Math 1. 2-hours test May 13, 2017

Transcription

1 Math. -hors test May, 7 JE/JKL.5.7 Problem restart:with(plots): A fnction f of two real variables is for x, y, given by f:=(x,y)-y/(x^+y^); f x, y y x y f(x,y); y x y (.) (.) Qestion In the x, y plane the three points A= (, ), B = (, ) and C= (, ) are considered. 'f(,)'=f(,); 'f(,-)'=f(,-); 'f(/,/)'=f(/,/); f f, = f, =, Since f(a) = f(c) =, A and C lie on the same level crve for f viz. on the level crve y f x, y =, x, y, x =, (x, y, y x y y =, x, y, x y =, (x, y),, 4 that is, the circle with centre at (, ) and radis except the point (,). Since f(b) =, B does not lie on this level crve for f, bt on the level crve f x, y =, x, y, x y =, (x, y),, 4 that is, the circle with centre at (, ) and radis except the point (,). The gradient for f is given by = (..) (..) (..) f x, y = f x (x, y), f y (x, y) = Qestion xy x y, x y x y, (x, y),.

2 If the first coordinate has to be zero, then either x of y mst be zero. This implies that if the second coordinate also has to be zero, then both x and y mst be zero. Bt f is not defind for (,). Since f x, y, for all x, y,, f has no stationary points. We consider the bonded and closed set of points M = x, y x y 4. I.e. M is the closed circlar disc between the two discs x y = (centre, and radis ) and x y = 4 (centre, and radis ). C:=implicitplot(x^+y^<4,x=-..,y=-..,scaling=constrained, linestyle=): C:=implicitplot(x^+y^,x=-..,y=-..,filled=tre, coloring=[gray,white],scaling=constrained,linestyle=): R:=implicitplot(x^+y^=,x=-..,y=-..,color=black,scaling= constrained,linestyle=,thickness=): R:=implicitplot(x^+y^=4,x=-..,y=-..,color=black,scaling= constrained,linestyle=,thickness=): C:=implicitplot(x^+y^4,x=-..,y=-..,scaling=constrained, filled=tre,coloring=[white,white],linestyle=): display(c,c,c,r,r,title="cirkelringen M"); Note that it is not a reqirement to the answer that yo can draw this figre of M in Maple. Qestion Since M is bonded and closed and since f is continos in M, then f has a global minimm and a global maximm in M. Since f has neither stationary points nor exception points in the interior

3 of M, these vales are attained at the bodary of M. Bondary investigation: (): x y = x = y, y ;. g y = f y, y = y, y ;. g y = for all y ;. I.e. g is increasing. g = f, = is smallest and g = f, = is largest. (): x y = 4 x = 4 y, y ;. h y = f 4 y, y = 4 y, y ;. h y = 4 for all y ;. I.e. h is increasing. h = f, = is smallest and h = f, = is largest. By nmerical comparison of these investigation we find that the global minimm is and is attained at point B and that the global maximm is, that is attained in point A. Note that since M is connected, the range is f M = ;. Problem restart: The approximating second-degree polynomial abot the point (, ) of a real fnction f x, y is given by P x, y = f, f x, x f y, y f xx, x f xy, xy f yy, y = x y = x y. while the approximating second-degree polynomial abot the point ( Q x, y = f = 4, f x f xx, x e e x, x f y f xy, x e y = 4 e, y e x,) is given by y f yy, y e y. Qestion From the given expression for P x, y we read that f, =, f x, =, f y, =, f xx, =, f xy, = and f yy, =.

4 Qestion If f has a proper local maximm or a proper local minimm in a point, then the point mst be a stationary point, since f has no exception points. Since f, = f x, ), f y (, ) =,, (,) is a stationary point for f. The Hessian matrix for f in the point (,) is H(, ) = f xx, f xy, f xy, f yy, = that has the eigenvales and. Since both eigenvales for H(, ) are positive, then f, = is a proper local minimm., Qestion From the given expression for Q x, y we read that f, = 4 e, f x, =, f y, =, f xx f yy, = Since f point for f. e., = f x The Hessian matrix for f in the point H, =,, f y f xx, f xy f xy, f yy, is, = e, f xy, =,, (,, = e, = and,) is also a stationary e, that has the eigenvales e and e. Since the two eigenvales for H, are of opposite sign, then f proper local minimm nor a proper local maximm., is neither a Problem restart:with(linearalgebra):with(plots): prik:=(x,y)-vectorcalcls[dotprodct](x,y):

5 kryds:=(x,y)-convert(vectorcalcls[crossprodct](x,y),vector) : A cylindrical srface F in the x, y, z -space is given by the parametric representation r:=(,v)-<^-,^+,v*: r(,v); v (.) where ; and v ;. plotd(r(,v),=..sqrt()/,v=..,orientation=[-4,8],axes= normal,view=[-...,..,-...]); Qestion r:=diff~(r(,v),); r v (..)

