Lecture 2 Intro to line integrls Dn Nichols nichols@mth.umss.edu MATH 233, Spring 218 University of Msschusetts April 12, 218 (2) onservtive vector fields We wnt to determine if F P (x, y), Q(x, y) is conservtive. Tht is, does there exist some f(x, y) such tht f(x, y) f x (x, y), f y (x, y) P (x, y), Q(x, y)? If P (x, y) xf(x, y), then f(x, y) P (x, y) dx F (x, y) + (y) If Q(x, y) y f(x, y), then f(x, y) Q(x, y) dy G(x, y) + D(x) Ide: compute these indefinite prtil integrls, see if they re comptible. Tht is, see if there s wy to choose (y) nd D(x) such tht the expressions bove re equl.
(3) onservtive vector fields For 3D field F P, Q, R, you need three prtil integrls: f(x, y, z) P (x, y, z) dx Q(x, y, z) dy R(x, y, z) dz When you compute n indefinite prtil integrl of function of three vribles, the constnt of integrtion depends on two vribles. f(x, y, z) dx F (x, y, z) + (y, z) where xf (x, y, z) f(x, y, z). (4) onservtive vector fields: exmple Exmple 1: Determine whether the vector field is conservtive. If it is conservtive, find potentil function for it. F(x, y, z) 2xe y yze xz, x 2 e y e xz, yxe xz + 1 Lbel the three component functions P, Q, nd R, so tht F(x, y, z) P (x, y, z), Q(x, y, z), R(x, y, z). P (x, y, z) dx ye xz + x 2 e y + (y, z) Q(x, y, z) dy ye xz + x 2 e y + D(x, z) R(x, y, z) dz ye xz + z + E(x, y) Let f(x, y, z) ye xz + x 2 e y + z. This is potentil for F.
(5) Derivtives nd integrls One gol of this clss: lern nd understnd true mening of the terms derivtive nd integrl. A derivtive is some wy of zooming in on specific point in spce or time nd mesuring how n object is chnging t tht point. f (): rte of chnge in f(x) t point x f(, b): vector describing chnge in multivrite function f(x, y) t point (, b) Derivtive of f t point P lets you construct liner pproximtion to f ner P, e.g. tngent line, tngent plne... An integrl is some wy of dding up the vlues of function over region of spce or time to find some kind of totl chnge. b f(x) dx: dd up vlues of f(x) on n intervl [, b] in R f(x, y) da: dd up vlues of f(x, y) on region D in R2 D Mny wys to do this depending on mening of region nd vlues of function. Whenever there s type of function for which we cn define the derivtive nd the integrl, there should be some kind of FTo-like theorem tht connects them. (6) The line integrl of sclr function The line integrl with respect to rc length lets us dd up vlues of sclr function f(x, y) long plne curve. Nottion: f(x, y) ds (ds mens with respect to rc length ) Geometric view: re of ribbon under f(x, y) on. Things you cn do with the rc length line integrl: Accumulte the vlue of some function over pth of motion. Let f(x, y) be the cost per unit distnce to trvel through the point (x, y). f(x, y) ds is the totl cost of trip long (t constnt speed). ompute verge vlue of some function over pth of motion. Let f(x, y) be the temperture t the point (x, y) The verge temperture you experience while wlking (t constnt speed) long the curve is 1 f(x, y) ds, where L is the rc length of. L
