Lecture 3: Curves in Calculus. Table of contents


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1 Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up the ply ground for the mthemticl study of curves in the dimensionl plne nd the 3dimensionl spce. In prticulr, we will discuss pproprite mthemticl representtion of curves, dene their tngent vectors nd rc lengths, nd derive formuls for these quntities. The required textook sections re.,.. The optionl textook sections re.3,.4,.5. The exmples in this note re dierent from exmples in the textook. Plese red the textook crefully nd try your hnds on the exercises. During this plese don't hesitte to contct me if you hve ny questions. Tle of contents Lecture 3: Curves in Clculus Prmetriztion of Curves Mthemticl representtions of curve Tngent Vectors nd Arc Length From possile velocities to ngent vectors Arc length of prmetrized curve Arc Length Prmetriztion Properties of curves from its derivtives
2 Dierentil Geometry of Curves & Surfces. Prmetriztion of Curves The most convenient wy (for ppliction of clculus) to represent sptil curve mthemticlly is to tret s the trjectory of moving prticle. This leds to the soclled prmetriztion of the curves. There re innitely mny possile prmetriztions of the sme curve... Mthemticl representtions of curve Level curves. Before the dvent of clculus, curve is usully dened through level sets: i. (in the plne) s level sets: f(x; y) = ; ii. (in the spce) s intersection of surfces (intersection of level sets): Exmple. A circle in R is represented s A stright line in R 3 is represented s f (x; y; z) = ; f (x; y; z) = : () f(x; y) = (x ) + (y ) r = : () x + x + 3 x 3 = ; x + x + 3 x 3 = : (3) The sic ide is to replce the study of the curve y the study of one or more simple surfces. Prmetrized curves. To pply clculus, the most convenient representtion of curve is through prmetriztion. Definition. (Definition.. in textook) A prmetrized curve in R n is mp : (; ) 7! R n, for some ; with 6 < 6. Note. Recll tht (; ) mens the set of ll numers > nd <, while [; ] mens the set of ll numers > nd 6. Exercise. Do you rememer/cn you guess wht (; ] nd [; ) mens? Exmple 3. A circle in R is represented s x(t) + r cos t (t) = = y(t) + r sin t for t (; ) with < ; >. (4). The reson for such trnsition is tht, without clculus, there is no wy to study curve directly. Insted one hs to look for simple functions tht could generte the curve nd turn the study of the curves to the study of those functions.. Another nme for function.
3 Mth 348 Fll 7 Exmple 4. A strightline in R 3 is represented s (t) x(t) x(t) x B y A+ x(t) z for t ( ; ). u v w C A (5) Exmple 5. (Cycloid) The trjectory of unit circle rolling in the plne long line. Figure. The cycloids We tke t = s the prmeter. We hve hgmmi(t) = (t sin t; cos t) for t [; ]. Exercise. Write down prmetrized representtion of unit circle rolling outside (epicycloid) or inside (hypocycloid) nother circle with rdius R centered t the origin (for the inside cse, ssume R > ). Plot the following two cses: Outside with R = (Crdioid); Inside with R = 4 (Astroid). 3 Wht hppens if R!? Exmple 6. (Vivini's curve) The intersection of sphere x + y + z = R with x + y = R x. It cn e prmetrized y (t) = (R cos t; R cos t sin t; R sin t); t [; ) (6) Exercise 3. Check Ex...8 in the textook for wht looks like. Compre (6) with the formul there. Cn you nd dierent prmetriztion? Exercise 4. Consider plne curve given y grph: y = f(x). Is it level curve or prmetrized curve? Exercise 5. Try to drw the prmetrized curve (t) = (cos t; sin t; t), t (; 3 ). Exercise 6. Try to drw (t) = (t 3 ; t ) for t ( ; ). Remrk 7. Is cler tht neither level set nor prmetrized representtions re unique. There re innitely mny pirs of plnes thntersect long the sme stright line; There re innitely mny (t): (; ) 7! R n such tht their imge set coincide. We emphsize tht this nonuniqueness is in fct good thing s it llows us to choose the most convenient of them. We will see how this works lter cos t R R cos t + t ; + sin t R R R R sin t + t. R R 3
