equaion of his ellipse as + 4 =. Thus he area inside his ellipse is 4 π. 4 + =
Appendi A: Area Rewrie he equaion of his ellipse in he form given b he area formula, as follows: = + = = +. + Thus he area inside he ellipse is π, which equals π. Mulipling numeraor and denominaor b, we see ha we could also epress his area as π. 49 + = To pu he equaion of he ellipse in he form given b he area formula, begin b dividing boh sides b, and hen force he equaion ino he desired form, as follows: = + = = +. + Thus he area inside he ellipse is π, which equals π. Mulipling numeraor and denominaor b, we see ha we could also epress his area as π. Find a posiive number c such ha he area inside he ellipse + c = is. To pu he equaion of he ellipse in he form given b he area formula, begin b dividing boh sides b, and hen force he equaion ino he desired form, as follows: = + c = = + c. c + Thus he area inside he ellipse is π c, which equals π c. We wan his area o equal, so we mus solve he equaion π c =. Squaring boh sides and hen solving for c gives c = π 8. Find numbers a and b such ha a > b, a + b =, and he area inside he ellipse is π. a + b = The area inside he ellipse is πab. Thus we need o solve he simulaneous equaions a + b = and ab =. The firs equaion can be rewrien as b = a, and his value for b can hen be subsiued ino he second equaion, giving he equaion a( a) =. This is equivalen o he equaion a a + = 0, whose s (which can be found hrough eiher facoring or he quadraic formula) are a = and a =. Choosing a = gives b = a =, which violaes he condiion ha a > b. Choosing a = gives b = =. Thus he onl o his eercise is a =, b =. Find a number such ha he area inside he ellipse is. 4 + 9 = Dividing he equaion above b, we have = 4 + 9 = 4 = ( + 9 ) + ). ( Thus he area inside his ellipse is π, which equals π. Hence we wan π =, which means ha = 0 π.