Prmetric Equtions Mth 6A Notes Written by Victori Kl vtkl@mth.ucsb.edu H 6432u Office Hours: R 12:30 1:30pm Lst updted 6/1/2016 If x = f(t), y = g(t), we sy tht x nd y re prmetric equtions of the prmeter t. The collection of points (x, y) = (f(t), g(t)) defines prmetric curve. If t is defined on domin t b, the initil point of the prmetric curve is (f(), g()) nd the terminl point is (f(b), g(b)). A very common prmetriztion of circle with rdius r (i.e. x 2 + y 2 = r 2 ) is given by x = r cos t, y = r sin t 0 t 2π. The esiest wy to prmetrize the curve y = f(x) is to let x = t, y = f(t). The derivtive dy dx is given by dy dx = dy dt dx dt if dx dt 0. If curve is described by the prmetric equtions x = f(t), y = g(t), α t β where f nd g re continuous on [α, β] nd is trversed exctly once s t increses from α to β, then the length of (rc length) is given by Vector Functions L = β α (dx ) 2 + dt ( ) 2 dy dt. dt A vector vlued function is vector whose components re functions. onsider r(t) given by r(t) = f(t), g(t), h(t) = f(t)i + g(t)j + h(t)k. r(t) is vector function from R (input ws t) to R 3 (output ws vector with three components). The limit of vector function is the limit of its components (provided they exist), i.e. lim t r(t) = lim t f(t), lim g(t), lim h(t). t t 1
The derivtive of vector r(t) is equl to the derivtive of ech of its components (provided they exist), i.e. r (t) = f (t), g (t), h (t). Recll tht velocity is given by r nd ccelertion is given by r. The speed is the mgnitude of velocity (see definition of mgnitude below). We cn lso integrte vector functions: ( b ) ( b ) ( b ) b r(t)dt = f(t)dt i + g(t)dt j + h(t)dt k. Recll the mgnitude (or length) of vector v = (v 1, v 2,..., v n ) is given by v = v1 2 + v2 2 +... + v2 n. The unit tngent vector to the curve r(t) is given by T = The rc length of r(t) from t b is given by L = b r (t) r (t). r (t) dt, which is similr to the rc length we defined erlier for prmetric equtions. The rc length function s is given by s(t) = t r (u) du, where is the initil point of r(t). ometimes we like to prmetrize curve with respect to rc length; to do so, we use the following steps: 1. Find the rc length function s. 2. Use your eqution from tep 1 to solve for t in terms of s, i.e. find t(s). 3. ubstitute your eqution from tep 2 into your function r(t); i.e. find r(t(s)). The curvture κ of r(t) is where T is the unit tngent vector of r(t). κ(t) = T (t) r (t) Two curves r 1 (t), r 2 (t) re sid to collide if they equl ech other t the sme time, i.e. r 1 (t) = r 2 (t) for some time t. Two curves intersect if they equl ech other but not t the time time, i.e. r 1 (t) = r 2 (s) for t s. 2
ot Product nd ross Product The dot product of two vectors = 1, 2,..., n nd b = b 1, b 2,..., b n is given by b = 1 b 1 + 2 b 2 +... + n b n. Notice tht the dot product is sclr (number). Here is lso useful theorem which llows us to clculte the ngle between two vectors: Theorem. The dot product of two vectors nd b is given by where θ is the ngle between the two vectors. b = b cos θ Two vectors re perpendiculr if nd only if their dot product is equl to 0. The cross product of two vectors = 1, 2, 3 nd b = b 1, b 2, b 3 is given by i j k b = 1 2 3 b 1 b 2 b 3. Notice the cross product is vector. The vector b is orthogonl (norml) to both nd b. We cn lso use the cross product to clculte the ngle between two vectors: Theorem 1. The cross product of two vectors nd b is given by where 0 θ π is the ngle between nd b. b = b sin θ Two vectors re prllel if nd only if their cross product is equl to the zero vector. Lines nd Plnes A line tht contins the point (x 0, y 0, z 0 ) in the direction v =, b, c is given by r(t) = r 0 + tv where r = x, y, z, r 0 = x 0, y 0.z 0. This cn be written in prmetric form s x = x 0 + t, y = y 0 + bt, z = z 0 + ct. Recll tht two lines r 1, r 2 intersect if they equl ech other t different times, i.e. r 1 (t) = r 2 (s) for t s. If two lines do not intersect but hve the sme direction, they re sid to be prllel; if two lines do not intersect nd do not hve the sme direction, they re sid to be skew. A plne tht contins the point (x 0, y 0, z 0 ) with norml vector n =, b, c is given by n (r r 0 ) = 0 where r = x, y, z, r 0 = x 0, y 0.z 0. This cn be written in the form (x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 or x + by + cz = d. 3
