ENGI 4 Fundmentls Prmetric Curves Pge 1.A.1 1.A Exmples for the Sketching of Prmetric Curves 3 A curve in is one-dimensionl object. To locte ny point on tht curve requires the vlue of just one prmeter ( rel number). The Crtesin prmetric equtions of ny curve re therefore x = x t, y = y t, z = z t, where t is ny rel number. () () ( ) The Crtesin vector prmetric eqution is r t = x t ˆ i + y t ˆ j + z t k ˆ, where t is ny rel number. () () () ( ) If the prmeter t is the time, then r(t) describes the loction of prticle t ny time t. dr The velocity of the prticle is just () dx ˆ dy ˆ dz v t = = i + j + kˆ. dt dt dt dt dv d r The ccelertion is () d x ˆ d y ˆ d z t = = = i + j + kˆ. dt dt dt dt dt Exmple 1.A.1 Sketch the curve in whose Crtesin eqution in prmetric form is x = t cos t, y = t sin t ( ) x + y = t cost + sin t = t (distnce from O) = t y x = tn t θ = ngle t (for t > 0)
ENGI 4 Fundmentls Prmetric Curves Pge 1.A. Exmple 1.A.1 (continued) Therefore the curve is spirl outwrds from O, with period π. Incorporting negtive vlues of the prmeter t yields mirror imge of this curve: ( ) ( ) t t x tcos t = x, y tsin t =+y reflection in y xis of t > 0 is t < 0. In polr coordintes (section 1.), the eqution of this spirl is just r = θ.
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.3 Exmple 1.A. The prmetric form of the Crtesin eqution of curve in is 3 x = t, y = t () Sketch the curve. (b) Wht hppens to the principl unit tngent ˆ dr dr T = dt dt t the origin? () dx dy = t, = 3t dt dt dx dy The only vlue of t t which ny of x, y,, is zero is t = 0. dt dt t 0 x 0, y 0, dx dy 0 nd dt dt t 0 x 0, y 0, dx dy 0 nd dt dt t 0 x 0, y 0, dx dy 0 nd dt dt Therefore, for t < 0, the curve is moving up nd left through the fourth qudrnt, rriving t the origin t t = 0. Therefter, the curve is moving up nd right through the first qudrnt. No prt of the curve is to the left of the y-xis. dy 3 3 = dy dx = t = t dy, 0 lim = 0 ( t ) dx dt dt t t 0 dx There must therefore be horizontl tngent (nd cusp) t the origin. Sketch of 3 x = t, y = t :
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.4 Exmple 1.A. (continued) Exmintion of concvity will help to confirm the behviour ner the origin. d y d dy d dy dt dt d dy dx dx = = = dx dx dx dt dt dx dx dt dt dt dt 3 3 = = = d y d t d ( t) () t ( t) dx dt t dt 4t The curve is therefore concve down everywhere in the fourth qudrnt (t < 0) nd concve up everywhere in the first qudrnt (t > 0). 3 (b) The tngent vector is d dx, dy t,3t t,3t d r r = = = = t 4+ 9t dt dt dt dt The unit tngent vector, everywhere on the curve except t the origin, is ˆ dr dr t 1 T = =, 3t dt dt t 4+ 9t t > 0 t = t t / t = +1 1 lim Tˆ = +, 0 = + 1, 0 = + ˆi t 0+ 4+ 0 t < 0 t = t t / t = 1 1 lim Tˆ = ( 1), 0 = 1, 0 = ˆi lim T ˆ t 0 4+ 0 t 0 + The unit tngent therefore is undefined t the origin. It flips direction bruptly, from i to +i, s the curve psses through the cusp t the origin; the curve reverses direction suddenly t the cusp.
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.5 Exmple 1.A.3 For the curve whose Crtesin eqution in prmetric form is 3 x = t, y = t 3t () Find the tngents t the point (x, y) = (3, 0). (b) Sketch the curve. () dx x = t = t dt 3 dy y = t 3t = 3t 3 dt x= 3 t =± 3 nd ( ) y = 0 t t 3 = 0 t = 0, ± 3 ( xy) ( ) t, = 3,0, t=± 3 The curve therefore psses through the point (3, 0) twice. ( ) ( ) 3 3 1 dy dy dx dy ± 3 = = = = ± dx dt dt dx ( 3,0) ± 3 ± 3 3 The two slopes re distinct. The curve therefore crosses itself t (3, 0). In Crtesin coordintes, the equtions of the two tngents to the curve t the point (3, 0) re y = ± 3 x 3 ( )
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.6 Exmple 1.A.3 (continued) (b) x = 0 t t = 0 x > 0 elsewhere. 3 ( ) y = t 3t = t t 3 = 0 t t = 0, ± 3 ( t ) dy 3t 3 = = 3 1 = 0 t t =± 1 dx t t dy dx is undefined t t = 0. dx dy The vlues of t, t which t lest one of x, y,, re zero, re dt dt t = 3, 1, 0, + 1, + 3 : Construct tble to id in sketching the curve: t t = ± 1 ( x, y) = ( 1, ) = ± 3 ( x, y) = ( 3,0 )
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.7 Exmple 1.A.3 (continued) Sketch of 3 x = t, y = t 3t: Concvity: d y d dy d dy dt dt d dy dx dx = = = dx dx dx dt dt dx dx dt dt dt dt d y d 3 3 1 3 3 = t t t t t = + dx dt d y = dx ( t + ) 3 1 4t 3 The curve is therefore concve up for t > 0 nd concve down for t < 0. This confirms the sketch. ( ) ( )
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.8 Exmple 1.A.4 Determine the shpe of the curve whose Crtesin eqution in prmetric form is x = cos t, y = sin t, z = t Exmine the projections of the curve onto the three coordinte plnes: In the x-y plne z = 0 nd x + y = cos t + sin t = 1 which is circle, centre O, rdius 1. Top view: In the y-z plne x = 0 nd y = sin t = sin z In the x-z plne y = 0 nd x = cos t = cos z Side views: The curve is therefore helix, centred on the z xis, rdius 1, rotting once round the z xis for every chnge of π in z.
ENGI 4 Fundmentls Prmetric Curves Pge 1.A.9 Exmple 1.A.4 (continued) Modified Mple plot: A Mple file tht genertes plot of this helix is vilble from the course web site, in the progrms directory: "http://www.engr.mun.c/~ggeorge/4/progrms/".
