MATHEMATICS 200 December 2014 Final Exam Solutions

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1 MATHEMATICS 2 December 214 Final Eam Solutions 1. Suppose that f,, z) is a function of three variables and let u 1 6 1, 1, 2 and v 1 3 1, 1, 1 and w 1 3 1, 1, 1. Suppose that at a point a, b, c), Find f at a, b, c). D u f D v f D w f 4 Solution. Write fa, b, c) F, G, H. We are told that D u f 1 6 1, 1, 2 F, G, H D v f 1 3 1, 1, 1 F, G, H D w f 1 3 1, 1, 1 F, G, H 4 so that F + G + 2H F G H F + G + H 4 3 E1) E2) E3) Adding E2) and E3) gives 2F 4 3 or F 2 3. Substituting F 2 3 into E1) and E2) gives G + 2H 2 3 G H 2 3 E1) E2) Adding E1) and E2) gives H 4 3 and substituting H 4 3 back into E2) gives G 6 3. All together fa, b, c) 3 2, 6, 4 2. Let fu, v) be a differentiable function of two variables, and let z be a differentiable function of and defined implicitl b fz, z). Show that z + z z Solution. We are told that the function z, ) obes f z, ), z, ) ) 1

2 for all and. Differentiating this equation with respect to and with respect to gives, b the chain rule, f u z, ), z, ) ) [ z, ) + z, ) ] + f v z, ), z, ) ) z, ) f u z, ), z, ) ) z, ) + f v z, ), z, ) ) [ z, ) + z, ) ] or, leaving out the arguments, f u [ z + z ] + fv z f u z + f v [ z + z ] Solving the first equation for z and the second for z gives so that as desired. z + z z f u f u + f v z z f u f u + f v z z f v f u + f v z f v z f u + f v ) f u + f v f u + f v z Remark: This is of course under the assumption that[ f u + f] v is nonzero. That is equivalent, b the chain rule, to the assumption that z fz, z) is non zero. That, in turn, is almost, but not quite, equivalent to the statement that fz, z) is soluble for z as a function of and. 3. Let z f, ) be given implicitl b a) Find the differential dz. e z + z +. b) Use linear approimation at the point 1, ) to approimate f.99,.1). Solution. B definition, the differential at a, b is f a, b) d + f a, b) d so we have to determine the partial derivatives f a, b) and f a, b). We are told that e f,) + f, ) + 2

3 for all and. Differentiating this equation with respect to and with respect to gives, b the chain rule, e f,) f, ) + f, ) 1 e f,) f, ) + f, ) + f, ) 1 Solving the first equation for f and the second for f gives So the differential at a, b is 1 f, ) e f,) + 1 f, ) f, ) e f,) + d 1 fa, b) + e fa,b) + b e fa,b) + b d Since we can t solve eplicitl for fa, b) for general a and b. There s not much more that we can do with this. b) In particular, when a 1 and b, we have e f1,) + f1, ) 1 + e f1,) 1 f1, ) and the linear approimation simpifies to f 1 + d, d ) f1, ) + Choosing d.1 and d.1, we have d 1 f1, ) + d d + d e f1,) + e f1,) + f.99,.1 ) Let S be the surface z Find all points P,, z ) on S with such that the normal line at P contains the origin,, ). Solution. The equation of S is of the form G,, z) z 1. So one normal vector to S at the point,, z ) is and the normal line to S at,, z ) is G,, z ) 2 îı )ĵj ˆk,, z),, z ) + t 2, 4 + 2, 1 For this normal line to pass through the origin, there must be a t with,, ),, z ) + t 2, 4 + 2, 1 3

4 or + 2 t )t z t E1) E2) E3) Equation E3) forces t z. Substituting this into equations E1) and E2) gives 1 + 2z ) )z E1) E2) The question specifies that, so E1) forces z 1 2. Substituting z 1 2 into E2) gives 1 1 Finall is determined b the requirement that,, z ) must lie on S and so must obe z ) ) So the allowed points P are 1 2, 1, 1 2) and 1 2, 1, 1 2). 5. Let f, ) + 3). a) Find all critical points of f, and classif each one as a local maimum, a local minimum, or saddle point. b) Find the location and value of the absolute maimum and minimum of f on the triangular region,, + 8. Solution. a) Thinking a little wa ahead, to find the critical points we will need the gradient of f and to appl the second derivative test of Theorem in the CLP III tet we will need all second order partial derivatives. So we need all partial derivatives of f up to order two. Here the are. f + 3) f f 2 f f f 2 f Of course, f and f have to be the same. It is still useful to compute both, as a wa to catch some mechanical errors.) The critical points are the solutions of f 2 + 3) f + 2 3) The first equation is satisfied if at least one of, 3 2 are satisfied. 4