6 rv:=diff~(r(,v),v); rv (..) The normal vector is N:=kryds(r,rv); N (..) The Jacobi fnction corresponding to r is Jacobi:=simplify(sqrt(prik(N,N))); Jacobi 4 4 (..4) Ar(F) = d = Jacobi, v d dv F Int(Int(Jacobi,=..sqrt()/),v=..)=int(int(Jacobi,=.. sqrt()/),v=..); 4 4 d dv = 7 (..5) Let L denote the directrix in the (x,y)-plane corresponding to F. Qestion L is the crve of intersection between the srface F and the x, y -plane. a parametric representation for L is then s:=r(,); s (..) where,. s:=diff~(s,); s (..) The Jacobi fnction corresponding to L hørende Jacobifnktion is Jacobi:=simplify(sqrt(prik(s,s))); Jacobi 8 (..)

7 Qestion L y x d = y x Jacobi d = Jacobi d Int(*Jacobi,=..sqrt()/)=int(*Jacobi,=..sqrt()/); Problem 4 8 d = 7 restart:with(linearalgebra):with(plots): prik:=(x,y)-vectorcalcls[dotprodct](x,y): kryds:=(x,y)-convert(vectorcalcls[crossprodct](x,y),vector) : div:=v-vectorcalcls[divergence](v): rot:=proc(x) ses VectorCalcls;BasisFormat(false);Crl(X)end proc: (..) A vector field in the x, y, z -space is given by V:=(x,y,z)-<x^,-*y*x,z: V(x,y,z); In the x, z -plane we consider a profile region A given by the parametric representation s:=(,v)-<,,v*sin(): s(,v); x y x z v sin (4.) (4.) where ; and v ;. plotd(s(,v),=..pi/,v=..,axes=normal,scaling=constrained) ;

8 A solid of revoltion Ω appears by rotation of A throgh the angle 4 abot the z-axis conterclockwise as seen from the positive end of the z-axis. Qestion The parametric representation for Ω r:=(,v,w)-<*cos(w),*sin(w),v*sin(): r(,v,w); cos w sin w v sin (4..) where ;, v ; and w ; 4. Qestion is the closed srface of Ω oriented with otward-pointing nit normal vector. From Gass' theorem we then get fås da

9 Flx(V, ) = Div (V) d = div(v)(x,y,z); 4 Div (V)(r(, v, w)) Jacobi(, v, w) d dv dw M:=<diff~(r(,v,w),) diff~(r(,v,w),v) diff~(r(,v,w),w); cos w sin w M sin w cos w v cos sin Jr:=simplify(Determinant(M)); Jr sin that is, since ; Jacobi:=-Jr;. The jacobi fnction corresponding to r is then Jacobi integranden:=*jacobi; integranden sin sin (4..) (4..) (4..) (4..4) (4..5) Int(Int(Int(integranden,=..Pi/),v=..),w=..Pi/4)=int(int (int(integranden,=..pi/),v=..),w=..pi/4); 4 sin d dv dw = 4 (4..) Qestion Let G denote the part of the srface of Ω, that bonds Ω above. Ths G is the srface of revoltion that appears by rotating the pper bondary crve for A in the x, z -plane throng the angle 4 abot the z-axis conter-clockwize as seen from the poitive side of the z-axis. A parametric representation G is then R:=r(,,w); R where ; and w ; 4 R:=diff~(R,); R Rw:=diff~(R,w);. cos w sin w sin cos w sin w cos (4..) (4..) (4..)

10 Rw sin w cos w (4..) Thenormal vector of the srface is N:=simplify(kryds(R,Rw)); cos cos w N cos sin w (4..4) N, w forms an acte angle with the z-axis, since its z-coordinat z, w = ;. for flade:=plotd(r,=..pi/,w=..pi/4,scaling=constrained): pil:=arrow(sbs(=,w=pi/8,r),sbs(=,w=pi/8,n)): display(flade,pil,orientation=[-,7]); With the chosen orientation of the closed bondary crve G for G we then have n G = N, w N, w for ; (the right-hand rle). From Stokes' theorem we then have

11 Cirk(V, G = Flx(Crl(V), G) = d dw rot(v)(x,y,z); n G Crl V d = G 4 N, w Crl V (R(,w)) (4..5) y The crl of the vector field on the srface is Crl:=<,,-**sin(w); Rot integranden:=prik(n,crl); integranden sin w sin w Int(Int(integranden,=..Pi/),w=..Pi/4)=int(int (integranden,=..pi/),w=..pi/4); 4 sin w d dw = 4 (4..) (4..7) (4..8)