(7) Prmeterized curves The formul for the line integrl requires prmetriztion of. By this we men pir of functions x(t), y(t) such tht is the set of points (x(t), y(t)) for ll t in some intervl [, b]. Or vector function r(t) x(t), y(t), where is the set of points with position vectors r(t) for ll t in [, b]. y x So if you wnt to integrte over curve nd you only know n implicit eqution for it (e.g. x 2 + y 2 1) or description, you need to prmetrize it first. Exmple 2: Prmetrize these curves. (8) Prmeterized curves Unit circle x 2 + y 2 1 Hyperbol (right hlf) x 2 + y 2 1 Line segment ( 1, 2) to (3, 1) x cos t y sin t t 2π x tn t y sec t π/2 < t < π/2 x 1 + 4t y 2 t t 1
(9) The line integrl of sclr function How should the line integrl of f(x, y) over be defined? f(x, y) ds should be the re of the ribbon under f(x, y) on the curve. Ide: brek this ribbon into smll rectngulr pieces nd dd res together Tke limit s number of rectngles. (1) The line integrl Suppose we integrte f(x, y) over the curve trced by r(t) from t to t b. Brek [, b] up into n subintervls t points t, t 1,..., t n. y Ech subintervl hs wih t b n. On subintervl [t i, t i+1 ], the curve is pproximted by the vector r (t i ) t. x
(11) The line integrl At ech point t i, drw rectngle with bse r (t i ) t nd height f ( r(t i ) ). z r(t i ) r (t i ) t x y (12) The line integrl Add up the res of ll these rectngles. This is pproximtely the re of the ribbon. z y x So the pproximtion with n subintervls is n 1 f ( r(t i ) ) r (t i ) t. i
Using limits, we obtin n expression for the exct re of the ribbon: n 1 f(x, y) ds lim f ( r(t i ) ) r (t i ) t n b i f ( r(t) ) r (t) We cll r (t) the rc length derivtive, sometimes lbeled ds. Definition (Line integrl with respect to rc length) (13) The line integrl Let be curve in R 2 prmetrized by the vector function r(t) x(t), y(t) on the intervl t b. The line integrl with respect to rc length of function f(x, y) on is defined s: f(x, y) ds b b f ( r(t) ) r (t) (dx f ( x(t), y(t) ) ) 2 + ( ) dy 2. (14) The line integrl: exmple Exmple 3: Let be the line segment between the points (1, ) nd (3, 4). ompute the line integrl x cos y ds. Remember, if r(t) on [, b] is prmetriztion of, the formul is f(x, y) ds b f ( r(t) ) r (t). Esy prmetriztion of line segment from (x 1, y 1 ) to (x 2, y 2 ): r(t) (1 t) x 1, y 1 + t x 2, y 2 x 1 + t(x 2 x 1 ), y 1 + t(y 2 y 1 ), t 1 So let r(t) 1 + 2t, 4t on the domin [, 1].
(15) The line integrl: exmple Exmple 3: (cont.) Integrting x cos y ds, where is the line segment from (1, ) to (3, 4). We prmetrized s r(t) 1 + 2t, 4t on [, 1]. The rc length derivtive is constnt: ds r (t) 2. So the integrl we need to compute is b f(r(t)) r (t) 1 (1 + 2t) cos (4t) 2 1 2 cos(4t) + 2 1 2 t cos(4t) 2 4 sin(4t) 1 + 2 ( 1 2 4 t sin(4t) + 1 ) 1 16 cos(4t) 5 sin 4 + 5 cos 4 5 5 sin 4 + 2 4 4 3 5 sin 4 5 + ( 1 + cos 4). 2 4 (16) Arc length Remember: the rc length of curve prmetrized by r(t) x(t), y(t) on [, b] is L b ompre with the line integrl formul: b r (t) f(r(t)) r (t) ds. f(x, y) ds So the line integrl of f(x, y) 1 over is the rc length of.
