MATHEMATICS 317 December 2010 Final Exam Solutions

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1 MATHEMATI 317 December 1 Final Eam olutions 1. Let r(t) = ( 3 cos t, 3 sin t, 4t ) be the position vector of a particle as a function of time t. (a) Find the velocity of the particle as a function of time t. (b) Find the arclength of its path between t = 1 and t =. olution. (a) As r(t) = ( 3 cos t, 3 sin t, 4t ), the velocity of the particle is r (t) = ( 3 sin t, 3 cos t, 4 ) (b) As ds, the rate of change of arc length per unit time, is dt ds dt (t) = r (t) = ( 3 sin t, 3 cos t, 4 ) = 5 the arclength of its path between t = 1 and t = is 1 dt ds dt (t) = 1 dt 5 = 5. Let be the upper half of the unit circle centred on (1, ) (i.e. that part of the circle which lies above the ais), oriented clockwise. ompute the line integral y dy. olution. We are being asked to evaluate the line integral F dr with being the specified semi circle and F = y ĵj. As F, the vector field F is not conservative. o we ll evaluate the integral directly. First, using the figure, y ( ) 1 cos θ, sin θ ( 1) + y = 1 θ ( ) 1, we parametrize by o the integral is y dy = r(θ) = ( (θ), y(θ) ) = (1 cos θ)îı + sin θ ĵj π (θ) y(θ) y (θ) dθ = π θ π (1 cos θ) sin θ cos θ dθ Making the substitution u = cos θ, du = sin θ dθ, u() = 1, u(π) = 1, y dy = (1 u) u ( du) = (u u ) du = u du = 13 3 =

2 3. Let be the surface given by r(u, v) = ( u + v, u + v, u v ), u, v (a) Find the tangent plane to the surface at the point (,, ). (b) This is a surface you are familiar with. What surface is it (it may be just a portion of one of the following)? sphere helicoid ellipsoid saddle parabolic bowl cylinder cone plane (c) In which direction does the parametrisation orient the surface? ircle the correct choices: In the ( positive / negative ) ( / y / z ) direction. olution. (a) To find the tangent plane we have to find a normal vector to the surface at (,, ). ince r = ( 1, u, 1 ) u r = ( 1, v, 1 ) v a normal vector to the surface at r(u, v) is r u r v = ( u v,, v u ) As r(u, v) = (,, ) when (the coordinate) u+v = and (the z coordinate) u v =, i.e when u = v = 1, a normal vector to the surface at (,, ) = r(1, 1) is ( 4,, ) or (, 1, ) and the equation of the specified tangent plane is (b) Our parametrization is ( ) + (y ) + z = or y = (u, v) = u + v y(u, v) = u + v z(u, v) = u v Adding (u, v)+z(u, v) = u and subtracting (u, v) z(u, v) = v so that u = 1 ( (u, v)+ z(u, v) ) and v = 1 ( (u, v) z(u, v) ). o on our surface y(u, v) = u + v = 1 4 = 1 (u, v) + 1 z(u, v) ( ) 1( ) (u, v) + z(u, v) + (u, v) z(u, v) 4 All points of our surface lie on y = + z. This is a parabolic bowl:

3 no points have y < and the y = Y (with Y > ) cross section is the circle + z = Y, y = Y the = cross section is the parabola y = z, = the z = cross section is the parabola y =, z = z y (c) This question was poorly worded. It should have specified that the surface was oriented by the normal vector r r. In part (a) we found that u v r r = ( u v,, v u ) u v The y coordinate of this vector, namely, is always positive. Depending on (u, v), with u, v, the and z coordinates can be either positive or negative. 4. Let F(, y) = ( y e y + sin, ye y + ) Let be the boundary of the triangle with vertices (, ), (1, ) and (1, ), oriented counter clockwise. ompute F dr olution. The integral that would be used for direct evaluation looks very complicated. o let s try Green s theorem. The curve is the boundary of the triangle T = { (, y) 1, y } y (1, ) y= (, ) (1, ) 3