ENGI 4 Fundmentls Prmetric Curves Pge 1.B.1 1.B Tngentil nd Norml Components of Velocity nd Accelertion dr The tngent vector to curve r(t) is () dx ˆ dy ˆ dz T t = = i + j + kˆ dt dt dt dt If the prmeter t is the time, then the tngent vector is lso the velocity vector v(t). The tngentil component v T of velocity v(t) is just the speed v(t). There is no component of velocity in the norml plne. dr dx dy dz The speed v(t) is sclr quntity: vt () = T() t = = + + dt dt dt dt But the rc length (distnce mesured long the curve) s is defined by ds dx dy dz = + + dt dt dt dt dr ds dx dy dz Therefore the speed () v t = = = + + dt dt dt dt dt nd the unit tngent vector is ˆ T v dr ds dr T = = = =. T v dt dt ds dr As seen in Exmple 1.A. bove, ˆT is ill-defined where = 0. dt 3 As curve trvels through, its tngent vector points stright hed, defining norml plne t right ngles to tht tngent vector. Imgine tht roller coster cr is trvelling long the curve, with the front in the direction of trvel nd oriented so tht the side doors re in the direction in which the cr is instntneously turning. Then the direction in which T ˆ is chnging defines the principl unit norml ˆN, (except where the curve is stright or hs point of inflexion).
ENGI 4 Fundmentls Prmetric Curves Pge 1.B. By definition, the mgnitude of ny unit vector is 1 nd therefore is bsolutely constnt. Only the direction of unit vector cn chnge. The nturl prmeter to use for ny curve (though usully not the most convenient in prctice) is the distnce trvelled long the curve: the rc length s. Therefore define the principl norml vector to be the derivtive of the unit tngent vector with respect to rc length: N dtˆ dtˆ ds = = ds dt dt from which it follows tht the unit principl norml vector is ˆ ˆ ˆ ˆ ˆ dt dt dt dt N = = ds ds dt dt The mgnitude of the principl norml vector is mesure of how shrply the curve is turning. It is therefore the curvture, dtˆ dtˆ ds κ = N = = ds dt dt nd N = κ Nˆ. The tngent nd principl norml vectors re orthogonl to ech other everywhere on the curve. A third unit vector, orthogonl everywhere to both ˆT nd ˆN, is the unit binorml vector, defined simply s Bˆ = Tˆ N ˆ These three unit vectors form n orthonorml set of vectors t every point on the curve where they re defined.
ENGI 4 Fundmentls Prmetric Curves Pge 1.B.3 Exmple 1.B.1 Find the unit tngent, norml nd binorml vectors everywhere on the helix r () t = x, y, z = cos t,sin t, t = dr d T cos t, sin t, t sin t, cos t,1 dt = dt = ds dr = = sin t + cos t + 1 = dt dt ˆ dr ds T = = dt dt sin t, cos t,1 dtˆ dtˆ ds d sin t, cos t,1 1 1 N = = = = cos t, sin t, 0 ds dt dt dt 1 ˆ N N = N = = cos t, sin t, 0 N [The unit principl norml therefore points directly towrds the z xis t ll times.] i j k ˆ ˆ ˆ 1 1 B = T N = sin t cost 1 = sin t, cos t,1 cost sin t 0 One cn esily show tht TN ˆ i ˆ = NB ˆ i ˆ = BT ˆ i ˆ = 0 [ll three vectors re orthogonl] nd tht TT ˆ i ˆ = NN ˆ i ˆ = BB ˆ i ˆ = 1 [ll three vectors re unit vectors]
ENGI 4 Fundmentls Prmetric Curves Pge 1.B.4 We know tht the velocity vector is purely tngentil: v = v Tˆ. The ccelertion vector is therefore dv dv ˆ ˆ dt = = T + v dt dt dt But ˆ ˆ ˆ ˆ ˆ ˆ dt dt dt dt ˆ dt ds N = = N = Nˆ = κv N ˆ dt dt dt dt ds dt dv = Tˆ + dt κv N ˆ The tngentil nd norml components of ccelertion re therefore T dv = nd dt N = κv An lterntive form for the norml component of ccelertion is ˆ d d d d d N v T r r r = = dt dt dt dt dt There is no binorml component of ccelertion.
ENGI 4 Fundmentls Prmetric Curves Pge 1.B.5 Exmple 1.B. Find the tngentil nd norml components of velocity nd ccelertion everywhere on the helix r () t = x, y, z = cos t,sin t, t dr vt = v = v = = sin t + cos t + 1 = dt T Tˆ dv = = dt v = = v 0 sin t, cos t,1 dtˆ cos t, sin t, 0 = dt ˆ cos sin 0 dt t + t + N = v = = 1 dt OR κ = dt v = dt = ˆ cos t + sin t + 0 1 1 ( ) N = κ v = = 1 OR d v = = cos t, sin t, 0 = = 1 dt + = nd = 0 = = 1 T N T N Therefore v = T ˆ nd = 0Tˆ + 1Nˆ
ENGI 4 Fundmentls Lines nd Plnes Pge 1-01 1.1 Eqution of Plne A non-zero vector n in ú 3 will fix the orienttion of plne, to be t right ngles to n. Let P be point tht is known to be on the plne. Let the Crtesin coordintes of P be (x o, y o, z o ). Its position vector, = OP = xo, yo, zo, llows us to pick out exctly one plne from the infinite set of prllel plnes tht shre the sme orienttion defined by the plne norml vector n. The two vectors, together, llow us to define single plne completely. Let r = OR = x, y, z be the position vector of generl point R, with Crtesin coordintes (x, y, z), in ú 3. But OR = OP + PR PR = OR OP = r Note tht the norm l vector n is t right ngles to ny vector lying in the plne Π. R on Π PR n ("perpendiculr to") PR i n = 0 r i n = OR ( ) 0 rn i = n i which is the vector eqution of plne.
ENGI 4 Fundmentls Lines nd Plnes Pge 1-0 Let n = ABC,, then n i rn i, (so tht A, B, C re the Crtesin components of the norml vector), ( ) = Axo + Byo + Czo = D constnt = Ax + By + C z The vector eqution of the plne, rn i = n i, then leds to the Crtesin eqution of the plne: Ax + By + Cz + D = 0 Exmple 1.1.1 Find n eqution for the plne through the point (3,, 4), which is norml to the vector ˆi + 3ˆj + k ˆ. n = ABC,, =,3,1 = x, y, z = 3,,4 o o o rn i = n i x + 3y + 1z = 3 + 3 + 1 4 ( ) Therefore the Crtesin eqution of the required plne is x + 3y + z = 4
ENGI 4 Fundmentls Lines nd Plnes Pge 1-03 Exmple 1.1. Find unit vector orthogonl to the plne x y + z = 7 From the Crtesin eqution of plne, one my red off the Crtesin coordintes of vector tht is t right ngles to tht plne. n = 1,, ( ) n = = 1 + + = 9 = n 3 Therefore unit norml is n ˆ = 1,, 1 3 [Note tht there is one other cceptble finl nswer, nmely the unit vector tht points in the opposite direction, n ˆ = 1 3 1,,.]
ENGI 4 Fundmentls Lines nd Plnes Pge 1-04 Equtions of Line A line L is determined uniquely by two vectors, line direction vector, v, which orients the line in ú 3, nd the position vector,, of point P known to be on tht line. Let R be generl point, with Crtesin coordintes (x, y, z), tht is constrined to lie on the line L. OR = OP + PR r = + PR But the line L is prllel to the direction vector v. PR = t v for some vlue of t. Therefore the prmetric vector eqution of the line is r = + t v ( t ) [Note tht there is one free prmeter, t, which cn be ny rel number. The number of free prmeters (one) mtches the dimension of the geometric object (the line is onedimensionl object).] The Crtesin equtions of the line cn be found from the vector prmetric form.