(17) Arc length Suppose we hve curve prmetrized by r(t) on [, b]. Define new function s(t) t r (z) dz. This function tells you the rc length of the curve from strting t the beginning,, nd ending t t. Using the FTo, we cn write ds r (t). Or insted ds r (t). This tells you the reltionship between two infinitely smll things (differentils): : n infinitely smll time intervl ds: length of curve segment trced in tht time intervl This is why we replce ds with r (t) when we compute the line integrl f(x, y) ds. (18) Arc length prmetriztion There re infinitely mny prmeteriztions for ech curve Different speeds, different directions Some hve constnt speed, others hve chnging speed When we compute line integrl over, it doesn t mtter which prmetriztion we use. The ds in f ds mens tht the speed is tken into ccount. There s lwys specil one clled the rc length prmetriztion where ds 1. This is often the best prmetriztion, but it might be hrd to find. Exmple: r(t) cos t, sin t ds r (t)
(19) Piecewise curves If is complicted, we cn brek it into pieces nd integrte seprtely over ech one. Exmple 4: Let be the tringle with vertices (, ), (1, ), (, 2). ompute the line integrl (x2 + 2y) ds. Brek the tringle into three line segments: 1, 2, 3. Line integrl over splits into three esier line integrls. (x 2 + 2y) ds (x 2 + 2y) ds + 1 (x 2 + 2y) ds + 2 (x 2 + 2y) ds. 3 y 2 3 1 x (2) Piecewise curves: exmple Exmple 4: (cont.) 1 : r(t) t,, t 1 r (t) 1 2 + 2 1, so ds. (x 2 + 2y) ds 1 1 t 2 1 3. 2 : r(t), t, t 2 r (t) 2 + 1 2 1, so ds. (x 2 + 2y) ds 2 2 2t 4. 3 : r(t) t, 2 2t, t 1. 3 (x 2 + 2y) ds 1 r (t) 1 2 + ( 2) 2 5, so ds 5. (t 2 + 2(2 2t)) 5 7 5 3. So the totl is (x 2 + 2y) ds (x 2 + 2y) ds + (x 2 + 2y) ds + (x 2 + 2y) ds 1 2 3 1 3 + 4 + 7 5 13 3 3 + 7 5. 3
(21) Line integrls with respect to x nd y We cn insted do line integrls which only mesure distnce horizontlly or verticlly. These re the line integrls with respect to x nd y. Definition (Line integrls with respect to x nd y) Let be curve in R 2 prmetrized by r(t) x(t), y(t) on [, b]. The line integrls with respect to x nd y of function f(x, y) on re defined: f(x, y) dx f(x, y) dy b b f(x(t), y(t)) dx f(x(t), y(t)) dy Note tht dx dy nd re the derivtives of the components of r(t), or equivlently the components of r (t). Remember, the line integrl with respect to rc length is written with ds; these re different. (22) Line integrls with respect to x nd y: exmple Exmple 5: Let be the line segment from (1, 1) to (4, 4). ompute the following integrls: (x 2 y) dx e y dy Let s prmetrize by r(t) 1 + 3t, 1 + 3t on [, 1]. Then dx 3 nd dy 3.
(23) Line integrls with respect to x nd y: exmple Exmple 5: (cont.) is r(t) 1 + 3t, 1 + 3t on [, 1]. (x 2 y) dx 1 1 1 (x(t) 2 y(t)) dx ( (1 + 3t) 2 (1 + 3t) ) 3 ( ) 9t (9t + 27t 2 2 1 ) 2 + 9t3 27 2. e y dy 1 1 y(t) dy e e 1+3t 3 1 e 1+3t e 4 e. (24) Line integrls with respect to x nd y Often we wnt to compute the sum of two integrls over the sme curve, one with respect to x nd one with respect to y. In this cse, we write f(x, y) dx + g(x, y) dy f(x, y) dx + g(x, y) dy. This mens integrte f with respect to x on nd integrte g with respect to y on, then dd the results. Exmple 6: ompute the line integrl e x dx + (x + y 1) dy, where is the prt of the prbol y + 8x 2 8 tht s bove the x-xis. (Need to prmetrize first)
(25) Line integrls with respect to x nd y: exmple Exmple 6: (cont.) Prmetrize s r(t) t, 8 8t 2 on [ 1, 1]. Then dx 1 nd dy 16t. e x dx + (x + y 1) dy 1 1 1 1 1 1 e t 1 x(t) dx e e t + e t + 1 + 1 1 1 1 1 1 (x(t) + y(t) 1) dy ( t + 8 8t 2 1 ) ( 16t) ( 128t 3 16t 2 84t ) ( ) 16 1 3 t3 1 e 1 e 1 32 3. (Integrl of n odd function on [, ] is zero, so we cn ignore terms x n with n odd.) (26) Line integrls with respect to x nd y An integrl over with respect to rc length (ds) is lwys the sme regrdless of the orienttion (direction) of the curve, becuse ds is lwys positive. y y x x But when you integrte with respect to x nd y, reversing the orienttion chnges the sign of the result. f(x, y) dx + g(x, y) dy f(x, y) dx + g(x, y) dy
(27) Line integrls in 3 dimensions Everything we ve done so fr extends netly to 3 (or more) dimensions: f(x, y, z) ds b b f(r(t)) r (t) (dx f ( x(t), y(t), z(t) ) ) 2 + ( ) dy 2 + ( ) dz 2 And f(x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz f(x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz (28) Force fields Remember: vector fields cn be used to describe forces (like grvity) tht ffect everything in region of spce, where the force vector on n object depends on the object s position. GMm F G (x, y, z) (x 2 + y 2 + z 2 x, y, z ) 3/2-1 1 1.5 -.5-1
(29) The line integrl of vector field Suppose there s n object moving through spce subject to force field, e.g. grvity. y x We wnt to mesure the work done on the object by this force field. Remember tht work is the energy chnge in physicl system cused by force. (3) Work If n object moves long the displcement vector d P Q with constnt force vector F, then the work done by tht force is F d. d Q P F Work formuls: 1 dimension 2 dimensions constnt force W F d W F d nonconstnt force W F dx W??