4 o F dr = {( y e y + sin ) d + ( ye y + ) dy } = { ( ye y + ) ( y y e y + sin )} ddy T {( = ye y + 1 ) ( )} y + ye y ddy = = T 1 1 d dy { 1 y } d { 4 } = = Let ( y ) F(, y, z) = + 1+, y 1+y, cos 5 (ln z) (a) Write down the domain D of F. (b) ircle the correct statement(s): (a) D is connected. (b) D is simply connected. (c) D is disconnected. (c) ompute F. (d) Let be the square with corners (3 ± 1, 3 ± 1) in the plane z =, oriented clockwise (viewed from above, i.e. down z ais). ompute F dr (e) Is F conservative? olution. (a) ince y is defined when and 1+ = e (1+ ) ln is defined when ln is defined, which is when > (assuming that we are not allowed to use comple numbers) and y 1+y = e (1+y ) ln y is defined when ln y is defined, which is when y > and cos 5 (ln z) is defined when ln z is defined, which when z > 4

5 the domain of F is D = { (, y, z) >, y >, z > } (b) The domain D is both connected (any two points in D can be joined by a curve that lies completely in D) and simply connected (any simple closed curve in D can be shrunk to a point continuously in D). (c) The curl of F is F = det y îı ĵj ˆk y z + 1+ y 1+y cos 5 (ln z) = ( 1 /)ˆk (d) The integrand for direct evaluation looks very complicated. On the other hand F is quite simple. o let s try tokes thoerem. Denote = { (, y, z) 4, y 4, z = } The boundary of is. Because of the clockwise orientation of, we assign the normal vector ˆk to. ee the sketch below z (,,) (,4,) (4,,) (4,4,) ˆn y Then, by tokes theorem, F dr = F ˆn d = = 4 d 4 dy ( 1 = [ 1 ln ] = ln 4 (e) ince F is not, F cannot be conservative. ( F ( ˆk) d = 1 ) 4 = d ( 1 ) d ) [ = ln ] 4 5

6 6. Let F(, y, z) = ( e y + y 1+ + cos(z), z, y ) Let be the surface which consists of two parts: the portion of the paraboloid y + z = 4( + 1) satisfying 3 and the portion of the sphere + y + z = 4 satisfying < and is oriented outward. ompute F ˆn d olution. First let s figure out what looks like. The = cross section of y + z = 4( + 1), namely the circle y + z = 4, =, and the = cross section of + y + z = 4, again the circle y + z = 4, =, coincide. o where = { (, y, z) 3, y + z = R() } R() = { 4( + 1) if 3 4 if Here is a sketch of the y = cross section of. z + z = 4 z = 4( + 1) The boundary of is the circle y + z = 16, = 3 oriented clockwise when viewed along the ais from large. o we can parametrize by r(t) = 3îı + 4 cos tĵj 4 sin t ˆk t π 6

7 onsidering the form of the flu integral that we are to evaluate, it is natural to use tokes theorem. F ˆn d = F dr 7. Let = π = 16 = 3π ( a mess, z(t) {}}{ 4 sin t, π Let be the portion of the surface y(t) {}}{ 4 cos t ) ( sin t + cos t ) dt F(, y, z) = ( 1 + z 1+z1+z, 1 + z 1+z1+z, 1 ) + y = 1 z 4 r (t) {}}{ (, 4 sin t, 4 cos t ) dt which is above the y plane. What is the flu of F downward through? olution. Let V = { (, y, z) + y 1 z 4, z 1 } Then the boundary, V, of V, with the orientation that is used in the divergence theorem, consists of two parts the surface, but with the upward pointing normal, and the disk D = { (, y, z) + y 1, z = }, with normal ˆk. o the divergence theorem gives F dv = V As F = and F(, y, ) = ( 1, 1, 1 ) 8. Let V F ˆn d = F ˆn d = F ˆn d + F ( ˆk) d D D F ( ˆk) d = d = π D F(, y) = ( 1, yg(y) ) and suppose that g(y) is a function defined everywhere with everywhere continuous partials. how that for any curve whose endpoints P and Q lie on the ais, distance between P and Q = F dr 7