ENGI 4 Fundmentls Lines nd Plnes Pge 1-05 Exmple 1.1.3 Find the Crtesin equtions of the line tht psses through the point (3, 1, ) nd tht is prllel to the vector ˆ i ˆ j + 3k ˆ. = 3, 1,, v = 1,, 3 The vector prmetric form is r = + t v or ( ) ( ) ( ) x, yz, = 3 + t, 1 t, + 3t Mking t the subject of ll three simultneous equtions x = 3 + t, y = 1 t nd z = + 3t, we obtin the Crtesin symmetric form for the equtions of the line: t = x 3 y 1 z = = 1 3 Generl cse: A line r = + t v which psses through the point (x o, y o, z o ) nd is prllel to the vector v = v1, v, v3, (where v 1, v, v 3 re ll non-zero) hs Crtesin symmetric form x xo y yo z zo = = v v v 1 If v 1 = 0, then seprte out the eqution x = x o. If v = 0, then seprte out the eqution y = y o. If v 3 = 0, then seprte out the eqution z = z o. 3
ENGI 4 Fundmentls Lines nd Plnes Pge 1-06 Exmple 1.1.4 Find the Crtesin equtions of the line through (1, 1, 1) tht is perpendiculr to the plne x + 3z = 1. The norml vector n to the plne is t right ngles to the plne Π. The line L is lso t right ngles to the plne Π. Therefore the line s direction vector v must be prllel to n. v = n =, 0, 3 = 1, 1, 1 The vector prmetric form is r = + t v or ( ) ( ) x, yz, = 1+ t, 1, 1+ 3t Mking t the subject of the equtions where possible (for x nd z ), the Crtesin symmetric form then follows: x 1 z+ 1 y = 1, = 3
ENGI 4 Fundmentls Lines nd Plnes Pge 1-07 More Equtions of Lines nd Plnes Three non-colliner points, A, B, C, define plne. The three vectors joining the three points to ech other lie in tht plne. Any two of them my be used s the bsis for coordinte grid tht cn be lid out on the entire plne. The vector prmetric eqution of the plne then follows: r = + su + tv s, t where = ny one of OA, OB, OC; uv, = ny two of AB, BC, CA; nd u v = n. The other vector eqution for the plne cn be recovered from this form: Vectors AB nd AC re both in the plne n = AB AC is norml to the plne. Let R be generl point in spce, with Crtesin coordintes (x, y, z). If nd o nly if po int R lies in the plne, then ARi AB AC = 0 r i n = 0 ( ) ( )
ENGI 4 Fundmentls Lines nd Plnes Pge 1-08 Exmple 1.1.5 Find the Crtesin eqution of the plne tht psses through the points A(1, 0, 0), B(, 3, 4) nd C( 1,, 1). AB = ( 1 ),( 3 0 ),( 4 0) = 1,3,4 AC =,, 1 u v = AB AC = n = 5, 9, + 8 ˆi ˆj kˆ 1 3 4 1 Let R be generl point in ú 3, with Crtesin coordintes (x, y, z). Then r = AR = ( x 1, ) ( y 0, ) ( z 0) = x 1, y, z For R to be on the plne, r i n = 0 AR i AB AC = 0 ( ) ( ) (x 1)( 5) + y( 9) + z(8) = 0 5x 9y + 8z = 5 or 5x + 9y 8z = Alterntive solution (next pge): 5
ENGI 4 Fundmentls Lines nd Plnes Pge 1-09 Alterntive Solution to Exmple 1.1.5: The plne Π is Ax + By + Cz + D = 0 This eqution must be true t ll three points on the plne. An under-determined liner system of three equtions for the four unknown coefficients then follows: A B C D (1, 0, 0) on Π: [ 1 0 0 1 0 ] (, 3, 4) on Π: [ 3 4 1 0 ] ( 1,, 1) on Π: [ 1 1 1 0 ] Row reduction leds to the reduced row echelon form [ 1 0 0 1 0 ] [ 0 1 0 9/5 0 ] [ 0 0 1 8/5 0 ] ( ) 9 8 ( ) ABCD,,, = D, 5D, 5D, D, D Select convenient non-zero vlue for D tht leves ll four coefficients s integers nd mkes t lest two of A, B nd C non-negtive. Therefore select D = 5: 5x + 9y 8z 5 = 0
ENGI 4 Fundmentls Lines nd Plnes Pge 1-10 Exmple 1.1.6 Find the ngle between the line L : x = 1, y z 3 = 3 nd the plne Π : x y + z = 4. nv i 1,,1 i 0,3, 4 sin θ = = =.459 nv 1+ 4+ 1 0+ 9+ 4 6 13 Therefore θ 6.9 Exmple 1.1.7 Find the intersection of the plnes Π 1 : x y + z = 1 nd Π : 3x 5y + z = 4. Solve the under-determined liner system x y z Π 1 [ 1 1 1 ] Π [ 3 5 4 ] RR R 3 R 1 : [ 1 1 1 ] [ 0 1 1 1 ] RR1 R 1 + R : [ 1 0 1 3 ] [ 0 1 1 1 ] x = t + 3, y = t + 1, z = t, t 0 ú
ENGI 4 Fundmentls Lines nd Plnes Pge 1-11 Exmple 1.1.7 (continued) The intersection is LINE. In vector prmetric form, it is x 3 y 1 z 0 t = = = 1 1 1 r = + t v, with = +3, 1, 0, nd v = +1, 1, 1,. Summry for Lines nd Plnes: Line Plne r Vector forms = + t v = point on line, v = direction vector r = + su + tv, = point on plne, u, v = vectors in plne OR rn i = n i, where n = u v Crtesin form x xo y yo z z = = v1 v v3 where = xo, yo, zo nd v = v, v, v 1 3 Ax + By + Cz + D = 0, n = A, BC, o
ENGI 4 Fundmentls Ares Pge 1-1 1.3 Are, Arc Length, Tngents nd Normls, Curvture With prmeteriztion: b ( ) A = f x dx tb dx A = y dt dt t where x(t ) =, x(t b ) = b, < b nd f (x) > 0 on [, b]. Exmple 1.3.1 Find the re enclosed in the first qudrnt by the circle x + y = 4. x = cos θ, y = sin θ. x = 0 θ = π x = θ = 0 dx x = cosθ = sinθ dθ
ENGI 4 Fundmentls Ares Pge 1-13 Exmple 1.3.1 (continued) 0 π / sinθ sinθ dθ 4 sin θ dθ ( )( ) A = = + π / 0 π / 1 1 = 4 ( 1 cosθ) dθ = θ sinθ 0 = π 0 ( 0 0) Therefore A = π Check: The re is the interior of qurter-circle. 1 1 ( π r ) = ( π ) 4 4 = π π / 0 Review of the Tngent: dx dy dz dr T =,, = dt dt dt dt The unit tngent is dr dr T = dt dt
ENGI 4 Fundmentls Arc Length, Tngent Pge 1-14 Arc Length In ú : ( s) ( x) ( y) Δ Δ + Δ In ú 3 : ( Δs) ( Δ x) + ( Δ y) + ( Δ z) ds dx dy dz dr = + + = dt dt dt dt dt The vector d r points in the direction of the tngent T to the curve defined prmetriclly dt by r = r(t). dr dr dr ds dr d T = = = t dt dt dt dt dt ds dr T = ds
ENGI 4 Fundmentls Arc Length, Tngent Pge 1-15 Exmple 1.3. () Find the rc length long the curve defined by r = 1 t sin t, 1cos t, sin t, from the point where t = 0 to the point ( ) 4 4 where t = 4π. (b) Find the unit tngent T. dx dt 1 1 cos 1 cos sin sin 4 4 () ( t) ( t) ( t) = = = = t dy dt = 1 ( sin t) = ( sin t cost) = sin tcost 4 4 dz = cost dt dr = dt sin t, sin tcos t, cos t ds dr = = + + = ( + ) + dt dt 4 sin t sin tcos t cos t sin t sin t cos t cos t = + = 1sin t cos t 1 ds dt 4π 0 [] = 1 s = 1dt = t 4π 0 Therefore s = 4π (b) dr dr T = dt dt d But r = 1 t ("for ll t" ) dt. Therefore T = sin t, sin tcos t, cost [Note tht, in this exmple, t itself is the rc length from the point (0, 1/4, 0), (where t = 0). There re very few curves for which the prmeteriztion is this convenient!]