4 Dierentil Geometry of Curves & Surfces In 348 we will only consider smooth curves, ths prmetrized curves (t) = (x (t); x (t); :::; x n (t)), t (; ), with ech x n (t) innitely dierentile for ll t (; ). Exercise 7. Show through exmples tht the imge of smooth curve my not look smooth.. Tngent Vectors nd Arc Length When curve is viewed s the trjectory of moving prticle, its velocities re represented mthemticlly s tngent vectors, nd the length of the trjectory is the rc length... From possile velocities to ngent vectors From possile velocities to tngent vectors. Let (t) R n e the trjectory of prticle. Then its velocity t t is given y (t) (t ) v(t ) = lim = d t!t t t (t ) = _(t ): (7) For prmetrized curve (t), we will view it s mthemticl model of trjectory of prticle nd dene its velocity t (t ) to e one tngent vector t (t ). Definition 8. (Tngent vector) Let C e curve in R n. Let x C. Then v R n is tngent vector of C t x if nd only if there is prmetriztion (t): (; ) 7! R n of C such tht i. there is t (; ) with (t ) = x ; ii. _(t ) = v. Exercise 8. Let C e curve in R n. Let x C. Let v e tngent vector of C t x. i. Prove tht v is lso tngent vector of C t x ; ii. Prove tht v is lso tngent vector of C t x ; iii. Convince yourself tht for ritrry c R, c v is tngent vector of C t x ; iv. (optionl) Prove tht for ritrry c R, c v is tngent vector of C t x. The tngent vector of (t) t x = (t ): _(t) Regulr curves. A prmetrized curve (t): (; ) 7! R n is regulr if nd only if is smooth nd _(t) =/ for ll t (; ). Exmple 9. The prmetrized curve (t) = check i. t; t ; t 3 re ll smooth; t t C A is regulr. To see this we t 3 4
5 Mth 348 Fll 7 ii. _(t) = B t 3 t A no mtter wht s. From now on curve mens regulr curve... Arc length of prmetrized curve 5 The rc length formul. Let (t): (; ) 7! R n e prmetrized curve. Let [; ] (; ). We try to otin its rc length L through the following: Pick = t < < t k =. Consider the stright line segments (t )(t ), (t )(t ), (t )(t 3 ), :::, (t k )(t k ). Then intuitively we hve L > the sum of the lengths of these line segments. Inspired y the ove, one denes L := sup =t <<t k = i= Xk k( ) ( )k (8) where we emphsize tht the supreme is tken over ll possile prtitions = t < < t k =. In prticulr, k N is not xed. The formul. Arc length from () to () = k_(t)k : Theorem. Let (t) e smooth prmetrized curve. Then its rc length from () to () is given y L = k_(t)k : (9) Proof. (optionl; You my wnt to red the ox fter the proof first) We prove in two steps.. L 6 R k_(t)k. By fundmentl theorem of clculus, we hve ( ) ( ) = _(t) : () Thus we hve Xk k( ) ( )k = X k t= t= k _(t) 6 X k_(t)k t= = k_(t)k : ()
6 Dierentil Geometry of Curves & Surfces. L > R k_(t)k. Let k N nd dene := + i ( ). Then we hve = t k < t < < t k =. By fundmentl theorem of clculus, we hve ( ) ( ) = _(t) = _( ) ( ) + R i () where " t R i := [_(t) _( )] = ti (s) ds # : (3) As x(t) is regulr curve, there is M > such tht Consequently k(s)k 6 M for ll s [; ]: (4) kr i k 6 M ds = M ( ) M ( ) = : (5) k Thus we hve Xk k( ) ( )k = X k i= i= k > X i= k > X i= k_( ) ( ) + R i k [k_( ) ( )k kr i k] k_( )k ( ) M ( ) k : (6) On the other hnd, we hve k_(t)k = [k_(t)k k_( )k] + +k_( )k ( ) 6 k_( )k ( ) + = k_( )k ( ) + t + k(s)kds 6 k_( )k ( ) + M ( ) = k_( )k ( ) + k_(t) _( )k M ( ) k : (7) 6
7 Mth 348 Fll 7 From this we conclude Xk i= k( ) ( )k s k!. This implies L := sup Xk =t <t <<t k = i= M ( ) k_(t)k >! (8) k k( ) ( )k > k_(t)k (9) nd ends the proof. Technicl Aside In the ove proof we hve used the following results from clculus. Supreme. Let A e collection (set) of numers. Then its supreme sup A is the smllest numer ths greter thn or equl to A. As consequence, to prove tht sup A 6, ll we need to do is to show tht for every A, there holds 6 ; to prove tht sup A >, ll we need to do is to nd one prticulr sequence of n A such tht lim n! ( n ) >. Exercise 9. Wht do we need to do to prove sup A > or sup A <? Tringle inequlity. Clssicl tringle inequlity: Vrint: kxk + kyk > kx + yk: () jkxk kykj 6 kx yk: () Generliztion: kx k + + kx k k > kx + + x k k: () Integrl: Together with the denite of Riemnn integrls, () yields the following inequlity for vector functions: kx(t)k > x(t) : (3) Exmples. Exmple. Clculte the circumference of the unit circle x + y =. 7