8 olution. The vector field F is conservative, with onsquently F = ϕ ϕ(, y) = + F dr = ϕ(q) ϕ(p ) = (Q) + = (Q) (P ) y(q) y ỹg(ỹ) dỹ ỹg(ỹ) dỹ (P ) y(p ) since y(q) = y(p ) =. Thus F dr = (Q) (P ) = distance between P and Q ỹg(ỹ) dỹ 9. (a) In the curve shown below (a heli lying in the surface of a cone), is the curvature increasing, decreasing, or constant as z increases? z (b) Of the two functions shown below, one is a function f() and one is its curvature κ(). Which is which? y y D 8

9 (c) Let be the curve of intersection of the cylinder + z = 1 and the saddle z = y. Parametrise. (Be sure to specify the domain of your parametrisation.) (d) Let H be the helical ramp (also known as a helicoid) which revolves around the z ais in a clockwise direction viewed from above, beginning at the y-ais when z =, and rising π units each time it makes a full revolution. Let be the the portion of H which lies outside the cylinder + y = 4, above the z = plane and below the z = 5 plane. hoose one of the following functions and give the domain on which the function you have chosen parametrizes. (Hint: Only one of the following functions is possible.) (a) r(u, v) = ( u cos v, u sin v, u ) (b) r(u, v) = ( u cos v, u sin v, v ) (c) r(u, v) = ( u sin v, u cos v, u ) (d) r(u, v) = ( u sin v, u cos v, v ) (e) Write down a parametrized curve of zero curvature and arclength 1. specify the domain of your parametrisation.) (Be sure to (f) If F is a constant on all of R 3, and is a cube of unit volume such that the flu outward through each side of is 1, what is? (g) Let F(, y) = ( a + by, c + dy ) Give the full set of a, b, c and d such that F is conservative. (h) If r(s) has been parametrized by arclength (i.e. s is arclength), what is the arclength of r(s) between s = 3 and s = 5? (i) Let F be a D vector field which is defined everywhere ecept at the points marked P and Q. uppose that F = everywhere on the domain of F. onsider the five curves R,, T, U, and V shown in the picture. Q R U P T V Which of the following is necessarily true? (1) F dr = T F dr 9

10 () R F dr = F dr = T F dr = U F dr = (3) R F dr + F dr + T F dr = U F dr (4) U F dr = R F dr + F dr (5) V F dr = (j) Write down a 3D vector field F such that for all closed surfaces, the volume enclosed by is equal to F ˆn d (k) onsider the vector field F in the y plane shown below. Is the ˆk th component of F at P positive, negative or zero? P olution. (a) The heli is approimately a bunch of circles stacked one on top of each other. The radius of the circles increase as z increases. o the curvature decreases as z increases. (b) Here are two arguments both of which conclude that f() is D. If were the graph y = f(), then f () would have two points of discontinuity. The curvature κ() would not the defined at those two points. The function whose graph is D is defined everywhere and so cannot be the curvature of the function whose graph is. The function whose graph is D has two inflection points. o its curvature is zero at two points. The function whose graph is is indeed zero at two points (that in fact correspond to the inflection points of D). o D is the graph of f() and is the graph of κ(). (c) For any fied y, + z = 1 is a circle of radius 1. o we can parametrize it by (θ) = cos θ, z(θ) = sin θ, θ < π. The y coordinate of any point on the intersection is determined by y = z. o we can use r(θ) = cos θ îı + sin θ ˆk + sin θ cos θ ĵj θ < π (d) We are told that the helical ramp starts starts with the y ais when z =. 1