ENGI 4 Fundmentls Arc Length, Tngent Pge 1-16 Review of Norml nd Binorml: T = 1 T i T = 1 dt dt 0 dt it + Ti = Ti = 0 ds ds ds dt d or ds = T 0 ds T (Note: unit vector cn never be zero). Select the principl norml N to be the direction in which the unit tngent is, instntneously, chnging: dt dt N ds = = ds dt dt The unit principl norml then follows: N dt = dt dt dt The binorml is t right ngles to both tngent nd principl norml: B = T N
ENGI 4 Fundmentls Curvture Pge 1-17 Curvture: The derivtive of the unit tngent with respect to distnce trvelled long curve is the dt principl norml vector, N =. Its direction is the unit principl norml N. Its ds mgnitude is mesure of how rpidly the curve is turning nd is defined to be the curvture: dt dt dr κ = N = = ds dt dt N = κ N 1 The rdius of curvture is ρ =. κ The rdius of curvture t point on the curve is the rdius of the circle which best fits the curve t tht point. Exmple 1.3. (continued) (c) Find the curvture κ(s) t ny point for which s > 0, for the curve r s = 1 s sin s, 1cos s, sin s ( ) ( ) 4 4 where s is the rc length from the point (0, 1/4, 0). From prt (b): T = sin s, sin scos s, cos s dt = sin scos s, cos s+ sin s, sin s = sin s, cos s, sin s ds 1 1 ρ = = κ + 1 sin s dt κ ( s) = = 1+ sin ds s
ENGI 4 Fundmentls Curvture Pge 1-18 Another formul for curvture: Let d nd ds d r r r = s = = = r, then dt dt dt r = s T dt r = st + s dt But dt dt dt ds = = ds dt ( κ N) s 3 r r = ( st) ( st + κs N) = 0 + κs B r r = κs 3, but s = r. Therefore κ = r r 3 r
ENGI 4 Fundmentls Curvture Pge 1-19 Exmple 1.3.3 Find the curvture nd the rdius of curvture for the curve, given in prmetric form by x = cos t, y = sin t, z = t. Assume SI units. Let c = cos t, s = sin t. r = c, s, t = r = s, c, 1 = r = c, s, 0 ( v ) ( ) r r ˆi ˆj kˆ = s c 1 = + s, c, s + c = + s, c, 1 c s 0 = s + c + = 1 r = s + c + = 1 r r 1 κ = = = 3 r Therefore nd κ = ( ) 1 1 m constnt 1 ρ = = m κ
ENGI 4 Fundmentls Conic Sections Pge 1-0 1.4 Conic Sections All members of the fmily of curves known s conic sections cn be generted, (s the nme implies), from the intersections of plne nd double cone. The Crtesin eqution of ny conic section is second order polynomil in x nd y. The only cses tht we shll consider in this course re such tht ny xis of symmetry is lined up long coordinte xis. For ll such cses, the Crtesin eqution is of the form A x + C y + D x + E y + F = 0 where A, C, D, E nd F re constnts. There is no xy term, so B = 0. The slope of the intersecting plne is relted to the eccentricity, e of the conic section. x + y = r A prmetric form is (x, y) = (r cos θ, r sin θ ), (0 < θ < π). x y + = 1 b where b = ( 1 e ) The circle is clerly specil cse of the ellipse, with e = 0 nd b = = r. The longest dimeter is the mjor xis (). The shortest dimeter is the minor xis (b). If mirror is mde in the shpe of n ellipse, then ll rys emerging from one focus will, fter reflection, converge on the other focus. A prmeteriztion for the ellipse is r( θ ) = cosθ ˆ i + bsin θ ˆj, ( 0 θ < π).
ENGI 4 Fundmentls Conic Sections Pge 1-1 y = 4x One vertex is t the origin. The centre nd the other vertex nd focus re t infinity. If mirror is mde in the shpe of prbol, then ll rys emerging from the focus will, fter reflection, trvel in prllel stright lines to infinity (where the other focus is). The primry mirrors of most telescopes follow prboloid shpe. x y = 1 b where b = e 1 ( ) The hyperbol hs two seprte brnches. As the curve retrets towrds infinity, the curve pproches the symptotes x y 0, bx = y. = ± b The distnce between the two vertices is the mjor xis (). If mirror is mde in the shpe of n hyperbol, then ll rys emerging from one focus will, fter reflection, pper to be diverging from the other focus. Circles nd ellipses re closed curves. Prbols nd hyperbols re open curves.
ENGI 4 Fundmentls Conic Sections Pge 1- A specil cse of the hyperbol occurs when the eccentricity is e = nd it is rotted 45E from the stndrd orienttion. The symptotes line up with the coordinte xes, the grph lies entirely in the first nd third qudrnts nd the Crtesin eqution is xy = k. This is the rectngulr hyperbol. Degenerte conic sections rise when the intersecting plne psses through the pex of the cone. Two cses re: 0 x y e < 1: + b = 0 point t the origin. x y e > 1: 0 b = line pir through the origin. Another degenerte cse is x y + = 1 nothing! b Exmple 1.4.1 Clssify the conic section whose Crtesin eqution is 3y = x + 3. Rerrnging into stndrd form, y x 3y x = 3 = 1 1 3 Compre with X Y = 1: X = y, Y = x, = 1, b= 3 b Therefore this is n hyperbol, rotted through 90, with vertices t (0, ±1) nd 3 symptotes x = ± 3y or y = ± x. 3 [The symptotes mke ngles of 30 with the x xis.]