5 If, the second equation reduces to 3), which is satisfied if either or 3. If 3 2, the second equation reduces to ) 3 3) which is satisfied if or 1. So there are four critical points:, ), 3, ),, 3), 1, 1). The classification is critical point f f f 2 f tpe, ) 3) 2 < saddle point 3, ) 6 3) 2 < saddle point, 3) 6 3) 2 < saddle point 1, 1) 2 2 1) 2 > 2 local min b) The absolute ma and min can occur either in the interior of the triangle or on the boundar of the triangle. The boundar of the triangle consists of the three line segments. L 1 {, ), 8 } L 2 {, ), 8 } L 3 {, ) + 8, 8 } An absolute ma or min in the interior of the triangle must also be a local ma or min and, since there are no singular points, must also be a critical point of f. We found all of the critical points of f in part a). Onl one of them, namel 1, 1) is in the interior of the triangle. The other three critical points are all on the boundar of the triangle.) We have f1, 1) 1. At each point of L 1 we have and so f, ). At each point of L 2 we have and so f, ). At each point of L 3 we have f, ) 8 )5) [8 2 ] with 8. As d d ) 4 1, the ma and min of on 8 must be one of 5 [ 8 2] or 5[ 8 2] 8 or 5[ 8 2] 4 8. So all together, we have the following candidates for ma and min, with the ma and min indicated. points) 1, 1) L 1 L 2, 8) 8, ) 4, 4) value of f 1 8 min ma p, 8q L 1 L 3 p1, 1q p4, 4q L 2 p8, q 5

6 6. In the plane, the disk is cut into 2 pieces b the line. Let D be the larger piece. a) Sketch D including an accurate description of the center and radius of the given disk. Then describe D in polar coordinates r, θ). b) Find the volume of the solid below z and above D. Solution. a) The inequalit is equivalent to 1) and sas that, ) is to be inside the disk of radius 1 centred on 1, ). Here is a sketch. D p1,q 1 p 1q 2 ` 2 1 In polar coordinates, r cos θ, r sin θ so that the line is θ π is r 2 2r cos θ or r 2 cos θ Consequentl D { r cos θ, r sin θ) π /2 θ π /4, r 2 cos θ } and the circle b) The solid has height z r above the point in D with polar coordinates r, θ. So the π/4 2 cos θ Volume r da r 2 dr dθ dθ dr r 2 D π/4 D 8 dθ cos 3 θ 8 3 π/2 3 8 [ ] π/4 sin θ sin3 θ 3 3 π/2 8 [ ) π/4 π/ π/2 dθ cos θ [ 1 sin 2 θ ] 7. The densit of hdrogen gas in a region of space is given b the formula ρ,, z) 6 z )]

7 a) At 1,, 1), in which direction is the densit of hdrogen increasing most rapidl? b) You are in a spacecraft at the origin. Suppose the spacecraft flies in the direction of,, 1. It has a disc of radius 1, centred on the spacecraft and deploed perpendicular to the direction of travel, to catch hdrogen. How much hdrogen has been collected b the time that the spacecraft has traveled a distance 2? [You ma use the fact that 2π cos 2 θ dθ π.] Solution. a) The direction of maimum rate of increase is ρ1,, 1). As ρ 4,, z) z + 22 ) ρ ) 2 1,, 1) ) 2 2) ρ,, z) 2z + 22 ) ρ ) 2 1,, 1) ρ z,, z) 1 ρ z 1,, 1) 1 2 So ρ1,, 1) 1 2 3,, 1). The unit vector in this direction is 1 1 3,, 1). b) The region swept b the space craft is, in clindrical coordinates, V { r cos θ, r sin θ, z) θ 2π, r 1, z 2 } and the amount of hdrogen collected is z + 2r 2 cos 2 θ ρ dv rdr dθ dz 1 + r 2 V V dz dz dz dz 2π 2π 2π 2π 1 dθ dr zr + 2 cos2 θ)r r 2 1 [ ] r dθ dr z 1 + r + 2r 2 cos2 θ cos 2 2r θ 1 + r 2 r 3 since 1 + r r + r3 r r r r r 2 [ z 1 dθ 2 ln1 + r2 ) + r 2 cos 2 θ ln1 + r 2 ) cos θ] 2 [ ] ln 2 dθ 2 z + cos2 θ ln2) cos 2 θ dz [π ln 2)z + π π ln 2] 2π ln 2 + 2π 2π ln 2 2π 7

