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# Analytic Geometry and Calculus 3 MATH 2400

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This 15 page Class Notes was uploaded by Cydney Conroy on Thursday October 29, 2015. The Class Notes belongs to MATH 2400 at University of Colorado at Boulder taught by Joshua Wiscons in Fall. Since its upload, it has received 178 views. For similar materials see /class/231821/math-2400-university-of-colorado-at-boulder in Mathematics (M) at University of Colorado at Boulder.

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## Reviews for Analytic Geometry and Calculus 3

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Date Created: 10/29/15

1x 9 03 1quot 0 7 00 D O Review Sheet for Midterm 2 Compute the following limits If a limit does not exist explain why 1 z 7 y a hm 7 mgwon 22 112 2 2 b hm L zya0gt0 1952 112 0 lim 7 mawon 96 y A point moves along the intersection of the elliptic paraboloid z x2 3y2 and the plane y 1 At what rate is z changing with respect to z when the point is at 217 Use total differentials to approximate the change in the value of fy ln 1 xy from P02 to Q7009 198 1 y leen x 2 f y yz P711 1 Q7099099 101 a Find the local linear approximation L to the function fxyz at the point P b Use part a to approximate fQ Let w zy yz 2x with x u2 7112 y uz 02 and z 112112 Compute 3 and 377 The sun is melting a rectangular block of ice When the blocks height is 1ft and the length of each edge of its square base is 2ft its height is decreasing at 2 inh and the length of the edges of the base is decreasing at Binh What is the rate of change of the volume of the block at that instant Let fyz xyz Find the directional derivative of this function at the point P1 11 in the direction lt111 Find a unit vector that is normal at P23 to the level curve of fy 3x2y2 7 xy through P Find an equation for the tangent plane and parametric equations for the normal line to the surface 2 y z at P244 Suppose that three quantities x y and z are constrained by the equation 2x23y2 22 20 The graph of this equation is a surface S in space a Verify that the point P2 1 3 is a point on S and nd the equation of the tangent plane to S at this point b Near P2 13 we can think of 2 as a function of z and y z fy explicitly determine its linear approximation Lf near fzy Without nding z 2 y 1 c Approximate the value of 2 corresponding to z 197 and y 112 11 Consider the function ay 23M4x By and the point P128 a Find an equation of the tangent plane to the graph of z ay at the point P b Find a normal vector to the tangent plane of the graph of z ay at the point P c Approximate f1118 12 Find the absolute extrerna of f on B7 where ay 5 7 3x 4y R is the closed triangular region with vertices 07 07 47 07 45 13 Find the points on the surface 22 zy 1 that are closest to the origin 1 Solution a Along then line y 07 i z r 1 Inn 7 Inn 7 7 410 x2 410 z 7 which does not exist Thus lirn 27112 NAHUM 9 b Covert to polar coordinates xzyz r4 cosz 6sin2 t9 3 lt r 2 y2 r i As 7 a 07 r3 a 07 so by the Squeeze Theorern7 2 2 y a 0 as Ly a 00 2 y2 z2y2 Thus lirn wgty 0gt0 x2 y 0 Along the line y x 4 962 112 W 52 W1 039 Along the curve y 3 6 1 i131 2x6 7 y We have different lirnits7 so lirn maFWD 2 2 Solution Need to nd 3732 1 Since7 2x the answer is 3 Solution my 111W w 1nlt1 w 1 y 1 x d7 d 77d f 21yx21xyy 1 2 1 0 Af N Emeow 7 0 Emugs i 2 4 Solution L7y72 fwlt717 17 1 fit717 17 71 f2lt717 17 71 1 so fm7111 E fz y w 1 fy i W so fy7111 i 2 fz SO f2lt717171 0 Putting everything together7 we get Ly7 z 1 7 1 b 1 1 f7099099 101 m L7099099 101 E7099 1 E099 i 1 Q 5 Solution 3w 7 6m 3m 6m 3y 311162 E T 5 671 5 y ZWU 95 ZWU y QWZ u2 112 u2022u u2 7 112 Uzi2 2u 2u22u12 6m 6m 3m 6m 3y 6m 32 E 5 mm 5 y Z21195 2W y 2u2 u2 112 u20272v u2 7 112 Uzi2 21 2u22u211 6 Solution We have V xzy We want 21 7 00 CD