11 In the cases of parametrisations (a) and (c), z = forces u = and u = forces = y =. That is only the origin, not the y ais. o we can rule out (a) and(c). In the case of parametrisation (b), z = forces v = and v = forces y = and = u. As u varies that sweeps out the ais, not the y ais. o we can rule out (b). In the case of parametrisation (d), z = forces v = and v = forces = and y = u. As u varies that sweeps out the y ais, which is what we want. Furthermore we are told that z = v runs from to 5 and that + y = u 4 o we want parametrisation (d) with domain u, v 5. (e) traight lines have curvature. o one acceptable parametrized curve is r(t) = t îı, t 1. (f) The cube has si sides. o the outward flu through is 6 and, by the divergence theorem, 6 = F ˆn d = F dv = dv = since has volume one. o = 6. (g) For the vector field F to be conservative, we need F 1 y = F y (a + by) = (c + dy) b = c When b = c, an allowed potential is a + by + d y. The specified set is { (a, b, c, d) a, b, c, d all real and b = c } (h) By the definition of arclength parametrisation, the arclength along the curve between r() and r(s) is s. In particular, the arclength between r() and r(3) is 3 and the arclength between r() and r(5), which is the same as the arclength between r() and r(3) plus the arclength between r(3) and r(5), is 5. o the arclength between r(3) and r(5) is 5 3 =. (i) In this solution, we ll use, for eample T to refer to the curve T, but with the arrow pointing in the opposite direction to that of the arrow on T. In parts (), (3) and (4) we will choose F to be the vector field G(, y) = y + y îı + + y ĵj 11

12 We saw, in Eample.3.14 of the LP IV tet, that G = ecept at the origin where it is not defined. We also saw, in Eample of the LP IV tet, that G dr = π for any counterclockwise oriented circle centred on the origin. (1) Let R 1 be the region between and T. It is the shaded region in the figure on the left below. Note that R 1 is contained in the domain of F, so that F = on all of R 1. The boundary of R 1 is T, meaning that the boundary consists of two parts, with one part being and the other part being T. o, by tokes theorem F dr F dr = F dr = F ˆk d = T R 1 R 1 and (1) is true. R U T R 1 R () False. hoose a coordinate system so that Q is at the origin and choose F = G. We saw, in Eamples.3.14 and of the LP IV tet, that the curl of G vanished everywhere ecept at the origin, where it was not defined, but that G dr. R (3),(4) False. Here is a counterample that shows that both (3) and (4) are false. hoose a coordinate system so that Q is at the origin and choose F = G. By tokes theorem G dr = G dr = because G = everywhere inside, including at P. o now both parts (3) and (4) reduce to the claim that G dr = G dr. U R We saw, in Eample of the LP IV tet, that G dr = π. R To finish off the countereample, we ll now show that G dr = π. Let R U be the region between U and R. It is the shaded region in the figure on the right above. Note that G = on all of R. including at P. The boundary of R is U R, meaning that the boundary consists of two parts, with one part being U and the other part being R. o, by tokes theorem G dr G dr = G dr = G ˆk d = U R R R and G dr = G dr = π U R 1 T

13 (5) False. For any conservative vector field F, with potential f, F dr is just the V difference of the values of f at the two end points of V. It is easy to choose an f for which those two values are different. For eample f(, y) = does the job. (j) Let be any closed surface and denote by V the volume that it encloses. Presumably the question assumes that is oriented so that = V. Then by the divergence theorem F ˆn d = F ˆn d = F dv V This is eactly the volume of V if F = 1 everywhere. One vector field F with F = 1 everywhere is F = îı. (k) Let be the counterclockwise boundary of a small square centred on P, like the blue curve in the figure below, but much smaller. all the square (the inside of ). V P By tokes theorem F ˆk d = F dr The contribution to F dr coming from the left and right sides of will be zero, because F is perpendicular to dr there. The contribution to F dr coming from the top of will be negative, because there F is a positive number times îı and dr is a negative number times îı. The contribution to F dr coming from the bottom of will be positive, because there F is a positive number times îı and dr is a positive number times îı. The magnitude of the contribution from the top of will be larger than the magnitude of the contribution from the bottom of, because F is larger on the top than on the bottom. o, all together, F dr <, and consequently (taking a limit as the square size tends to zero) F ˆk is negative at P. 13

