MATHEMATICS 317 April 2017 Final Exam Solutions

Transcription

1 MATHEMATI 7 April 7 Final Eam olutions. Let r be the vector field r = îı + ĵj + z ˆk and let r be the function r = r. Let a be the constant vector a = a îı + a ĵj + a ˆk. ompute and simplif the following quantities. Answers must be epressed in terms of a, r, and r. There should be no s, s, or z s in our answers. a r b r c r a d r olution. a b c ince r = r a = det r = + + z z = îı + ĵj + ˆk z + + z = îı + ĵj + ˆk = r îı ĵj ˆk z = îı a a z + ĵj a z a + ˆk a a a a a we have îı ĵj ˆk r a = det z = a îı a ĵj a ˆk a a z a z a a a = a d ince r = îı + ĵj + ˆk + + z / z = îı + + z + ĵj / + + z + ˆk z / + + z / we have r = + + z + / + + z / + z z + + z / = + + z + + z / + + z = / + + z = / r

2 . a how that the planar vector field is conservative. b Find a potential function for F. F, = cos, sin sin c For the vector field F from above compute F dr, where is the part of the graph = sin from = π/ to = π. olution. a, b The function f, is a potential for F, if and onl if it obes Integrating the first of these equations gives f, = cos f, = sin sin f, = sin + g ubstituting this into the second equation gives which integrates to sin + g = sin sin or g = sin g = cos + with an arbitrar constant. Hence f, = sin + cos + is a potential for an constant. Because F has a potential, it is conservative. c We ma parametrize b rt = sintîı + tĵj π t π As f, = sin + cos is a potential for F F dr = f rπ f r π / = f, π f π, π / = sin = π sin. A race track between two hills is described b the parametric curve rθ = 4 cos θ, sin θ, 4 cosθ, θ π a ompute the curvature of the track at the point 4,, 4.

3 b ompute the radius of the circle that best approimates the bend at the point 4,, 4 that is, the radius of the osculating circle at that point. c A car drives down the track so that its position at time t is given b rt. Note the relationship between t and θ is θ = t. ompute the following quantities. i The speed at the point 4,, 4. ii The acceleration at the point 4,, 4. iii The magnitude of the normal component of the acceleration at the point 4,, 4. olution. a For the specified curve rπ = 4,, 4 and rθ = 4 cos θ, sin θ, 4 cosθ vθ = r θ = 4 sin θ, cos θ, sinθ aθ = r θ = 4 cos θ, sin θ, cosθ vπ =,, aπ = 4,, vπ aπ =,, 8 o the curvature at θ = π is b The radius is κπ = c et Rt = rt. Then In particular, vπ aπ vπ = κπ = 4 7 R t = t r t,, 8,, = R t = r t + 4t r t 7 R π = 4,, 4 R π = π vπ =, 4 π, speed = R π = 4 π acceleration = R π = vπ + 4π aπ = 6π, 4, 4π The normal component of the acceleration has magnitude ds 7 κ = 4 π = 4 7 π dt 4 4

4 4. ompute the flu integral F ˆn d, where F =,, z and is the part of the paraboloid z = 5 ling inside the clinder + 4, with orientation pointing downwards. olution. We ll start b parametrizing. Note that as + runs from to 4, z runs from 5 to, and that, for each fied Z 5, the cross section of with z = Z is the circle + = 5 Z, z = Z. o we ma parametrize b rθ, Z = 5 Z cos θ îı + 5 Z sin θ ĵj + Z ˆk θ π, Z 5 ince θ = 5 Z sin θ îı + 5 Z cos θ ĵj Z = cos θ îı 5 Z sin θ ĵj + ˆk 5 Z so that θ îı ĵj ˆk Z = det 5 Z sin θ 5 Z cos θ = 5 Z cos θ, 5 Z sin θ, /, cos θ 5 Z sin θ 5 Z.. in the LP IV tet gives ˆn d = ± θ Z dθdz = ± 5 Z cos θ, 5 Z sin θ, / dθdz hoosing the minus sign to give the downward pointing normal F ˆn d = = ince 5 5 dz dz π π dθ {}}{ {}}{ [5 Z] / cos θ [5 Z] / cos θ sin θ, z {}}{{}}{ [5 Z] / sin θ, Z 5 Z cos θ, 5 Z sin θ, / dθ [5 Z] cos 4 θ [5 Z] cos θ sin θ [5 Z] sin 4 θ + Z cos4 θ + cos θ sin θ + sin4 θ = cos θ + sin θ = 4

5 the flu F ˆn d = = π 5 dz π dθ [5 Z] Z = π [ [5 Z] Z ] 5 5 dz [ 4 = π 5 + ] [5 Z] Z = π 5. Let = + + be the curve given b the union of the three parameterized curves r t = cos t, sin t,, t π/ r t =, cos t, sin t, t π/ r t = sin t,, cos t, t π/ a Draw a picture of. learl mark each of the curves,, and and indicate the orientations given b the parameterizations. b Find and parameterize an oriented surface whose boundar is with the given orientations. c ompute the line integral F dr where F = + sin, z + log +, + e z olution. a Observe that the curve is one quarter of a circle in the plane, centred on the origin, of radius, starting at,, and ending at,, and the curve is one quarter of a circle in the z plane, centred on the origin, of radius, starting at,, and ending at,, and the curve is one quarter of a circle in the z plane, centred on the origin, of radius, starting at,, and ending at,,. Here is a sketch. z,,,,,, 5