dVialdx dt 7 9x dt lt271gt 2lt2gtlt1gt 22 72 Our answer is 7 3h 111 7 Solutlon Set u 7 51 3y dti L 31171 We want Duf111 Vf111 u Now7 Vf lty27z27xy so Vf111 lt17171 Thus Duf111 A Solution VfP is normal to the level curve of f through P Vf lt6zy2 7 y762y 7 z so VfP lt105770 We need a unit vector7 so we use 0577 7 1 lt105 70 7 1 lt21 14 Hlt10570H 7 51637 T M637 39 Solution The tangent plane at P will have n ltQP 371P 71 as a normal vector Since7 31 7 3y 6 7 6 6 7 72 i and 72 7i we calculate that n lt287 727 71 Thus the tangent plane at P 3m 2y2 3y 3 has equation 28x 7 2 7 2y 7 4 7 42 7 4 0 The normal line at P has n as a direction vector7 so the normal line has vector equation lt27474 tlt287 727 71 Parametrically7 we get 228ty472t7247t Answers a The tangent plane to S at the point P is given by 8z 7 2 6y 71 62 7 3 0 b The linear approximation ofz ay near 21 is Lfxy 3 7 7 2 7 y 71 c z197112 292 11 Answers a Tangent plane is given by z 8 25x 71 3y 7 2 b A normal vector to the graph of z ay at the point P is n lt725731gt c f11 18 m 99 Solution First nd the critical points f1 3 fy 4 Since the rst partials are never zero there are no critical points Next nd extrema on the boundary a One leg of the triangle is y 0 0 S x S 4 Along this path the function values are fx0573 0 zlt4 Since this is a decreasing function the absolute maximum is f0 0 5 and the absolute minimum is f4 0 77 A C7 V f47y74y7 03135 Since this is an increasing function the absolute maximum is f45 13 and the absolute minimum is f40 77 A O V values are 5 5 fx1z57341z52x 0 4 Since this is an increasing function the absolute maximum is f45 13 and the absolute minimum is f00 5 Therefore the absolute maximum is f45 and the absolute minimum is f40 Solution We can minimize the square of the distance The square of the distance to the origin is d2 2 yz 22 x2y2xy1 To nd critical points set the rst derivatives equal to zero d2w2y0 d2y2yz0 Another leg of the triangle is z 4 0 S y S 5 Along this path the function values are Along the hypotenuse of the triangle y Ex 0 S x S 4 Along this path the function which implies y 72x and z 72y The only critical points occur at z 07 y 0 Now use the second derivative test to con rm this is a minimum d2m 2gt0 2 1 3gt0 Therefore the points on the surface 07 07 i1 are closest to the origin D H 03 flaw Review Sheet for Midterm 1 Let v1 lt111 v2 lt7111 Find the angle between v1 and v2 in terms of arccosw A wagon is pulled horizontally by exerting a force of 10 lbs on the handle at an angle of 60 How much work is done by moving the wagon 50 feet Let v lt111gt b lt220gt a Find the orthogonal projection of V onto b b Find the component of V orthogonal to b Show that 12 gtlt 12 O for any vector 12 Let 131 lt1 23 132 lt321 Find a vector that is orthogonal to both 131 and 132 Find the volume ofthe tetrahedron de ned by P10 0 0 P23 72 75 P31 4 74 P40 3 2 Consider the points A4 723 B8 776 and C6 746 7 7 Find the area of the parallelogram spanned by the vectors AB and AC AA C793 VV Find an equation of the plane containing the points A B 0 Find an equation of the sphere centered at the origin and tangent to the plane containing the points A B C A O V A D V Find the area of the triangle with vertices A1 0 0 B0 1 0 and C00 1 and com pare it to the area of its projection on the zy plane A C7 V Find the distance from the point P14 48 to the plane 6x 7 2y 32 2 A O V Find an equation of the sphere centered at the point P1448 and touching the plane 6x 7 2y 32 2 A D V Find an equation of the plane containing the line L I39t 23ti1 72tj 71 tk and the point P131 Find a parametric equation of the line of intersections of the planes z 7 y 22 2 and 3x y 7 z 4 b Consider the points A2 771 B654 C6 23 and D7 744 a Find the area of the parallelogram de ned by E and AC 7 7 7 b Find the volume of the parallelipiped de ned by AB AC and AD c Find the distance