December 2005 MATH 217 UBC ID: Page 2 of 11 pages [12] 1. Consider the surface S : cos(πx) x 2 y + e xz + yz =4.

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1 ecember 005 MATH 7 UB I: Page of pages ]. onsider the surface : cos(πx) x y + e xz + yz =4. (a) Find the plane tangent to at (0,, ). (b) uppose (0.0, 0.96,z) lies on. Give an approximate value for z. (c) uppose a>0 is very small. Then the circular cylinder x +(y ) = a cuts a tiny disk from the surface. Approximately what is the area of this disk? (a) Let G(x, y, z) =cos(πx) x y + e xz + yz. ThenG(x, y, z) =4on, so a normal to at (0,, ) is n = G(0,, ) = π sin(πx) xy + ze xz, x + z, xe xz + y (0,,) =,,. Thus the plane tangent to at (0,, ) is 0=n x 0,y,z =x +(y ) + (z ), or z = x (y ). (b) The point (0.0, 0.96,z) lies on the tangent plane just found when z = ] x (y ) = (0.0) ( 0.04) =.0..0,y=0.96 This is a good approximation to the z-coordinate on the surface as well. (Accurate approx: z =.0658.) (c) Let be the area of the patch of described in the question, and let A = πa be the area of the disk produced by projecting the patch into the xy-plane. Then A d da = G G/ z =, so A =πa. (0,,) ontinued on page

2 ecember 005 MATH 7 UB I: Page of pages ]. how that each critical point of this function gives a local minimum: f(x, y) = (x y x ) + (x ). First-Order Analysis: For f(x, y) = (x y x ) + (x ),wehave f x =(x y x )xy ] + (x )x] =x y x y xy +x x +, f y =(x y x )x ]=x 4 y x x, Find critical points (P s) starting with 0 = f y (x, y) =x (x y x ). Two cases arise: (i) ase x =0: Heref x (0,y) = for all y, so there are no P s with x =0. (ii) ase x y x = 0. In this case f x =0iff0=(x )x], i.e., x =0orx =. We already know there are no P s when x = 0, so we retain only the case x =. This gives y = x + above, and two solutions: x = = (x, y) =(, 0), x =+ = (x, y) =(, ). Further Investigation (option ): alculation shows that f(, 0) = 0 and f(, ) = 0. vidently f(x, y) 0 for all (x, y), so both critical points found above are global minimizers for f. Of course, that makes them local minimizers as well. Further Investigation (option ): tart with the second derivatives of f. f xx = 6x y 6xy y +6x, f xy = 4x y x x, f yx = 4x y x x, f yy = x 4. ] ] fxx f P (, 0): Here H(, 0) = xy 5 =. f yx f yy The determinant and diagonal entries are all ] positive, ] so this P gives a local minimum. fxx f P (, ): Here H(, ) = xy =. f yx f yy Again the determinant and diagonal entries are all positive, so this P gives a local minimum. ontinued on page 4

3 ecember 005 MATH 7 UB I: Page 4 of pages ]. Find the centroid (x, y, z) of the solid inside the cylinder x + y = 4, above the plane z =0,and below the paraboloid z =+x + y. Name the given solid. learly has rotational symmetry around the z-axis, so x =0=y. Also, has a simple projection along the z-axis onto the disk = { (x, y) :x + y 4 }. The projection fibre associated with point (x, y) runs from z min =0toz max =+x + y. Using polar coordinates in leads to this iteration, valid for any scalar field f on : If] = fdv = In particular, the volume of is given by = ] +x +y f(x, y, z) dz da(x, y) z=0 π +r θ=0 r=0 z=0 f(r cos θ, r sin θ, z) dz r dr dθ. I] = π +r θ=0 r=0 z=0 dz r dr dθ = π θ=0 r=0 r (r + r ) dr dθ =π + r4 4 ] r=0 =π + 4] = π and the moment of across the plane z =0isgivenby Iz] = = π θ=0 r=0 π θ=0 r=0 +r z=0 zdzrdrdθ ( + r ) rdrdθ r +r + r 5 =π r=0 = π +8+ ] = π r dr = π ( ) 6. + r4 4 + r6 6 ] r=0 Hence z = zdv dv = π(6/) π = 8. ontinued on page 5

4 ecember 005 MATH 7 UB I: Page 5 of pages ] 4. Let I = y/ y dx dy. x + y (a) Rewrite I as an iterated integral in polar coordinates. (b) valuate I. Hints: sec(at) dt = a ln sec(at)+tan(at), csc(at) dt = a ln csc(t) cot(t). The outer limits on I show that the domain lies in the horizontal strip y ; the inner limits show that it lies between the lines x = y and x = y/. Thus I = da(x, y), x + y where is a quadrilateral in the upper half-plane. Two sides of are provided by lines through the origin: these have polar equations θ = π/ andθ =π/4. The horizontal sides of are y =, i.e., r = =csc(θ), and y =, i.e., r = sin(θ) sin(θ) =csc(θ). These sides define r min and r max for use in the polar iteration of I: π/4 cscθ π/4 I = θ=π/ r=csc θ r rdrdθ= csc θ csc θ] dθ = ln csc(θ) cot(θ) θ=π/ =ln + ln / / ( ) ( ) =ln + ] =ln + 6. π/4 θ=π/ ontinued on page 6