Level urves nd ontour Mps The level curves (or contour curves) of function f of two vribles re the curves with equtions f(x, y) = k where k is constnt in the rnge of f. Let P (x, y) be point on contour mp. We cn estimte prtil derivtives t the point P using our contour mp. Using the following strtegy: f x t P : Move to the right of your point. If the level curves decrese (i.e. k decreses), then f x < 0; if the level curves increse (i.e. k increses), then f x > 0. f y t P : Move up from your point. If the level curves decrese (i.e. k decreses), then f y < 0; if the level curves increse (i.e. k increses), then f y > 0. We cn lso use contour mps to estimte second prtil derivtives by looking t the spreding or bunching of the level curves. If the level curves bunch together s you move up or down, then the rte of the prtil derivtive is incresing. If the level curves spred prt s you move up or down, then the rte of the prtil derivtive is decresing. If the level curves sty constnt s you move up or down, then rte of the prtil derivtive is constnt. If... The prtil derivtive ws incresing t n incresing rte, or decresing t decresing rte, then the second prtil derivtive is > 0. The prtil derivtive ws decresing t n incresing rte, or incresing t decresing rte, then the second prtil derivtive is < 0. If the prtil derivtive hd constnt rte then the second prtil derivtive is = 0. ee the contour exmple for more informtion. Limits of Multivrible Functions To show limit of multivrible function does not exist, we tke different pths to show the limits re different on ech pth. For exmple, suppose we wnted to show does not exist. If we tke the pth x = y, then But, if we tke the pth x = 0, then x 2 y 2 lim (x,y) (0,0) x 2 + y 2 y 2 y 2 lim y 0 y 2 + y 2 = 0. 0 2 y 2 lim y 0 0 2 + y 2 = lim y 2 y 0 y 2 = 1. These two limits re clerly not equl, nd so the limit does not exist. 4
Prtil erivtives If f is function of two vribles, its prtil derivtives re the functions f x, f y defined by f(x + h, y) f(x, y) f x (x, y) = lim h 0 h f(x, y + h) f(x, y) f y (x, y) = lim. h 0 h When we tke the prtil derivtive with respect to x, y remins constnt. When we tke the prtil derivtive with respect to y, x remins constnt. We cn tke higher prtil derivtives s well. The second prtil derivtives of function f(x, y) re: f xx = (f x ) x, f xy = (f x ) y, f yx = (f y ) x, f yy = (f y ) y. We tke the derivtive left to right using this nottion. Theorem (lirut s Theorem). uppose f is defined on disk tht contins the point (, b). If the functions f xy, f yx re both continuous on, then f xy (, b) = f yx (, b). Tngent Plnes nd Liner Approximtions If f hs continuous prtil derivtives, n eqution of the tngent plne to the surfce z = f(x, y) t the point P (x 0, y 0, z 0 ) is z = z 0 + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). The lineriztion (or liner pproximtion) L(x, y) of function f(x, y) t the point (x 0, y 0 ) is the eqution L(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 ), which is just the eqution of the tngent plne. If z = f(x, y), the totl differentil dz is dz = z z dx + x y dy. dz is pproximtely equl to z, the chnge in height of our function z. irectionl erivtives nd Grdient If f is function of vribles x, y, z, the grdient of f is f = f x, f y, f z. 5