from the point D to the plane passing through A B and C Find the vector component of v along b and the vector component of v orthogonal to b where v lt3 72 76 and b lt1 72 2 Find the area of the triangle with vertices P1 10 Q101 and R0110 13 Find parametric equations for the line a through 3 718 and parallel to V 235 b through 31 71 and 32 76 14 Where does the line parallel to zyz 1 2t3t5 7 7t and through the point 02 71 intersect the coordinate planes 15 Where does the line zyz 2tt 7 1 73t intersect the hyperboloid 27i1 19 9 16 Determine whether the lines L1 and L2 are parallel skew or intersecting If they intersect nd the point of intersection L1x1t y27t 23t L2z27t y12t z4t 17 Show that the lines L1 and L2 are not paralleland do not intersect one another L1z27t y3t z4t L2x573t yt z23t 18 Find an equation of the plane 75 1 2 with normal vector n 3 752 123 with normal vector n 159 712 10 73 0 72 74 4 1 6 2 1 73 5 71 4 2 72 4 e passing through 71 73 2 and containing the line zyz 71 7 2t4t 2 t passing through AAAA c passing through and 2z7y321 19 Determine whether the planes are parallel perpendicular or neither If neither nd the angle between them a snug7321 73z6y720 b 2z2y724 73z76y3z10 20 a Determine whether the line L and the plane 73 intersect or are parallel Lx774t y36t 295t 734xy2217 b Do the same for these Lx32t y675t z23t 733x2y7421 2 H 2 D 23 2 a 2 U 2 2 2 00 2 CD 3 O 31 7 Derive an equation in s y and z for the plane that contains the point P0z0y020 and is perpendicular to an arrowvector represented by the cartesian vector n lta bcgt Show the distance between the parallel planes cw by 02 d1 and ax by 02 d2 is ldl dzl D Given rectangular coordinates convert them to cylindrical coordinates and spherical coordi nates a 07270 17 717 Given cylindrical coordinates convert them to rectangular coordinates and spherical coordi nates a xi 73 0 27 07 2 Given spherical coordinates convert them to cylindrical coordinates and rectangular coordi nates a 0 3 b 57 07 0 Convert the given equation in cylindrical or spherical coordinates to an equivalent equation in cartesian coordinates and identify the surface represented by it a p 2 sec b b r2 cos 20 2 Match the equations Match each rectangular equation from the rst column with an equiva lent cylindrical equation in the second column Then match each cylindrical equation in the second column with an equivalent spherical equation in the third column 1 z 1x2y2 a rcos02sin052 10 15 2x2y5210 b 27 Let I39t cost i sintj tk and to E Find the vector Iquott0 Find a parametric equation of the line tangent to the graph of I39t 62ti 7 2 cos3tj at the point where t 1 1 Evaluate 62 i e tj tk dt 0 Solve the vector initial value problem for I39t by integrating antidifferentiating and using the initial conditions to nd the constants of integration I39 t 9sin ti 9cos tj 4k I390 3i 4j r 0 2i 7 7j ii cos0sin 2sin0sin 5cos 10 3 3 3 D 03 F Find an arc length parametrization of the curve that has the same orientation as the given parametrization and has It 1 as the reference point rt2t2it3 193 Find the arc length of I39t tZi cost tsin tj sint 7 tcos tk for 0 S t S 7139 Find an arc length parametrization of the curve that has the same orientation as the given parametrization and has It 0 as the reference point I39t QCOSStiSin3tj 0 S t S H D 03 F 0 7 00 CD Review Sheet for Midterm 3 edA R where R is the triangle bounded by the line z y 2 and the coordinates axes Evaluate Find xz ydA where R is the parallelogram bounded by the lines R y0 xy1 2z7y0 227173 Find the area of the region bounded by the curves y47 y2z4 y1 zy6 For the following integrals a sketch the region of integration b reverse the order of integration and evaluate it 2 1 3 i yem dx dy 0 1 4 271 n 6x 2 dy dz 6x 2 dy dz 0 E 1 E Find the volume ofthe solid that lies above the zy plane below 2 x and within the triangle