5 ecember 005 MATH 7 UB I: Page 6 of pages ] 5. Let be a simple closed curve in the plane x +y + z =, oriented counterclockwise when viewed from high on the z-axis. (a) how that I() def = ydx+zdy xdz depends only on the area of the region enclosed by and not on the position or shape of. (b) Let be the triangular path from (, 0, 0) to (0,, 0) to (0, 0, ) to (, 0, 0). Find I() by calculating a cross product and using part (a). Observe that I() = F dr for F = y, z, x, and notice that i j k F = / x / y / z = i (0 ) j ( 0) + k (0 ) =,,. y z x Thus, by tokes s Theorem, every capping surface for obeys I() = ( F) N d =,, N d. Let be the plane region inside. Then the unit normal to is the same at every point, namely, N = n n =,, + + =,,. (This comes from reading a normal n =,, from the coefficients in the plane equation, and making sure it points upward for compatibility with the given orientation of.) onsequently I() =,,,, d = ( 6) d = Area(). As announced, this value depends only on the area inside and not on its position or shape. The triangular path in (b) lies in the given plane, and it encloses an area we can find using the cross product: Area() = u v, where u =(,, 0), v =(, 0, ). Thus i j k I() = Area() =( ) 0 =,, =. 0 ontinued on page 7

6 ecember 005 MATH 7 UB I: Page 7 of pages ] 6. Let be the surface cut from the parabolic cylinder z = y by the planes x =0,x =,andz =0. valuate y z I = 4y + d and I y z = 4y + d. The projection of along the z-axis into the xy-plane fills the rectangle =0, ], ]. ince every point of obeys 0 = G(x, y, z) def = y + z, area elements on are related to area elements in via d = G(x, y, z) G/ z Thus, for any constant p>0, y p z I p = 4y + d = da(x, y) = 0, y, da(x, y) = +4y da(x, y). y p ( y ) ( +4y da(x, y) = y 4y + p y +p) da(x, y). The region has reflection symmetry across the line y =0,soI p = 0 whenever p is an odd integer: thus When p is even, we get I p > 0: in particular, I = y= I =0. ( y y 4) dy dx = y y5 5 ] y= = 5 ] = ontinued on page 8

7 ecember 005 MATH 7 UB I: Page 8 of pages ( ) ( ) ( ) def e ] 7. For each a>0, evaluate I a = e x x ln(y) dx + y + sin(z) dy + y cos(z) dz, given a a : x = a cos(t), y = a, z = a sin(t), 0 t π. Recognize I a = F dr for F = e x ln(y), ex + sin(z), ycos(z). a y (The curve a is not closed, so it is unsafe to guess I a = 0.) ould F be conservative? Investigate i j k F = / x / y / z e x ln(y) e x /y +sin(z) y cos(z) = i (cos(z) cos(z)) j (0 0) + k (ex /y e x /y) =0. ince F passes the screening test at every point of the half-space where y>0, there must be some scalar field φ that satisfies F = φ there, i.e., () φ x = ex ln(y), () φ y = ex y + sin(z), φ () z = y cos(z). From (), φ = e x ln(y)+(y, z) forsome independent of x. This implies φ y = e x /y + y, and substitution in () gives e x y + y(y, z) =φ y = ex y + sin(z), so y(y, z) = sin(z), so (y, z) =y sin(z)+(z) for some independent of y and x. Thus φ = e x ln(y) +y sin(z) +(z) andφ z = y cos(z) + (z). ubstitution in () gives for some constant K. We arrive at y cos(z)+ (z) =y cos(z), so (z) =0, so (z) =K F = φ for φ(x, y, z) =e x ln(y)+y sin(z)+k. Any real K will work; we choose K = 0 and conclude that I a = φ dr = φ(r(π)) φ(r(0)) = φ( a, a, 0) φ(a, a, 0) = e a ln(a) e a ln(a) = ( e a e a) ln(a). a Alternative (irect): The given parametrization, x, y, z = a cos(t), a, asin(t), implies dx, dy, dz = a sin(t) dt, 0 dt, a cos(t) dt = a sin(t), 0, acos(t) dt. Making these substitutions gives I a = I a () + I a () + I a (),where π π I a () = e x ln(y) dx = ln(a) e a cos(t) ( a sin(t)) dt = ln(a) e a cos(t) a I () a = I () a = a ( ) e x y + sin(z) dy = a (y cos(z)) dz = a π 0 π t=0 ( e a cos(t) t=0 ) + sin(a sin(t)) 0 dt =0, t=0 a cos(a sin(t))(a cos(t)) dt = a sin(a sin(t)) =ln(a) e a e a], (The substitutions u = a cos(t) andw = a sin(t) helped evaluate I a () and I a ().) onsequently I a = I a () = ln(a) e a e a]. π t=0 =0. ontinued on page 9