The grdient vector gives the direction of the fstest increse of f. mximum rte of chnge. Its mgnitude gives us the If f is differentil function, f hs directionl derivtive u f(x, y) in the direction of unit vector u given by u f(x, y, z) = f(x, y, z) u. We cn lso use the grdient to write equtions of the tngent plne to surfce f(x, y, z) = k t the point P (x 0, y 0, z 0 ): f (x x 0, y y 0, z z 0 ) = 0. Mximum nd Minimum Vlues A function of two vribles hs locl mximum t (, b) with locl mximum vlue f(, b) if f(x, y) f(, b) when (x, y) is ner (, b). If f(x, y) f(, b) when (x, y) is ner (, b), then f hs locl minimum t (, b) nd f(, b) is locl minimum vlue. In other words, locl mximum hs function vlue bigger thn everything round it, nd locl minimum hs function vlue smller thn everything round it. Theorem. If f hs locl mximum or minimum t (, b) nd the first-order prtil derivtives of f exist there, then f x (, b) = 0 nd f y (, b) = 0. We set f x = f y = 0 to find the criticl points (, b). We then use the second derivtive test to clssify the criticl points. Theorem (The econd erivtive Test). uppose the second prtil derivtives of f re continuous on disk with center (, b) nd suppose tht f x (, b) = 0 nd f y (, b) = 0 (i.e. (, b) is criticl point of f). Let = f xx (, b)f yy (, b) [f xy (, b)] 2. (i) If > 0 nd f xx (, b) > 0, then f(, b) is locl minimum. (ii) If > 0 nd f xx (, b) < 0, then f(, b) is locl mximum. (iii) If < 0, then f(, b) is sddle point. It my be helpful to remember the formul to s the following determinnt: = f xx f xy = f xxf yy (f xy ) 2. f yx f yy The mtrix bove is clled the Hessin mtrix. You my lso use the eigenvlues to clssify whether criticl point is mx, min, or sddle. The Extreme Vlue Theorem tells us tht function f on bounded set cn ttin its mximum inside or on the boundry of. To find the bsolute mximum nd minimum vlues of continuous function f on closed, bounded set : 1. Find the vlues of f t the criticl points of f in. 6
2. Find the extreme vlues of f on the boundry of. 3. The lrgest vlues from (1) nd (2) re the bsolute mximum vlue; the smllest vlues is the bsolute minimum vlue. Vector Fields A vector field is function F tht ssigns point (x, y, z) to vector F(x, y, z). The grdient f forms vector field, we cll it the grdient vector field. A vector field F is clled conservtive vector field if there exists potentil function f such tht f = F. url nd ivergence The curl of function F = P i + Qj + Rk is defined s curl F = F. Written out more explicitly: i j k curl F = F = x y z P Q R = y z Q R i x z P R j + x P ( R = y Q ) ( P i + z z R x The divergence of function F = P i + Qj + Rk is defined s Q k y ) j + div F = F = P x + Q y + R z. ( Q x P y Notice tht the curl of function is vector nd the divergence of function is sclr (just number). Line Integrls If f is function defined long smooth curve defined by x = x(t), y = y(t), t b, then the line integrl of f long is b (dx ) 2 ( ) 2 dy f(x, y)ds = f (x(t), y(t)) + dt. dt dt ) k 7
If is defined by the prmetriztion r(t) = x(t), y(t), then this formul cn lso be written s b f(x, y)ds = f(r(t)) r (t) dt. Another form of line integrl is the following: P (x, y)dx + Q(x, y)dy = P (x(t), y(t))x (t)dt + Q(x(t), y(t))y (t)dt. The previous definition ws for line integrls of sclr functions. We cn lso evlute line integrls with vector functions. Let F be continuous vector field defined on smooth curve given by vector function r(t), t b. Then the line integrl of F long is b F dr = F(r(t)) r (t)dt. This lso known s work or circultion. If given grph of vector field F nd pth, we cn determine the sign of the line integrl F dr: If the pth trvels ginst the vector field, then the integrl is negtive. If the pth trvels long the vector field, then the integrl is positive. If the pth ppers to both in symmetric wy, or if the vector field