Mth 35 Review heet, pring 2014 For the finl exm, do ny 12 of the 15 questions in 3 hours. They re worth 8 points ech, mking 96, with 4 more points for netness! Put ll your work nd nswers in the provided booklets. To get ll 8 points for question it is very importnt tht you show clerly ll your working out nd resoning. Min Topics: Types of derivtives. Let f be function from Eucliden n-spce to Eucliden m- spce, ie f : R n R m. For x = (x 1, x 2,..., x n ) we cn write f(x) = (f 1 (x), f 2 (x),..., f m (x)). () Then the derivtive of f is n m n mtrix f 1 f 1 x 1 x 2 f 2 f 2 x f(x) = 1 x 2..... f m x 2 f m x 1 (b) For the cse m = n, the Jcobin of f t x is det f(x). Other nottions for the Jcobin re f 1 f 1 f x 1 x 2 1 x m (f 1, f 2,, f m ) f 2 f 2 f x nd 1 x 2 2 x m (x 1, x 2,, x m )...... f m f m f x 2 m (c) For m = 1 we hve f : R n R nd the derivtive becomes 1 n mtrix. This vector is clled the grdient: ( f f(x) =, f,, f ). x 1 x 2 x n Lgrnge multipliers. Let f : R n R nd g : R n R hve continuous prtil derivtives. To find the mximum nd minimum vlues of f(x) subject to the constrint g(x) = k, solve the system x 1 f(x) = λ g(x) g(x) = k. Inverse Function Theorem. uppose g : R m R m hs continuous prtil derivtives. If g(y 0 ) = t 0 nd the Jcobin of g t y 0 is not zero then g hs n inverse ner y 0. This mens there is function h : R m R m so tht ll solutions to re given by y = h(t). f 1 x n f 2 x n f m x n. g(y) = t with y ner y 0 nd t ner t 0 x m
Riemnn ums. ouble integrls f(x, y) da nd triple integrls f(x, y, z) dv B re defined s the limits of double nd triple Riemnn sums. ouble integrls. () Fubini s Theorem tells us tht, if f is continuous, double integrl f(x, y) da over rectngle = [, b] [c, d] cn be evluted s the iterted integrl b d c f(x, y) dydx or d b c f(x, y) dxdy. (b) For more complicted type I regions we use b II regions we use d h2 (y) c h 1 f(x, y) dxdy. (y) g2 (x) g 1 f(x, y) dydx nd for type (x) (c) For circulr region R we chnge from rectngulr coordintes (x, y) to polr (r, θ) with (x, y) x = r cos θ, y = r sin θ, (r, θ) = r. If (x, y) R corresponds to (r, θ) then f(x, y) dxdy = f(r cos θ, r sin θ) (x, y) (r, θ) drdθ = R Triple integrls. f(r cos θ, r sin θ) rdrdθ. () Fubini s Theorem tells us tht, if f is continuous, triple integrl f(x, y, z) dv B over box B = [, b] [c, d] [r, s] cn be evluted s the iterted integrl b d s or in ny other order of integrtion. c r f(x, y, z) dzdydx (b) A type 1 solid region E is one tht lies between two grphs u 1 (x, y) nd u 2 (x, y) with (x, y). Then [ ] u2 (x,y) f(x, y, z) dv = f(x, y, z) dz da E nd similrly for the other xes x nd y. u 1 (x,y) (c) For cylindricl solid region E we chnge from rectngulr coordintes (x, y, z) to cylindricl (r, θ, z) with x = r cos θ, y = r sin θ, z = z. with obvious vrints if the cylindricl xis is in the x or y direction. (d) For sphericl solid region E we chnge to sphericl coordintes (ρ, θ, φ) with x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, (x, y, z) (ρ, θ, φ) = ρ2 sin φ.
If (x, y, z) E corresponds to (ρ, θ, φ) then f(x, y, z) dxdydz = f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) (x, y, z) E (ρ, θ, φ) dρdθdφ = f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ 2 sin φ dρdθdφ. Vector Fields. () A vector field F ssigns vector to ech point in spce. For exmple F(x, y) = 2i+xyj is 2-dimensionl field nd F(x, y, z) = z 2 i 3j+y/zk is 3-dimensionl. (b) If f(x, y, z) is function then its grdient vector field is f(x, y, z) = f x i + f y j + f z k nd vector field F is conservtive if F = f for some f (its potentil function). (c) For vector field F = P i + Qj + Rk, we define the new vector field i j k curl F = F = x y z P Q R nd the divergence ( function) div F = F = P x + Q y + R z. pce curves. Let be spce curve, prmeterized by r(t) = (x(t), y(t), z(t)) t b. The unit (length) tngent vector T t point r(t) on is given by T(t) = r (t) r (t) for r (t) = (x (t), y (t), z (t)). For exmple the circle of rdius, centered t the origin in the xy-plne with positive (counter-clockwise) orienttion cn be prmeterized by r(t) = ( cos t, sin t), 0 t 2π. For nother exmple, the line segment from r 0 to r 1 my be prmeterized by r(t) = (1 t)r 0 + tr 1, 0 t 1. Line Integrls. Let be spce curve s bove. () The line integrl of function f long is b f ds = f(r(t)) r (t) dt. (b) The line integrl of function f long w.r.t. x is b f dx = f(r(t))x (t) dt. nd similrly w.r.t. y nd z.