(31) Work Let r(t) be the object s pth of motion from t to t b. For ny whole number n, we cn brek this curve into n segments t the points t i + i t, where t (b )/n. Ech segment cn be pproximted by tngent vector. y x Look t the i-th segment, from t i to t i + t. For smll t... the curve is pproximtely the tngent vector r (t i ) t the force is pproximtely the constnt vector F(x(t i ), y(t i )) y (32) Work x So the work done over the i-th segment is pproximtely F ( x(t i ), y(t i ) ) r (t i ) t.
(33) Work If we dd up the pproximte work done over ech of n sections, we get n 1 F ( x(t i ), y(t i ) ) r (t i ) t. i Tking the limit s n, this becomes definite integrl: W b b F ( x(t), y(t) ) r (t) F ( r(t) ) r (t). (34) The line integrl of vector field Integrting vector field over curve is useful in other pplictions besides work too. Definition (Line integrl of vector field) Let be curve prmetrized by the function r(t) on [, b] nd let F be vector field. The line integrl of F long is written F dr b F ( r(t) ) r (t). Different from the line integrl of sclr function w.r.t. rc length, x, or y This definition works in ny number of dimensions.
(35) The line integrl of vector field: nottion The textbook occsionlly uses different nottion: (I hve no ide why) F T ds T is the tngent vector field: T(x(t), y(t)) r (t) for points on the curve. Awkwrd. Just keep this in mind in cse you see it on the homework. (36) The line integrl of vector field: exmple Exmple 7: ompute the integrl F dr, where F(x, y) 1, 2y nd is the curve x y 2 between y 2 nd y 2. Remember, we need to find prmetriztion r(t) on [, b] nd then compute F dr b F(r(t)) r (t). We cn rewrite the implicit eqution defining s x y 2. For prmeteriztion of, tke r(t) t 2, t for 2 t 2. F dr b 2 2 2 2 ( dx F(x(t), y(t)), dy ) ( 1, 2t 2t, 1 ).
(37) The line integrl of vector field Remember tht if r(t) x(t), y(t) is curve in R 2, then dx r (t), dy. If we write F s F(x, y) P (x, y), Q(x, y), then the integrl becomes F dr F r (t) b [ P (x(t), y(t)) dx P (x, y) dx + Q(x, y) dy ] + Q(x(t), y(t))dy So the line integrl of vector field is relly the sme s the combined line integrl with respect to x nd y tht we sw erlier. (38) The line integrl of vector field: exmple Exmple 8: ompute the integrl by the vector function F dr, where is the spce curve prmetrized nd F is the vector field r(t) sin t, cos t, t 2 t 4π. F(x, y, z), z, 1. In this problem we ve been given prmetriztion, which mkes things esier. F dr b 4π 4π ( dx F(x(t), y(t), z(t)), dy, dz (, t, 1 cos t, sin t, 2t ) ( t sin t + 2t) 16π 2 + 4π. )
(39) Grvity Let s go bck to the grvity exmple we did lst time: very smll mss orbiting round very lrge mss in circulr pth of motion (rdius R). F(x, y, z) GMm x, y, z (R 3 ) 3/2 r(t) R cos t, R sin t, The force field is lwys orthogonl to the pth of the object, so the work done by grvity is lwys zero. Even though grvity is cusing the object to chnge direction, the mount of kinetic energy in the system remins constnt. r (t) F F r (t) (4) Homework No clss Tuesdy UMss follows Mondy schedule Pper homework #2 due Thursdy WebAssign homeworks 16.1, 16.2 due Wednesdy night Finl exm: Mondy 5/7, 8: AM. Fully cumultive!