is perpendiculr to the pth t ll times, then the integrl is zero. The Fundmentl Theorem of Line Integrls Theorem (Fundmentl Theorem of Line Integrls). Let be smooth curve given by the vector function r(t), t b. Let f be differentible function whose grdient vector f is continuous on. Then f dr = f(r(b)) f(r()). In other words: the line integrl of conservtive vector field is independent of the pth. We cn use the following theorem to determine when vector field is conservtive. Theorem. Let F be vector field. The following re conservtive: (i) F is conservtive (ii) There exists potentil function f such tht f = F (iii) curl F = 0 (iv) F is pth independent (v) F dr = 0 for ALL closed pths Note: If F = P i + Qj, F is conservtive if Q x = P y. 8
ouble Integrls If f(x, y) 0 then the double integrl f(x, y)da represents the volume of the solid tht lies R bove the rectngle R nd below the surfce z = f(x, y). We cn estimte double rectngles using Riemnn sums: R f(x, y)da m i=1 j=1 n f(x ij, y ij ) A where A = x y, m, n re the number of prtitions of x, y intervls, respectively, nd x, y re the width of the prtitions. Theorem (Fubini s Theorem). If f is continuous on the rectngle R = {(x, y) : x b, c y d}, then b d d b f(x, y)da = f(x, y)dydx = f(x, y)dxdy. R c Fubini s Theorem sys tht the order of integrtion doesn t mtter when we integrte over rectngles. Fubini s Theorem lso pplies to other chnge of coordintes (like polr coordintes). We cn lso integrte over regions tht re not defined by rectngles. If f is continuous on = {(x, y) : x b, g 1 (x) y g 2 (x)} then f(x, y)da = b g2(x) g 1(x) c f(x, y)dydx. If f is continuous on = {(x, y) : c y d, h 1 (y) x h 2 (y)}, then f(x, y)da = d h2(y) c h 1(y) f(x, y)dxdy. The order of integrtion my mtter depending on the function nd region we re integrting over (we sw severl exmples of this in section). To find the re over region, we cn integrte A = where da is n re element (could be dxdy, dydx, or rdrdθ s seen in the next section). Polr oordintes We use polr coordintes whenever we integrte over circulr regions. We use the prmetriztion x = r cos θ, y = r sin θ. It follows tht x 2 + y 2 = r 2. da 9
If f is continuous on polr region = {(r, θ) : α θ β, h 1 (θ) r h 2 (θ)}, then f(x, y)da = β h2(θ) α h 1(θ) f(r cos θ, r sin θ)rdrdθ. In other words, we substitute the equtions bove in for x nd y nd use the re element da = rdrdθ. Green s Theorem We sy tht closed curve is positively oriented if it is oriented counterclockwise. It is negtively oriented if it is oriented clockwise. Theorem (Green s Theorem). Let be positively oriented, piecewise smooth, simple closed curve in the plne nd let be the region bounded by. If F(x, y) = (P (x, y), Q(x, y)) with P nd Q hving continuous prtil derivtives on n open region tht contins, then F dr = P dx + Qdy = ( Q x P y ) da. In other words, Green s Theorem llows us to chnge from complicted line integrl over curve to less complicted double integrl over the region bounded by. Triple Integrls To find the volume of solid E, we cn integrte V = where dv is volume element. Triple integrls re very similr to double integrls. Not only do we need to find bounds for x nd y, but we lso need to find bounds for z. The bounds my hve dependence on zero, one, or two vribles. ylindricl oordintes ylindricl coordintes re n ppliction of polr coordintes into three dimensions. In fct, we use the sme substitutions for x nd y, nd just leve z the sme: It follows tht r 2 = x 2 + y 2. E dv x = r cos θ, y = r sin θ, z = z. 10