14 1. ay whether the following statements are true or false. (a) If F is a 3D vector field defined on all of R 3, and 1 and are two surfaces with the same boundary, but 1 F ˆn d F ˆn d, then F is not zero anywhere. (b) If F is a vector field satisfying F = whose domain is not simply-connected, then F is not conservative. (c) The osculating circle of a curve at a point has the same unit tangent vector, unit normal vector, and curvature as at that point. (d) A planet orbiting a sun has period proportional to the cube of the major ais of the orbit. (e) For any 3D vector field F, ( F) =. (f) A field whose divergence is zero everywhere in its domain has closed surfaces in its domain. (g) The gravitational force field is conservative. (h) If F is a field defined on all of R 3 such that F dr = 3 for some curve, then F is non-zero at some point. (i) The normal component of acceleration for a curve of constant curvature is constant. (j) The curve defined by is the same as the curve defined by r 1 (t) = cos(t 4 )îı + 3t 4 ĵj, < t <, r (t) = cos tîı + 3t ˆk, < t < olution. (a) False. We could have, for eample, F zero at one point and strictly positive elsewhere. One eample would be F = 3 îı + y 3 ĵj + z 3 ˆk, with 1 and being the upward oriented top and bottom hemispheres, respectively, of the unit sphere +y +z = 1. (b) False. The conditions that (1) F = and () the domain of F is simply-connected, are sufficient, but not necessary, to imply that F is conservative. For eample the vector field F =, with any domain at all, is conservative with potential. Another eample (which does not depend on choosing a domain that is smaller than the largest possible domain) is F = 1 with domain { (, y, z) } +y (, y) (, ). That is, the domain is R 3 with the z ais removed. (c) That s true. onsider any point r(t ) on a parametrized curve r(t). That s the blue point in the figure below. The centre of curvature for the curve at r(t ) is c = 14

15 r(t ) ˆT ˆN c r(t ) + ρ(t ) ˆN(t ). It is the red dot in the figure. The radius of the osculating circle is the distance from its centre, c, to any point of the circle, like r(t ). That s r(t ) c = ρ(t ) ˆN(t ) = ρ(t ). The curvature of the osculating circle is one over its radius. o its curvature is 1 ρ(t ) = κ(t ). The unit normal to the osculating circle at r(t ) is a unit vector in the opposite direction to the radius vector from the centre c to r(t ). The radius vector is r(t ) c = ρ(t ) ˆN(t ), so the unit normal is ˆN(t ). The osculating circle lies in the plane that best fits the curve near r(t ). (ee the beginning of 1.4 in the LP IV tet.) o the unit tangents to the osculating circle at r(t ) are perpendicular to both ˆN(t ) and ˆB(t ) and so are either ˆT(t ) or ˆT(t ), depending on how we orient the osculating circle. (d) False. Kepler s third law is that a planet orbiting a sun has the square of the period proportional to the cube of the major ais of the orbit. (e) True. That s part (a) of Theorem in the LP IV tet. (f) True. Every domain contains closed surfaces. This has nothing to do with vector fields. (g) True. We saw this in Eample.3.4 in the LP-IV tet. (h) False. Let F be an everywhere defined conservative vector field with potential ϕ. Then F = everywhere. If P and Q are two points and if ϕ(p ) ϕ(q) = 3 and if is a curve from Q to P, then F dr = 3. One eample would be ϕ(, y, z) =, F = îı, P = (3,, ), Q = (,, ). (i) False. The normal component of acceleration depends on speed, as well as curvature. (j) False. The curve r 1 contains only points in the y plane. Every r (t) with t has a nonzero z coordinate. 15