whose vertices are 10 02 and 12 Compute the area of the region enclosed by the rose 7 sin 26 given in polar coordinates Find the volume of the solid below the cone 2 x2 yz inside the cylinder 2 y2 2y and above the plane 2 0 Compute 197x27y2dA R where R is the region in the rst quadrant within the circle x2 y2 9 Consider the parametric surface zuv fooltuvltoo Find an equation of the tangent plane to the surface at the point 1 4 3 Find the area of the portion of the plane 2x 2y z 8 in the rst octant Find the area ofthe portion ofthe cone I39u 1 u cosv iu sinvju k for which 0 S u S 21 and0 v g Find the volume of the solid bounded by z 2 7 2 2 2 y 0 and y 3 03 a 0 51 00 D O H Find the volume if the tetrahedron bounded by 2x By z 67 z 07 y 07 and z 0 Express the following integral as an equivalent integral with z integration rst7 y integration next7 and x integration last 2 W x47y2iz2 fltzyzdzdydz 0 0 0 Find the volume of the region that lies inside the sphere 2 y2 22 4 and the cylinder x2 y2 1 Use cylindrical or spherical coordinates to evaluate 3 x97m2 Qizziy2 z dz dy dx 0 0 0 Use cylindrical or spherical coordinates to evaluate 2 x47m2 Sizziy2 dz dy dx 0 0 75m2y2 Evaluate x2 d5 along 0 o rt6ti3 t2j213k7 09 1 Evaluate F dr where Fxyz y z i 7 zzj 7 4y2 k7 and C O rttit2jt4k 091 Find the work done by the force eld Fxy zzi zyj on a particle that moves around the circle x2 y2 4 in the counter clockwise direction Given that Fxy 2y3i 3ij and o rtsintit21j 09 a nd the potential function for F b nd the work done moving a particle along 0 in the force eld E H D 03 F 0 Sa 7 00 CD Answer Answer Answer Answer i a The region is a triangle with vertices 00 10 and 12 0 2 1 3 1 2m 3 yem dzdy yemdydx 0 g 0 0 ii a The region is bounded by z y2 and z 2 7 y Note that these curves intersect at 11 and 4 72 b 1 4 27 1 21 99 6z2y2dydz 6z2y2dydz 6z2y2dxdy i 0 1 xi 72 12 2 1 2 2 AnswerV zdyd 0 2721 3 I sin29 Answer A42 rdrd0 Note sin2291cos49 0 0 7r 25in9 39r Answer V rdzdrd0 0 0 0 Note sin3 6 17 cos2 9 sir10 Answer Use polar coordinates 3 97139 97x27y2dA 97r2rdrd0 R 0 0 Answer The point 143 is obtained when u 1 and 1 2 Then 3r 3r 712 712 w 6ultgtx6ltgt is normal to the surface where ruv uzi 112j u 1 k We compute 6 672uik 3r 72 39 k 31 DJ so 37212 2i k and 12 uj k Thus w 74i 7 2j 8k Therefore the equation of the tangent plane is 74x 7 1 7 2y 7 4 82 7 3 0 which simpli es to 2x y 7 42 76 10 Answer We have 2 8 7 22 7 2y so 72 and 37 72 Thus 32 2 32 2 A i i 1 CIA 3 dA 24 ax 6y R R where R is the triangle with vertices 07 07 47 07 and 07 4 Note R is a triangle with base and height equal to 4 so dA 8 11 Answer We have a lcosvisinvjk 3U 6r 1 77us1nvlucosvg 31 Thus7 3r 3r 7 7 7 7 i 2 auxav ucosvl usianukH 2u Now7 a a 21 A ix dudv 2u2dudv 3U 31 0 R 3 1 242 12 Answer V dzdsdy 0 71 m2 3 27 z 6722731 13 Answer V dzdydz 0 0 0 14 Answer This is the 1st octant part of the sphere 2 y2 22 4 so 2 W x47y2722 2 W W fy2d96dydz fy2d2dydz O O O O O O 27r 1 W 32W 15 AnswerV2 rdzdrd6 7747ng 0 0 0 3 16 Answer 3 W 9427 g 3 9772 162 xdzdydz rzcos dzdrd0 0 0 0 0 0 0 5 2 M 3427212 I 2 342 dzdydz2 rdzdrd6 0 0 7 0 0 7275 17 Answer 5z2y2 2 18 Answer 1 1 zzds 6t2t3H6i6 2tj6t2klldt72 t4t21dt C 0 0 19 Answer 1 1 Fdr t2t4i7t2ji4t4ki2tj4t3kdt t2t472t3716t7dt C 0 0 20 Answer 27r WFdr 4cos2ti4costsintj72sinti2costjdt C 0 21 Answer 2 a 3 2 a 2 2 2 i i a Since ailQxy 62y and 82 y 62y F is conservative If we let f be the 6 6 potential function for F we have Vf F so 67f Qxyg and 61 32y2 Integrate 9 6 6 61 with respect to z to nd that ay gs 9y Since 67f 322y2 we nd that 9 g y 0 so gy is just a constant K Thus f is of the form ay gs K Any value for K will satisfy the equation Vf F so we might as well choose K 0 Thus ay xzys is a potential function for F of course ay gs K is as well for any constant b WCFdr0Vfdr f1217f01

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