8 Dierentil Geometry of Curves & Surfces Solution. p Method. We clculte the curve length l of the grph y = x ; 6x6. Then the circumference is L = l. q l = + p x dx = p dx x / x=sin t = = = : (4) So the circumference is L=. / Method. We prmetrize x(t) = cos t; y(t) = sin t, 6 t <. Then L = p x (t) + y (t) Exmple. Clculte the rc length of the spce curve = : (5) for t from to. Solution. We hve L = x = cos t; y = sin t; z = t (6) = p x (t) + y (t) + z (t) p p = : (7) Exmple 3. Clculte the rc length of the cycloid (t sin t; cos t) from t = to. Solution. We hve p L = ( cos t) + (sin t) p = cos t r = sin t sin t = = 8: (8) Exmple 4. Clculte the rc length of the Limcon of Pscl (( + cos t) cos t; ( + cos t) sin t), t (; ). 8
9 Mth 348 Fll 7 Solution. We hve (t) = (cos t + cos t; sin t + sin t). p L = ( sin t sin t) + (cos t + cos t) p = cos t : (9) It turns out tht this integrl cnnot e clculted explicitly. Exercise. Clculte the rc length of x = cos 3 t; y = sin 3 t; t [; ). Exercise. (Optionl) Is cler tht rc length should e independent of prmetriztion. Prove this. 3. Arc Length Prmetriztion Among the innitely mny possile prmetriztions of the sme curve, one prticulr prmetriztion, clled rc length prmetriztion stnds out s the most convenient due to its ility to simplify clcultions. Motivtion. For generl prmetrized curve (t): (; ) 7! R n, in generl we hve k_(t)k vrying with t. As we will see lter, mny clcultions could e gretly simplied if k_(t)k = for ll t (; ). Such prmetriztion is clled the rc length prmetriztion of the curve. Exercise. In this cse we hve the rc length etween x(t ) nd x(t ) to e t t. Existence of rc length prmetriztion. Theorem 5. Let (t): (; ) 7! R n e regulr curve. Then there is strictly incresing function T : (; ) 7! (; ) such tht the curve (s) := (T (s)) is prmetrized y its rc length. Nottion. The convention is tht, when the prmetriztion is rc length prmetriztion, we use s s the vrile. Ths, when we write curve s x(s), we ssume is lredy prmetrized y rc length. Proof. (Optionl) See Proposition.3.6 in the textook. Exmples. Exmple 6. Consider the circle ( cos t; sin t) in the plne. We hve k_(t)k = k( sin t; cos t)k = : (3) Thus the rc length prmetriztion is ( cos(s/); sin(s/)). Exmple 7. Consider the spce curve x = cos t; y = sin t; z = t: (3) 9
10 Dierentil Geometry of Curves & Surfces The rc length prmetriztion is cos p p p s/ ; sin s/ ; s/. 4. Properties of curves from its derivtives Exmple 8. k(t)k = constnf nd only if _(t) (t) = for ll t. Proof. We hve The conclusion now follows. d k(t)k = d ((t) (t)) = _(t) (t): (3) Exmple 9. Let (t) e curve in R 3. Then the following two re equivlent. ) There is nonzero constnt vector such tht (t) = f(t) for some sclr function f(t) =/. ) _(t) (t) = for ll t. Proof. )=)). We hve _(t) (t) = f _ (t) f(t) = f _ (t) f(t) ( ) = : (33) )=)). Let (t) := (t) f(t) with f(t) := k(t)k. Then we hve = _(t) (t) = d (f(t) (t)) (f(t) (t)) = f _ (t) (t) + f(t) _ (t) (f(t) (t)) = f _ (t) (t) (f(t) (t)) + f(t) _ (t) (f(t) (t)) = f _ (t) f(t) ((t) (t)) + f (t) _ (t) (t) = f (t) _ (t) (t) : (34) As f(t) =/, we hve _ (t) (t) =, or _ (t) is prllel to (t). On the other hnd, s k(t)k = (t) = k(t)k =, y Exmple 8 there holds _ (t) (t) = or _ (t)?(t). k(t)k k(t)k Consequently _ (t) =, or (t) = is constnt vector. Exercise 3. Wht hppens if we drop the ssumption f(t) =/? Exmple. Let (t) e curve in R 3. Then the following re equivlent. ) There is nonzero constnt vector such tht?(t) for ll t; ) ((t) _(t)) (t) = for ll t.
11 Mth 348 Fll 7 Proof. )=)). We hve _(t) = d ( (t)) = ; (t) = d ( _(t)) = : (35) Thus if we set the 33 mtrix =( _ ), then =. Conseqeuently, reclling =/, ( _) = det = : (36) )=)). Cse. (t) _(t) = for ll t. Then y Exmple 9 (t) = f(t) for some constnt vector. Tke? nd the conclusion follows. Cse. (t) _(t) =/ for ny t. Let (t) := (t) _(t). As ( _) =, is contined in the plne spnned y nd _. Consequently is perpendiculr to the sme plne, nd Now let (t) := (t) k(t)k. Exercise 4. Prove tht (t) _ (t) =. (t) _(t) = ( _) ( ) = : (37) By similr rgument s in Exmple 9, we conclude tht _ (t) =, ths is constnt vector. Now the conclusion follows from (t) = for ll t.
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