ENGI 4 Fundmentls Conic Sections Pge 1-3 Exmple 1.4. Clssify the conic section whose Crtesin eqution is 1x + 8y = 168x + 168y. Rerrnging into stndrd form, first complete the squre. 1x + 8y 168x 168y = 0 1(x 8x) + 8(y 6y) = 0 1((x 4) 16) + 8((y 3) 9) = 0 1(x 4) + 8(y 3) = 588 ( x 4) ( y 3) + 8 1 = 1 Compre this with the stndrd form X Y + b = 1: X = x 4, Y = y 3, = 8, b= 1 The conic section is therefore n ellipse, semi mjor xis 8, semi minor xis 1, (from which the eccentricity cn be found to be exctly e = 1/), centre (4, 3). It hppens to pss exctly through the origin. Moving up to three dimensions, we hve the fmily of qudric surfces.
ENGI 4 Fundmentls Qudric Surfces Pge 1-4 1.5 Clssifiction of Qudric Surfces Agin, we shll consider only the simplest cses, where ny plnes of symmetry re locted on the Crtesin coordinte plnes. In nerly ll cses, this elimintes crossproduct terms, such s xy, from the Crtesin eqution of surfce. Except for the prboloids, the Crtesin equtions involve only x, y, z nd constnts. The five min types of qudric surfce re: The ellipsoid (xis lengths, b, c) x + y b + z c = 1 The xis intercepts re t (±, 0, 0), (0, ±b, 0) nd (0, 0, ±c). All three coordinte plnes re plnes of symmetry. The cross-sections in the three coordinte plnes re ll ellipses. Specil cses: = b > c : oblte spheroid ( squshed sphere ) = b < c : prolte spheroid ( stretched sphere or cigr shpe) = b = c : sphere Hyperboloid of One Sheet (Ellipse xis lengths, b ; ligned long the z xis) x + y b z c = 1 For hyperboloids, the centrl xis is ssocited with the odd sign out. In the cse illustrted, the hyperboloid is ligned long the z xis. The xis intercepts re t (±, 0, 0) nd (0, ±b, 0). The verticl cross sections in the x-z nd y-z plnes re hyperbole. All horizontl cross sections re ellipses.
ENGI 4 Fundmentls Qudric Surfces Pge 1-5 Hyperboloid of Two Sheets (Ellipse xis lengths b, c ; ligned long the x xis) x y b z c = 1 For hyperboloids, the centrl xis is ssocited with the odd sign out. In the cse illustrted, the hyperboloid is ligned long the x xis. The xis intercepts re t (±, 0, 0) only. Verticl cross sections prllel to the y-z plne re either ellipses or null. All cross sections contining the x xis re hyperbole. Elliptic Prboloid (Ellipse xis lengths, ligned long the z xis) b ; z c = x + y b For prboloids, the centrl xis is ssocited with the odd exponent out. In the cse illustrted, the prboloid is ligned long the z xis. The only xis intercept is t the origin. The verticl cross sections in the x-z nd y-z plnes re prbole. All horizontl cross sections re ellipses (for z > 0).
ENGI 4 Fundmentls Qudric Surfces Pge 1-6 Hyperbolic Prboloid (Hyperbol xis length ; ligned long the z xis) z c = x y b For prboloids, the centrl xis is ssocited with the odd exponent out. In the cse illustrted, the prboloid is ligned long the z xis. The only xis intercept is t the origin. The verticl cross section in the x-z plne is n upwrd-opening prbol. The verticl cross section in the y-z plne is downwrd-opening prbol. All horizontl cross sections re hyperbole, (except for point t z = 0). The plots of the five stndrd qudric surfces shown here were generted in the softwre pckge Mple. The Mple worksheet is vilble from link t "http://www.engr.mun.c/~ggeorge/4/progrms/index.html". Degenerte Cses: x + y b + z c = 0 : A single POINT t the origin. x + y b z c = 0 : Elliptic CONE, ligned long the z xis; [symptote to both types of hyperboloid]. x + y b + z c = 1 : NOTHING x + b y = 1 : ELLIPTIC CYLINDER, ligned long the z xis. x b y = 1 : HYPERBOLIC CYLINDER, ligned long the z xis. y x = : PARABOLIC CYLINDER, vertex line on the z xis. b
ENGI 4 Fundmentls Qudric Surfces Pge 1-7 x + b y = 0 : LINE (the z xis) x + b y = 1 : NOTHING x b y = 0 : PLANE PAIR (intersecting long the z xis) x = 1 : Prllel PLANE PAIR x = 0 : Single PLANE (the y-z coordinte plne) x = 1 : NOTHING Exmple 1.5.1 Clssify the qudric surfce, whose Crtesin eqution is x = 3y + 4z. Rerrnging into stndrd form, x y z = + 6 4 3 Compre to the stndrd form Z X Y = + : Z = x, X = y, Y = z, =, b= 3, c =6 c b The qudric surfce is therefore n elliptic prboloid, ligned long the x xis, with its vertex t the origin.
ENGI 4 Fundmentls Qudric Surfces Pge 1-8 Exmple 1.5. Clssify the qudric surfce, whose Crtesin eqution is z = 1 + x. Rerrnging into stndrd form, z x = 1 Compre to the stndrd form X Y = 1: X = z, Y = x, = b= 1 b The qudric surfce is therefore one of the degenerte cses. It is hyperbolic cylinder, centre t the origin, ligned long the y xis. Exmple 1.5.3 Clssify the qudric surfce, whose Crtesin eqution is x y + z + 1 = 0. Rerrnging into stndrd form, y x z = 1 Compre to the stndrd form X Y Z = 1: X = y, Y = x, Z = z, = b= c = 1 b c The qudric surfce is therefore n hyperboloid of two sheets, centre t the origin, ligned long the y xis.