6 b lies completel on the sphere + + z = 4. o it is natural to choose = {,, z + + z = 4,,, z } and to parametrize using spherical coordinates ince so that rθ, ϕ = cos θ sin ϕ îı + sin θ sin ϕ ĵj + cos ϕ ˆk, θ π, ϕ π = sin θ sin ϕîı + cos θ sin ϕĵj θ = cos θ cos ϕîı + sin θ cos ϕĵj sin ϕ ˆk ϕ θ îı ĵj ˆk ϕ = det sin θ sin ϕ cos θ sin ϕ cos θ cos ϕ sin θ cos ϕ sin ϕ.. in the LP IV tet gives = 4 cos θ sin ϕ îı 4 sin θ sin ϕ ĵj 4 sin ϕ cos ϕ ˆk ˆn d = ± θ ϕ dθdϕ = 4 cos θ sin ϕ îı + sin θ sin ϕ ĵj + cos ϕ ˆk sin ϕ dθdϕ We want ˆn to point outward, for compatibilit with the orientation of. o we choose the + sign. ˆn d = 4 cos θ sin ϕ îı + sin θ sin ϕ ĵj + cos ϕ ˆk sin ϕ dθdϕ = rθ, ϕ sin ϕ dθdϕ c The vector field F looks too complicated for a direct evaluation of the line integral. o, in preparation for an application of tokes theorem, we compute îı ĵj ˆk F = det z + sin z + log + + e z = 4 ˆk o, b tokes theorem Theorem 4.4. in the LP IV tet, F dr = F ˆn d = π/ = 6 = 4π dϕ π/ π/ dϕ dθ 4ˆk cos θ sin ϕ îı + sin θ sin ϕ ĵj + cos ϕ ˆk 4 sin ϕ π/ dθ cos ϕ sin ϕ = 6 π 6 cos ϕ π/

7 6. onsider the vector field F = P îı + Qĵj, where P = a ompute and simplif Q P. + +, Q = + b ompute the integral R F dr directl using a parameterization, where R is the circle of radius R, centered at the origin, and oriented in the counterclockwise direction. c Is F conservative? arefull eplain how our answer fits with the results ou got in the first two parts. d Use Green s theorem to compute F dr where is the triangle with vertices,,,,, oriented in the counterclockwise direction. e Use Green s theorem to compute F dr where is the triangle with vertices,,,,, oriented in the counterclockwise direction. olution. a If,,, we have b Parametrize R b Q P = = = rθ = R cos θ îı + R sin θ ĵj θ π o R F dr = = π π = π Frθ {}}{ { îı cos θ + sin θ + sin θ cos θ ĵj } r θ {}}{ R sin θ îı + R cos θ ĵj dθ R dθ c If F were conservative, the line integral F dr would be for an closed curve, b Theorem.4.6.b in the LP IV tet. o F is not conservative. Note that F is not defined at, =, and so fails the screening test F = at, =,. d Denote b R the interior of the triangle. It is the gre region in the figure 7

8 ,, R, Note that, is not in R. o Q P is defined and zero throughout R. o, b Green s theorem Theorem 4.. in the LP IV tet, F dr = Q P dd = R e Note that, is in the interior of triangle specified for this part. o Q P is not defined in that interior and we cannot appl Green s theorem precisel as we did in part d. We can work around this obstruction b picking a number r > that is small enough that the circle r, of radius r centred on,, is completel contained in the interior of the triangle. Then we work with the region R defined b removing the interior of the circle r from the interior of the triangle. It is the gre region sketched below., R r,, The boundar of R consists of two parts the triangle, oriented counterclockwise, and the circle r. That is, the circle r, but oriented clockwise, rather than counterclockwise. 8

9 Then Q P is well defined and zero throughout R and, b Green s theorem, = Q P dd R = F dr + F dr r = F dr F dr r o F dr = r F dr. B part b, with R = r, r F dr = π, so F dr = π. 7. onsider the vector field F,, z = z arctan, z ln +, z Let the surface be the part of the sphere + + z = 4 that lies above the plane z = and be oriented downwards. a Find the divergence of F. b ompute the flu integral F ˆn d. olution. a The divergence is F = z arctan + z ln + + z z = b The compleit of F and the simplicit of F strongl suggest that we use the divergence theorem to evaluate F ˆn d. However, is not a closed surface and is not the boundar of a solid. The figure on the left below is a sketch of the part of in the first octant. z,, ˆn z,, V ˆn V D ˆn V On the other hand is part of the surface of the solid V = {,, z + + z 4, z } which is sketched on the right above. The boundar of V consists of two parts: 9

10 the original surface, but with upward, rather than downward, normal and the disk D = {,, z +, z = } with normal ˆk. o the divergence theorem Theorem 4..9 in the LP IV tet gives F ˆn d = F dv = V V = F ˆn d + F ˆk d = VolumeV D Thus V dv F ˆn d = VolumeV + D F ˆk {}}{ d = VolumeV AreaD = VolumeV 9π since D is a circular disk of radius. To compute the volume of V, we slice V into thin horizontal pancakes each of thinkness dz. The pancake at height z has cross section the circular disk + 4 z. As this disk has area π4 z, the pancake has volume π4 z dz. All together and VolumeV = dz π4 z = π F ˆn d = 5π ] [ [4z z = π 4 7 ] = 5π 9π = 4π