8 8. Consider the iterated integral I a a 2 2 where a is a positive constant. a z 2) 214 dz d d a) Write I as an iterated integral in clindrical coordinates. b) Write I as an iterated integral in spherical coordinates. c) Evaluate I using whatever method ou prefer. Solution. The main step is to figure out what the domain of integration looks like. The outside integral sas that runs from a to. The middle integrals sas that, for each in that range, runs from a 2 2 to. We can rewrite a 2 2 in the more familiar form a 2,. So, ) runs over the third quadrant part of the disk of radius a, centred on the origin. p a,q 2 ` 2 a 2 Finall, the inside integral sas that, for each, ) in the quarter disk, z runs from to a We can also rewrite z a in the more familiar form z 2 a 2, z. So the domain of integration is the part of the interior of the sphere of radius a, centred on the origin, that lies in the octant,, z. V {,, z) a, a2 2, z a } {,, z) z 2 a 2,,, z } 8

9 z 2 ` 2 ` z 2 a 2 a) Note that, in V,, ) is restricted to the third quadrant, which in clindrical coordinates is π θ 3π. So, in clindrical coordinates, 2 { V r cos θ, r sin θ, z) r 2 + z 2 a 2, π θ 3π } 2, z { r cos θ, r sin θ, z) z a, π θ 3π 2, r a 2 z 2 } and I a V dz z 2) 214 dv 3π/2 π a 2 z 2 dθ dr r r 2 + z 2) 214 V r 2 + z 2) 214 r dr dθ dz b) The spherical coordinate ϕ runs from when the radius vector is along the positive z ais) to π /2 when the radius vector lies in the plane) so that I z 2) 214 dv ρ ρ 2 sin ϕ dρ dθ dϕ V π/2 dϕ 3π/2 π dθ a c) Using the spherical coordinate version I π/2 a a431 π 862 a431 π 862 dϕ π π/2 dρ ρ 43 sin ϕ 3π/2 π/2 dϕ dθ a 3π/2 π dϕ sin ϕ 9 V dρ ρ 43 sin ϕ dθ sin ϕ

10 9. Let E be the region bounded b z 2, z 2, and 3. The triple integral f,, z) dv can be epressed as an iterated integral in the following three orders of integration. Fill in the limits of integration in each case. No eplanation required. z z f,, z) dz d d z z f,, z) d dz d z z f,, z) d d dz Solution. The hard part of this problem is figuring out what E looks like. First here are separate sketches of the plane 3 and the plane z 2 followed b a sketch of the two planes together. z z z z 2 z Net for the parabolic clinder z 2. It is a bunch of parabolas z 2 stacked side b side along the ais. Here is a sketch of the part of z 2 in the first octant. z z 2 Finall, here is a sketch of the part of E in the first octant. E does have a second half gotten from the sketch b reflecting it in the z plane, i.e. b replacing. 1

11 z 3, z 6 p3,? 6,6q z 2 z 2 3 So 1 E {,, z) 3, 6 6, 2 z 2 } Order dz d d: On E, runs from 6 to 6. For each fied in this range, z) runs over E {, z) 3, 2 z 2 }. Here is a sketch of E. z p3,6q z 2 3 p 2 {2, 2 q p3, 2 q z 2 From the sketch and the integral is E {, z) 2 /2 3, 2 z 2 } 6 3 z2 6 2 /2 z 2 f,, z) dz d d 1 The question doesn t specif on which side of the three surfaces E lies. When in doubt take the finite region bounded b the given surfaces. That s what we have done. 11

12 Order d dz d: and the integral is Also from the sketch of E above E {, z) 2 z 6, z/2 3 } 6 z6 3 6 z 2 z/2 f,, z) d dz d Order d d dz: From the sketch of the part of E in the first octant, we see that, on E, z runs from to 6. For each fied z in this range, ) runs over E z {, ) 3, 6 6, 2 z 2 } {, ) z/2 3, 2 z } {, ) } z/2 3, z z So the integral is z6 3 z z z/2 z f,, z) d d dz 12