ENGI 4 Fundmentls Surfces of Revolution Pge 1-9 1.6 Surfces of Revolution Consider curve in the x-y plne, defined by the eqution y = f (x). If it is swept once round the line y = c, then it will generte surfce of revolution. At ny prticulr vlue of x, thin cross-section through tht surfce, prllel to the y-z plne, will be circulr disc of rdius r, where r = f ( x) c Let us now view the circulr disc fce-on, (so tht the x xis nd the xis of rottion re both pointing directly out of the pge nd the pge is prllel to the y-z plne). Let (x, y, z) be generl point on the surfce of revolution. From this digrm, one cn see tht r = (y c) + z Therefore, the eqution of the surfce generted, when the curve y = f (x) is rotted once round the xis y = c, is ( ) ( ) ( ) y c + z = f x c Specil cse: When the curve y = f (x) is rotted once round the x xis, the eqution of the surfce of revolution is + = ( ( )) or y + z = f ( x) y z f x
ENGI 4 Fundmentls Surfces of Revolution Pge 1-30 Exmple 1.6.1 Find the eqution of the surfce generted, when the prbol y round the x xis. = 4x is rotted once The solution is immedite: y + z = 4x, which is the eqution of n elliptic prboloid, (ctully specil cse, circulr prboloid). The Curved Surfce Are of Surfce of Revolution For rottion round the x xis, the curved surfce re swept out by the element of rc length Δs is pproximtely the product of the circumference of circle of rdius y with the length Δs. ΔA π y Δ s Integrting long section of the curve y = f (x) from x = to x = b, the totl curved surfce re is x b π = x= ( ) A = f x ds For rottion of y = f (x) bout the xis y = c, the curved surfce re is x= b b ds ds dy A = π f ( x) c ds = π f ( x) c x= dx nd 1 = + dx dx dx Therefore b A = π f x c 1 + f x ( ) ( ( )) dx
ENGI 4 Fundmentls Surfces of Revolution Pge 1-31 Exmple 1.6. Find the curved surfce re of the circulr prboloid generted by rotting the portion of the prbol y = 4cx (c > 0) from x = ( > 0) to x = b bout the x xis. A x b π = x= = y ds y = 4cx yy = 4c y = c y ds 4 4 1 dy 1 c 1 c x + = + = + = + = c dx dx y 4cx x b x+ c b A = π cx dx = 4π c x+ c x Therefore ( ) 1/ dx ( x+ c) 3/ = 4π c 3 b (( ) 3/ ( ) 3/ ) 8π c A = b+ c +c 3
ENGI 4 Fundmentls Hyperbolic Functions Pge 1-3 1.7 Hyperbolic Functions When uniform inelstic (unstretchble) perfectly flexible cble is suspended between two fixed points, it will hng, under its own weight, in the shpe of ctenry curve. The eqution of the stndrd ctenry curve is most concisely expressed s the hyperbolic cosine function, y = cosh x, where x x e + e cosh x = The solutions to some differentil equtions cn be expressed conveniently in terms of hyperbolic functions. The other five hyperbolic functions re x x e e sinh x =, x x sinh x e e 1 tnh x = =, sech x = =, x x x x cosh x e + e cosh x e + e x x 1 e + e 1 coth x = = nd csch x = =. x x x x tnh x e e sinh x e e Unlike the trigonometric functions, the hyperbolic functions re not periodic. However, prity is preserved: Of the six trigonometric function, only cos θ nd sec θ re even functions. Of the six hyperbolic functions, only cosh θ nd sech θ re even functions. The other functions re odd.
ENGI 4 Fundmentls Hyperbolic Functions Pge 1-33 There is close reltionship between the hyperbolic nd trigonometric functions. From the Euler form for e jθ, e jθ = cos θ + j sin θ, e jθ = cos θ j sin θ jθ jθ e e 1 sinθ = = sinh ( jθ) = jsinh ( jθ) j j jθ jθ e + e cosθ = = cosh( jθ ) ( jθ ) ( θ) sinθ sinh 1 tnθ = tnh θ tnh θ cosθ = jcosh j = j = ( j ) j ( j ) Identities: Let x = jθ : sin θ + cos θ 1 sinh x + cosh x 1 j cosh x sinh x 1 tnh x 1 1 + tn θ sec θ 1 + j cosh x 1 tnh x sech x etc.
ENGI 4 Fundmentls Hyperbolic Functions Pge 1-34 Derivtives d dx ( sinh x) = x x x d e e e + e = dx x Therefore d dx ( sinh x) = cosh x d dx ( cosh x) = x x x d e + e e e = dx x Therefore d ( cosh x) = + sinh x dx [Note the different sign from the trigonometric version, (cos x) ' = sin x.] d dx ( tnh x) Therefore OR = ( cosh x)( cosh x) ( sinh x)( sinh x) d sinh x 1 = = dx cosh x cosh x cosh x d dx ( tnh x) = sech x Let x = jθ, then θ = jx tnh(jθ) = j tn θ tnh x = j tn( jx) d ( tnh x) = j( sec ( jx) ) ( j) = + sec ( jx) = sec dx = sech (jθ) = sech x. d dx Therefore ( ) tnh x = sech x θ
ENGI 4 Fundmentls Hyperbolic Functions Pge 1-35 d dx ( csch x) d dx 1 ( ) = ( ) ( = ( ) sinh x sinh x cosh x) 1 coshx = sinh x sinh x Therefore etc. d dx ( ) csch x = csch xcoth x A list of identities nd derivtives for hyperbolic functions is presented in the suggestions for formul sheet, in Appendix A to these lecture notes.
ENGI 4 Fundmentls Integrtion by Prts Pge 1-36 1.8 Integrtion by Prts Review: Let u(x) nd v(x) be functions of x. Then, by the product rule of differentition, Integrting with respect to x : [ ] uv = u v dx + uv dx d du dv ( uv) = v + u dx dx dx This leds to the formul for integrtion by prts: Exmple 1.8.1 [ ] uv dx = uv u v dx Find 3 x I = xe dx. [v' must be identified with fctor of the integrnd tht cn be ntidifferentited esily.] u = x v = x e x 1 u = x v = e x [ ] I = uv u v dx = xe xe 1 x x dx x ( 1 x 1 ) = e + C Therefore ( ) I 1 x 1 e x = + + C
ENGI 4 Fundmentls Integrtion by Prts Pge 1-37 Exmple 1.8. [Repeted use of integrtion by prts] Find I = x cos xdx. u = x v' = cos x u' = x v = sin x I = x sin x x sin x d x u v Using integrtion by prts gin, with u = x nd v' = sin x, u' = v = cos x ( ) I = x sin x xcos x cos x dx = x sin x + x cos x sin x + C Therefore ( ) I = x cos x dx = x sin x + xcos x Check: I' = x sin x + (x ) cos x + cos x x sin x + 0 = x cos x T + C
ENGI 4 Fundmentls Integrtion by Prts Pge 1-38 Shortcut ( tbulr form for repeted integrtions by prts): I = x cos xdx : Reding off the digonls, I = x sin x + x cos x sin x + C Exmple 1.8.3 Find I = xe dx. 4 x Note: There re three wys the tble cn end: 1) column 'D' reduces to 0 (s in the exmples on this pge); ) the product cross row is esy to integrte; 3) the product cross row is constnt multiple of the originl integrnd. (ex. 1.8.4 next pge) Therefore I = (x 4 4x 3 + 1x 4x + 4) e x + C Check: I' = (x 4 4x 3 + 1x 4x + 4 + 4x 3 1x + 4x 4 + 0) e x = x 4 e x T
