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Test 2 Study Guide

by: Wesley Hunt

Test 2 Study Guide MATH 213

Wesley Hunt
GPA 2.88
Analytic Geometry and Calculus III
Neil Epstein

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About this Document

12-1-12.9 inclusive. Contains equations and some images.
Analytic Geometry and Calculus III
Neil Epstein
Study Guide
math calc iii
50 ?




Popular in Analytic Geometry and Calculus III

Popular in Mathematics (M)

This 0 page Study Guide was uploaded by Wesley Hunt on Wednesday January 20, 2016. The Study Guide belongs to MATH 213 at George Mason University taught by Neil Epstein in Spring 2016. Since its upload, it has received 85 views. For similar materials see Analytic Geometry and Calculus III in Mathematics (M) at George Mason University.


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Date Created: 01/20/16
Test ll Study Guide 121 Planes and Surfaces N is orthogonal The dot product of nr r0 O But for a 3D gure we need to have the dot product of all three axes c4x xdbbj ck 1JO Just like with vectors planes are parallel if they have the same or opposite directions They are orthogonal if the dot product equals zero HOW to Structure an Equation for a Plane 60 1 and normal vector 3 4 6 Normal vector is a b c and internal points are of the points on normal line 3x 64y 0 6z 1 3x 184y 6z6 3x4y 6z12 How do you nd the Normal Vector P2 1 3 Q 14 0 RO1 5 Take the cross product of P0 and OR Subtract P from Q and R P39QxQR 15 3x 202 153 202 n10i8j10k THEN take the dot product nP10810x 2y1z 3O 5x4y5z21 What s a Face A trace is a parallel intersection across any of the axes Determine Surfaces 2 2 Ellipsoidzx 2y 2Z 21 a b C 2 2 Parabaloid x 2y 2 z a b 2 2 2 Hyperboloid one Sheet x 2 1 a 2 2 2 Hyperbolid Two Sheets x 1 2 a N N 2 2 z x Elllptle Cone 2 2 Y 2 X 2 2 Hyperbollie Paraboloid Z W a For all values of ya values point down or up depending on the sign Intersection Q X 2y z 5 R 2X y z 7 Both planes need to intersect xy Z cancels out X 3 and y1 Therefore 3 10 is your intersection vector Intersection of a line is the same as everything else Parallel vector with 3 10 parallel to 3 3 3 Rt 3 1 0 3 3 3t Rt 33t 13t 3t 122 Graphs and Level Curves As x and y change elevation 2 will change glxyll4x2y2 It will always increase by x or y places With 2 just add another dimension going outward 123 Limits and Continuity A limit of f x y is continuous at a b if the limit as x y approaches a b of x y equals a b You can also add a 2 lim 3x2zyzcosHx Hz xyzlgt21 1 This is a continuous function so plug and chug 322 112Cos 2 1 I 1 14 But what about if the function is not continuous at a point THEN you check at xy Find the two paths of the limit In this case we get two different values therefore the limit does not exist 124 Partial Derivatives The most important thing is When taking the derivative of one variable the other is just a constant fixyl3x2y4 Fx 3 Fy 8y3 fix Zlxy2z33yz FX y223 This is a constant paired with X Remember that Fy xz32y32 Derivative of y paired with a constant F2 322xy23y fixylxlnx2y2 Fxlnx2y2r1x2y22x Fyx1x2y22y Some are constants some are variables but the rest are normal For e lower the constant and keep everything else xy xy fxe cosz xe cosz 125 Multivariable Chain Rule Make a tree diagram follow the branches multiply respective branches and add products of other branches yaxy2tX2 2y22t22t2t am 23 28t42t4 8t10t4 8t ELIE H H 126 Directional Derivatives and the Gradient If F is differentiable of X y that function has a directional derivative for any unit vector Dufixyyfx xyafyxyb Take partial of x rst multiply by a rinse and repeat for y Find FX y x2y3y4 at 2 1 for angle pi4 Unit vector in direction of angleltcos pi4 sin pi4gt 2 gtZ 2 2 Dx 2xy3a 3y2x24y3b D2 1 2213root22 32212413root22 2 4 6 What s a Gradient Vector fxlxyl lxylgt It s what we just solved for This is just the notation for it 127 Tangent Planes and Approximation Suppose f has continuous partial derivatives An equation of the tangent plane to the surface 2 fx y at point PxO yO zO z z0fxx0y0x x0 lx0yoyyo EX fx y 20 x2 7y2at 21 appr0ximate f195108 Estimate zcoordinate z20 22 71253 Use 2 1 3 as your new vector z 3fxlt21gtix 2ifylt21gtlty 1gt 7322 20 x2 Z 5 1 FX 2 1 23 fng 7322 20 x2 Z lt31 Fy 2 1 73 D255 PIugandchug from here x 27y 1 Approximate using the last function In this case the answer is about 288 128 Maxima and Minima Let D Da b Fxxa bFyya b fxya b2 A If D gt 0 AND Fxxa b gt O l Equation is a minimum B If D gt O and Fxxa b lt O l Equation is a maximum C If D lt O this is a saddle point Fx y Xy1Xy Xyx2yxy2 Fx y2xyy2 Fy xx22xy Fxx 2yO Fyy 2x Solve for D y2Xyy20 12xy 0 OR 12xy O Y 12x 00and 10 Y 12x O xx22x12x xx22x4x2 O x3x1 X O or 13 0 1 and 13 13 are your new points Set up the equation Take the D equation evaluated at the critical points DO O 1 D1 O 1 DO 1 1 D13 13 13 Last one is not a saddle point FxxD in that case is 23 Local max 129 Lagrange Multiplier It s a way to nd a minmax of f x y 2 Form a new function f x y z A Just add lambda flxyzl v8lxyzlkl Take partial derivatives with respect to all variables They all equal zero Cxy6x212y2 If there must be a value of 90 what is your minimum cost Constraint X Y 90 6x812y2 yxy 90FXBY 6x812y2 yx yy90y 39y F 8 O x 4x y x 8 v F24 Ogt y y v y 24 Fy x y900 1900 48 24 90 48 130 48 Then plug and chug You should get either max or min A 1440 X 120 y 60 You need to nd lambda and plug that back into x and y Then return to original equation 6120812602 The only problem is with multiple variables Solving the systems could mess you up but take your time on them


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