8 ecember 005 MATH 7 UB I: Page 9 of pages ] 8. A particle travels from (, ) to (, ) along the curve y = x,thenbackto(, ) along the curve y = x 4 +, under the influence of the force Find the work done, i.e., W = F =(y + e x ln(y)) i +(e x /y) j. F dr, for the curve described above. Match W = (y + e x ln(y)) dx +(e x /y) dy with the left side in Green s Theorem by choosing P = y + e x ln(y), Q = ex y. Then if is the plane region inside the particle s path, ( Q W = x P ) ( ]) e x da(x, y) = y y + ex y A single integral is enough to find this area (using even symmetry): W = x= ( ( x ) ( + x 4 ) ) dx = 0 da(x, y) = da(x, y) = Area(). ( x x 4) dx = ] = Alternative (onservative Reduction): plit F = yei+ φ with φ(x, y) =e x ln(y). ince the path is a closed curve, W = ydx+ φ dr = ydx+0. Parametrizing gives (using even symmetry) ] ] W = ( x ) dx + (x 4 +)dx = x x x 5 + x= x= 5 + x = ++ ] = Alternative (irect): Parametrize both branches of directly, writing W = I + I.HereI is the line integral along,wherey = x implies dy = xdx,andx runs from right to left: I = ( x )+e x ln( x ) ] ( ) e x dx + ( xdx) = x= x= ( x )+ d ( e x ln( x ) ) dx dx = 6+ (e e ) ln(). x= x ] = 6+ e x ln( x ) imilarly, I is the line integral along,wherey = x 4 + implies dy =4x dx, so I = (x 4 +)+e x ln(x 4 +) ] ( ) e x dx + x= x 4 (4x dx) + = (x 4 +)+ d ( e x ln(x 4 +) ) ] dx =+ dx 5 + e x ln(x 4 +) ] x= =+ 5 +(e e ) ln(). ancellation occurs, leaving W = I + I = = ] x= ontinued on page 0

9 ecember 005 MATH 7 UB I: Page 0 of pages ] 9. Let denote the part of the surface z = e x selected by the simultaneous inequalities y 0, x, y x, and let F = x y xy, xy xy, z( + x + y 4xy). Let Φ be the upward flux of F through. (a) xpress Φ as a double integral over a suitable region in xy-space. (b) Use the ivergence Theorem to express Φ as a different double integral over. uggestion: Imagine as the top surface of a solid, whose bottom is z = 0 and whose sides are vertical planes. (c) valuate Φ. Let be the triangle in the xy-plane where y 0, x, y x. (a) The surface has a simple projection along the z-axis into. ince 0 = G(x, y, z) def = z e x at each point of, the vector area element on is related to the plane area element in by G(x, y, z) d = da(x, y) = xe x, 0, da(x, y). G/ z (Notice that this is directed upwards.) It follows that Φ= F d = F xe x, 0, da(x, y) ( ) = x(x y xy)e x + e x ( + x + y 4xy) da(x, y) = e ( x x y x y ++x + y 4xy ) da(x, y). (b) Notice that F =(xy y)+(xy x)+( +x + y 4xy) =. Thus, by the divergence theorem, the outward flux of F through any closed surface equals the volume inside that surface: F d = FdV = dv =Vol(). Use this observation on the solid defined by { = (x, y, z) :(x, y), 0 z e x}. The solid has as its top surface, and four other sides made from patches of various planes. On the bottom, z =0,wehave N = k so F N = 0 at every point. o there is no flux through the bottom. On the side where y =0,wehave N = j, sof N = 0 at every point. There is no flux through this side. On the side where x =,wehave N = i, sof N = 0 at every point. There is no flux through this side. On the vertical plane where y = x, wehave N =(j i)/ so F N = ( x y xy]+xy xy] ) = ( x x ]+x x ] ) =0. There is no flux through this side. Thus all the flux of F through comes through its top surface,! o e x Φ= dv = dz da = e x da. (c) The double integral from part (b) is much more convenient than the one from part (a). x ] e x Φ= e x dy dx = xe x dx = = e. y=0 z=0 The nd

10 This examination has pages including this cover The University of British olumbia essional xamination ecember 005 Mathematics 7 Multivariable and Vector alculus losed book examination Time: hours Name ignature tudent Number pecial Instructions: alculators may NOT be used. A formula sheet has been provided. If you need more space than is provided for a question, use the back of the previous page. Rules governing examinations. All candidates should be prepared to produce their library/am cards upon request.. Read and observe the following rules: No candidate shall be permitted to enter the examination room after the expiration of one half hour, or to leave during the first half hour of the examination. andidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. AUTION - andidates guilty of any of the following or similar practices shall be immediately dismissed from the examination and shall be liable to disciplinary action. (a) Making use of any books, papers or memoranda, other than those authorized by the examiners. (b) peaking or communicating with other candidates. (c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness shall not be received.. moking is not permitted during examinations Total 08