ENGI 4 Fundmentls Integrtion by Prts Pge 1-39 Exmple 1.8.4 (recursive use of integrtion by prts) Find x I = e sin bx dx. Either or x cosbx x sinbx I = e + e b b b I x x e e b I = + sin bx bcosbx I 1 + I = b x e sin bx bcosbx b x b I e sin bx bcosbx [ ] 1 b + I = x e sin bx bcosbx x + b I = e sin bx bcosbx [ ] ( + ) = [ ] ( ) [ ] Therefore Check: x e x sin bx dx e = ( sin bx bcosbx) C + b + Let s = sin bx nd c = cos bx, then s' = bc, c' = bs nd I = x e ( s bc) + C + b x di e = ( ( s bc ) + ( ( bc) b( bs) )) dx + b = x x e e ( s bc + bc + b s) = ( + b ) s = + b + b x e sinbx T
ENGI 4 Fundmentls Leibnitz Differentition Pge 1-40 1.9 Leibnitz Differentition of Definite Integrl ( ) ( ( ) ( ) ) ( ) d y = g x dg df y = g( x) H H x, y dy = H( x, g x ) H( x, f ( x) ) + dy dx y = f x dx dx y = f( x) x If the limits of integrtion re both constnt, then just differentite the integrnd with respect to x, treting ll other terms s constnts. d dx b ( (, ) ) b f f x t dt = dt x Exmple 1.9.1 di t Evlute, where I () t = zt. dt dz t Using Leibnitz differentition: di d d t = (( t) t) ( t) (( t) t) () t + z() 1 dz dt dt dt t z = t z 3t 9t = 4t t + = 3t + = z = t Directly: t t z 4t t 3t I() t = zt dz = t = t = t t 3 di d 3t 9t = = dt dt 3 See Problem Set 3 nd Section 5.10 for more prcticl exmples of Leibnitz differentition. End of Chpter 1
ENGI 4 Fundmentls Polr Coordintes Pge 1-41 1. Polr Coordintes The description of the loction of n object in ú reltive to the observer is not very nturl in Crtesin coordintes: the object is three metres to the est of me nd four metres to the north of me, or (x, y) = (3, 4). It is much more nturl to stte how fr wy the object is nd in wht direction: the object is five metres wy from me, in direction pproximtely 53 north of due est, or (r, θ) = (5, 53 ). Rdr lso opertes more nturlly in plne polr coordintes. r = rnge θ = zimuth O is the pole OX is the polr xis (where θ = 0) Anticlockwise rottions re positive. [Nuticl berings re very different: positive rottion is mesured clockwise, from zero t due north!] Exmple 1..01 The point P with the polr coordintes (r, θ) = (4, π / 3) lso hs the polr coordintes or ( 4, π + π ) ( 4, π + π ) 3 ( n π ) or 4, π +, n = ny integer 3 3
ENGI 4 Fundmentls Polr Coordintes Pge 1-4 Exmple 1..01 (continued) The point ( 4, π π ) + is t P. So lso is π ( π nπ) 3 4, + +, n = ny integer. 3 In generl, if the polr coordintes of point re (r, θ), then ( π ) ( r, θ + nπ ) nd r, θ + ( n+ 1) (n = ny integer) lso describe the sme point. The polr coordintes of the pole re (0, θ) for ny θ. In some situtions, we impose restrictions on the rnge of the polr coordintes, such s r > 0, π < θ < +π for the principl vlue of complex number in polr form. Conversion between Crtesin nd polr coordintes: x = rcosθ y = rsinθ Inverse: x + y = r cos θ + r sin θ = r r = x + y r = ± x + y
ENGI 4 Fundmentls Polr Coordintes Pge 1-43 y x rsinθ = = r cosθ tnθ Therefore tnθ = y x More informtion is needed in order to select the correct qudrnt. Exmple 1..0 Find the polr coordintes for the point whose Crtesin coordintes re ( 3, 4). ( 3, 4) is in the second qudrnt. x = 3, y = 4 r = 9 + 16 = 5 r = ± 5 4 tnθ = 3 = If we choose r > 0, then one vlue of θ is θ = Tn 1 (4/3) + π.1 rd. One possibility: (r, θ) = (5,.1) (to 3 s.f.) Therefore, to 3 s.f., (r, θ) = (5,.1 + nπ) or ( 5,.1 + (n+1)π), ( n ) 4 3
ENGI 4 Fundmentls Polr Coordintes Pge 1-44 Exmple 1..03 Find the Crtesin coordintes for (, θ ) = (, π ) ( 1 1) 11π π π = = + ( ) π 3 3 3 11 3 r. x = rcosθ = cos π = 1 y = rsinθ = sin π = 3 3 3 Therefore ( xy, ) = ( 1, 3 ) Polr Curves r = f (θ) The representtion (x, y) of point in Crtesin coordintes is unique. For curve defined implicitly or explicitly by n eqution in x nd y, point (x, y) is on the curve if nd only if its coordintes (x, y) stisfy the eqution of the curve. The sme is not true for plne polr coordintes. Ech point hs infinitely mny possible r, θ nπ r, θ + n+ 1 π (where n is ny integer). A ( ) representtions, ( + ) nd ( ) point lies on curve if nd only if t lest one pir (r, θ) of the infinitely mny possible pirs of polr coordintes for tht point stisfies the polr eqution of the curve. It doesn t mtter if other polr coordintes for tht sme point do not stisfy the eqution of the curve.
ENGI 4 Fundmentls Polr Coordintes Pge 1-45 Exmple 1..04 The curve whose polr eqution is r = 1 + cos θ is crdioid (literlly, hert-shped curve). { r =, θ = nπ } (where n is ny integer) stisfies the eqution r = 1 + cos θ. (r, θ) = (, nπ) is on the crdioid curve. But (, nπ) is the sme point s (, (n+1)π). θ = (n+1)π 1 + cos θ = 0 r. Yet the point whose polr coordintes re (, (n+1)π) is on the curve! Exmple 1..05 Convert to polr form the eqution x + y = x + y + 3y. r = r + 3r sin θ r ( r 1 3 sin θ ) = 0 r = 0 or r = 1 + 3 sin θ But ( ( )) 1 3 0, rcsin is solution of r = 1 + 3 sin θ r = 0 is included in r = 1 + 3 sin θ. Therefore the polr eqution of the curve is r = 1 + 3 sin θ (which is limçon). Note tht there is no restriction on the sign of r ; it cn be negtive.
ENGI 4 Fundmentls Polr Coordintes Pge 1-46 Exmple 1..06 Convert to Crtesin form the eqution of the crdioid curve r = 1 + cos θ. r = 1 + x r r = r + x r x = r (r x) = r Therefore (x + y x) = x + y Tngents to r = f (θ) x = r cos θ = f (θ) cos θ y = r sin θ = f (θ) sin θ By the chin rule for differentition: dy dy d dy dx = θ = dx dθ dx dθ dθ This leds to generl expression for the slope nywhere on curve r = f (θ) : dy dx = dr sinθ + r cosθ dθ dr cosθ r sinθ dθ dy dx = 0 nd 0 dθ dθ t ( r, θ ) horizontl tngent t (r, θ). dx dy = 0 nd 0 dθ dθ t ( r, θ ) verticl tngent t (r, θ).