If f is continuous nd E = {(x, y, z) : (x, y), u 1 (x, y) z u 2 (x, y)} where is given in polr coordintes by = {(r, θ) : α θ β, h 1 (θ) r h 2 (θ)}, then β h2(θ) u2(r cos θ,r sin θ) f(x, y, z)dv = f(r cos θ, r sin θ) rdzdrdθ. E α h 1(θ) u 1(r cos θ,r sin θ) In other words, we substitute the equtions bove in for x nd y nd use the volume element dv = rdrdθdz. phericl oordintes phericl coordintes re nother ppliction of polr coordintes in three dimensions, but now we hve three prmeters insted of two. Our prmeters re ρ (r in polr coordintes), θ (sme s in polr coordintes), nd φ, known s the zimuthl ngle. The zimuthl ngle is mesured from the positive z-xis to the negtive z-xis. The substitutions for x, y, nd z re s follows: It follows tht ρ 2 = x 2 + y 2 + z 2. x = ρ cos θ sin φ, y = ρ sin θ sin φ, z = ρ cos φ. If f is continuous nd E = {(ρ, θ, φ) : ρ b, α θ β, c φ d}, then d β b f(x, y, z)dv = f(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ)ρ 2 sin φ dρ dθ dφ. E c α In other words, we substitute the equtions bove in for x, y, z nd use the volume element dv = ρ 2 sin φ dρ dθ dφ. o when do we use cylindricl coordintes versus sphericl coordintes? As the nmes of these methods imply, use cylindricl coordintes for cylinders, nd sphericl coordintes for spheres. But these certinly ren t the only shpes we integrte over. My rule of thumb is to use cylindricl coordintes for everything other thn spheres (this includes prboloids, cylinders, nd so on). If spheres re involved, then I use sphericl coordintes. hnge of Vribles uppose we wnt to integrte f(x, y) over the region R, but R might be complicted in terms of x nd y. If we cn find trnsformtion x = g(u, v), h(u, v) such tht the region R becomes, we cn rewrite our integrl s follows: f(x, y)da = f(g(u, v), h(u, v)) J dudv where J is the Jcobin ( fudge fctor ) given by x J = u y u R 11 x v y. v
urfce Are If the prmetriztion of surfce is given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k where is the region of u, v, then the surfce re of over is A = r u r v da. Note tht n = r u r v is the norml to the surfce. urfce Integrls We cn evlute two types of surfce integrls: surfce integrls over sclr functions nd surfce integrls over vector fields. If is given by prmetriztion r(u, v) nd f is sclr function defined on, then f(x, y, z)d = f(r(u, v)) r u r v da where is the rnge of the prmeters u, v. If F is continuous vector field defined on n oriented surfce with unit norml vector n, then the surfce integrl of F over is F d = F n d. If is given by vector function r(u, v), then the bove integrl cn be written s F d = F (r u r v )da where is the prmeter domin (the rnge of the prmeters u, v). If we re given is the function z = g(x, y), we cn define the prmetriztion of to be r(x, y) = (x, y, g(x, y)). Then the surfce integrl of F over is given by F d = F (r x r y )da where is the projection of z = g(x, y) onto the xy plne. 12
tokes Theorem Theorem (tokes Theorem). Let be n oriented piecewise-smooth surfce tht is bounded by simple, closed, piecewise-smooth boundry curve with positive orienttion. Let F be vector field whose components hve continuous prtil derivtives on n open region in R 3 tht contins. Then F dr = curl F d. In other words, the line integrl round the boundry curve of surfce of the tngentil component of F is equl to the surfce integrl of the norml component of the curl of F. Note: This looks very similr to Green s Theorem! In fct, Green s Theorem is specil cse of tokes Theorem; it is when F is restricted to the xy plne. tokes Theorem cn then be thought of s the higher-dimensionl version of Green s Theorem. ivergence Theorem Theorem (ivergence Theorem). Let E be simple solid region nd let be the boundry surfce of E, given with positive (outwrd) orienttion. Let F be vector field whose component functions hve continuous prtil derivtives on n open region tht contins E. Then F d = div FdV. In other words, the divergence theorem sttes tht the flux of F cross the boundry surfce of E is equl to the triple integrl of the divergence of F over E. This theorem llows us to chnge from double integrl to triple integrl. E 13