(c) The line integrl of vector field F long is b F dr = F(r(t)) r (t) dt. Note the reltions nd F dr = F T ds = b b = = F(r(t)) T(t) r (t) dt F(r(t)) r (t) dt F dr P dx + Q dy + R dz for F = P i + Qj + Rk. Fundmentl Theorem for Line Integrls. For smooth curve (prmeterized s bove) nd f with continuous prtil derivtives then f dr = f(r(b)) f(r()). It follows from this theorem tht integrls of conservtive vector fields re independent of the pth tken between the endpoints. It lso follows tht the integrl of conservtive field round closed curve is zero. Tests for when vector field is conservtive. We cn use the following tests. For 2-dimensionl field F(x, y) = P i + Qj with continuous prtil derivtives on domin (x, y) then F conservtive = P y = Q on, x P y = Q on ( open, simply connected) = F conservtive. x For 3-dimensionl field F with continuous prtil derivtives on domin then F conservtive = curl F = 0 on, curl F = 0 on = R 3 = F conservtive. For exmple, if you compute curl F nd find it is not zero, then F is not conservtive. Another wy to prove field is conservtive is to try to prtilly integrte it w.r.t. x, y (nd z) to find the potentil function f. Prmetric urfces. Let be surfce, prmeterized by r(u, v) = (x(u, v), y(u, v), z(u, v)) (u, v). The tngent vectors to the surfce t the point r(u, v) re r u = x u i + y u j + z u k, r v = x v i + y v j + z v k.
This gives norml vector r u r v to the surfce t the point r(u, v) nd the eqution of the tngent plne to the surfce there is We cn define the unit norml to to be ((x, y, z) r(u, v)) (r u r v ) = 0. n = r u r v r u r v. hnging the sign of n gives the opposite direction for the norml. A surfce is clled orientble if you cn choose the norml n so tht it vries continuously over. (For exmple, the Möbius strip is not orientble, but sphere is.) Every orientble surfce hs two possible orienttions. urfce Integrls. Let be surfce prmeterized s bove. () The surfce integrl of function f over is f d = f(r(u, v)) r u r v da. (b) The surfce integrl of vector field F over n oriented surfce is F d = ± F (r u r v ) da with the sign depending on the choice of orienttion. Note the reltion r u r v F n d = F r u r v r u r v da = F (r u r v ) da = F d nd this surfce integrl is lso clled the flux of F cross. When the surfce is grph. A nice cse is when surfce is given by the set of points (x, y, z) where (x, y) nd z = g(x, y). We give this surfce the upwrd orienttion. Let F = P i + Qj + Rk be vector field. You should know how to derive the useful formul F d = ( P g x Q g y + R Green s Theorem. Let be positively oriented, piecewise-smooth, simple closed curve in the xy-plne with the region bounded by. If P nd Q hve continuous prtil derivtives on n open region contining then ( Q P dx + Q dy = x P ) da. y ) da.
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 with continuous prtil derivtives on n open region contining then F dr = curl F d. The ivergence Theorem. Let E be simple solid region nd let be the boundry surfce of E with the outwrd orienttion. Let F be vector field with continuous prtil derivtives on n open region contining E then F d = div F dv. Theorems. You should know how to stte precisely nd pply the following theorems: (i) The Inverse Function Theorem (ii) Fubini s Theorem (iii) The Fundmentl Theorem for Line Integrls (iv) Green s Theorem (v) tokes Theorem (vi) The ivergence Theorem Mesurement. A nice ppliction of our work is to compute lengths, res nd volumes. 1 ds = length of curve 1 da = re of flt surfce 1 d = re of surfce 1 dv = volume of solid B. B We lso sw how to compute the re of using line integrl: choose P nd Q so tht Q P = 1, for exmple Q = x nd P = 0, then by Green s Theorem x y re of flt surfce = 1 da = x dy. enter of mss. Another ppliction is to find the mss m nd center of mss (x, y) of n object with possibly vrying density. () Let thin wire with liner density ρ(x, y) t ech point be shped like the curve in the xy-plne. Then m = ρ(x, y) ds, x = 1 xρ(x, y) ds, y = 1 yρ(x, y) ds. m m E
(b) Let be lmin (thin flt shpe) in the xy-plne with ρ(x, y) giving the mss per re t ech point. Then m = ρ(x, y) da, x = 1 xρ(x, y) da, y = 1 yρ(x, y) da. m m (c) Let thin sheet be curved like the prmetric surfce with ρ(x, y, z) giving the mss per re t ech point. Then m = ρ(x, y, z) d nd the center of mss (x, y, z) is found by x = 1 xρ(x, y, z) d, y = 1 yρ(x, y, z) d, m m z = 1 zρ(x, y, z) d. m