ENGI 4 Fundmentls Polr Coordintes Pge 1-47 At the pole (r = 0): dr sinθ + 0 dy dθ dr = = tnθ provided 0 dx d r cosθ 0 dθ dθ dr If r 0 but dθ 0 s θ θ o, then the rdil line θ = θ o is tngent t the pole. [This cn be of some help when sketching polr curves.] Exmple 1..07 Sketch the curve whose eqution in polr form is r = cos θ. Two methods will be demonstrted here. The first method is direct trnsfer from Crtesin plot of r ginst θ (s though the curve were y = cos x). The second method is systemtic tbulr method, involving investigtion of the behviour of the curve in intervls of θ between consecutive criticl points (where r nd/or its derivtive is/re zero or undefined). Method 1.
ENGI 4 Fundmentls Polr Coordintes Pge 1-48 Method. r = cos θ = 0 t θ = (ny odd multiple of π/) r = 0 t θ = (ny odd multiple of π/4) dr = sinθ = dθ 0 t θ =(ny integer multiple of π) r' = 0 t θ = (ny integer multiple of π/) Therefore tbulte in intervls bounded by θ = (consecutive integer multiples of π/4). θ 0 π/ π/ π π 3π/ 3π/ π π 5π/... θ 0 π/4 π/4 π/ π/ 3π/4 3π/4 π π 5π/4... r 1 0 0 1 1 0 0 1 1 0... Region in sketch (1) () (3) (4) (5) This leds to the sme sketch s in Method 1 bove. You cn follow plot of r = cos nθ by Method 1 (for n = 1,, 3, 4, 5 nd 6) on the web site. See the link t "http://www.engr.mun.c/~ggeorge/4/progrms/". The distinct polr tngents re π θ = ± 4
ENGI 4 Fundmentls Polr Coordintes Pge 1-49 Length of Polr Curve If r = f (θ) (for α < θ < β), then x = f (θ) cos θ nd y = f (θ) sin θ Let r = f (θ), r' = f ' (θ), c = cos θ nd s = sin θ, then dx dy = r c rs nd = r s + rc dθ dθ (( ) ) ( ) ( r c ) dx dy + = r c rcsr + r s + r s + rcsr + dθ dθ = (r') (c + s ) + 0 + r (s + c ) dr = r + dθ Therefore the length L long the curve r = f (θ) from θ = α to θ = β is L β dr = r + α dθ dθ Exmple 1..08 Find the length L of the perimeter of the crdioid r = 1 + cos θ. α = 0, β = π. r = 1 + cos θ = 1 + c dr = sinθ = s dθ dr r + = dθ 1 + c + c + s = + c
ENGI 4 Fundmentls Polr Coordintes Pge 1-50 Exmple 1..08 (continued) But 1 + cos x = cos x. Set θ = x. dr θ r + = cos dθ π θ θ L = cos dθ = cos dθ 0 0 Using symmetry in the horizontl xis, π π π θ θ θ θ θ 0 0 0 π ( ) L = cos d = 4 cos d = 4 sin = 8 1 0 Therefore the perimeter of the crdioid curve is L = 8. Note: For π < θ < π, nd π 0 θ cos 0! d θ = cos θ θ = cos Exmple 1..09 Find the rc length long the spirl curve r = e θ ( > 0), from θ = α to θ = β. θ dr r = e = e dθ θ ( ) ( ) ds = e + e = e dθ θ θ θ β s = e dθ = e α θ β α ( ) L = e e θ β α
ENGI 4 Fundmentls Polr Coordintes Pge 1-51 Are Swept Out by Polr Curve r = f (θ) ΔA Are of tringle ( ) = 1 r r +Δr sin Δ θ But the ngle Δθ is smll, so tht sin Δθ Δθ nd the increment Δr is smll compred to r. Therefore 1 Δ Δ A r θ A β r 1 = α dθ Exmple 1..10 Find the re of circulr sector, rdius r, ngle θ. r = constnt, β = α + θ 1 α + θ 1 α + θ 1 α α A r dφ r φ r α = = = + (( θ) α) Therefore A = 1 θ r Full circle: θ = π nd A = πr.
ENGI 4 Fundmentls Polr Coordintes Pge 1-5 Exmple 1..11 Find the re swept out by the polr curve r = e θ over α < θ < β, (where > 0 nd α < β < α + π ). The condition (α < β < α + π ) prevents the sme re being swept out more thn once. If β > α + π then one needs to subtrct res tht hve been counted more thn once [the red re in the digrm] β 1 1 A = e d = α Therefore ( ) θ e θ θ β α ( β α ) A = e e 4 In generl, the re bounded by two polr curves r = f (θ) nd r = g(θ) nd the rdius vectors θ = α nd θ = β is β (( ( )) ( ( )) ) 1 A = f θ g θ dθ α See the problem sets for more exmples of polr curve sketching nd the clcultion of the lengths nd res swept out by polr curves.
ENGI 4 Fundmentls Polr Coordintes Pge 1-53 Rdil nd Trnsverse Components of Velocity nd Accelertion At ny point P (not t the pole), the unit rdil vector ˆr points directly wy from the pole. The unit trnsverse vector ˆθ is orthogonl to ˆr nd points in the direction of incresing θ. These vectors form n orthonorml bsis for ú. Only if θ is constnt will ˆr nd ˆθ be constnt unit vectors, (unlike the Crtesin i nd j). The derivtives of these two non-constnt unit vectors cn be shown to be drˆ dt dθ = ˆ θ nd dt d ˆ θ dt dθ = ˆ dt r Using the overdot nottion to represent differentition with respect to the prmeter t, these results my be expressed more compctly s r ˆ = θ ˆ θ nd ˆ θ = θ rˆ The rdil nd trnsverse components of velocity nd ccelertion then follow: r = r rˆ v = r = r rˆ + rr ˆ = r rˆ + r θ ˆ θ v = r nd v = rθ rdil trnsverse ( ) ˆ ˆ rrˆ + rrˆ + ( r θ + r θ r θ ˆ) ˆ ˆ ( ) ( ( ) ) ( ) = v = r = θ θ + θ = rrˆ + r θ θ + r θ θ r θ rˆ = r r θ rˆ + r θ + r θ ˆ θ trnsverse The trnsverse component of ccelertion cn lso be written s tr = ( r θ ) rdil 1 d rdt
ENGI 4 Fundmentls Polr Coordintes Pge 1-54 Exmple 1..1 A prticle follows the pth r = θ, where the ngle t ny time is equl to the time: θ = t > 0. Find the rdil nd trnsverse components of ccelertion. r = θ = t r = θ = r = θ = 1 0 v = r rˆ + r θ ˆ θ = rˆ + t ˆ θ ( v = 1 + t ) ( θ ) ˆ ( θ θ) = r r r + r + r ˆ θ = trˆ + ˆ θ ( = 4 + t ) Therefore r = t nd tr = Exmple 1..13 For circulr motion round the pole, with constnt rdius r nd constnt ngulr velocity θ = ω, the velocity vector is purely tngentil, v = r ω ˆ θ, nd the ccelertion vector is ( ) rˆ ( ) = + + ˆ = r rθ r θ rθ θ rω r which mtches the fmilir result tht centrifugl or centripetl force = r ω, directed rdilly inwrd. ˆ