Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

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1 Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio, with P (x) =x ad Q(x) =e x x. We calculate R R P (x) dx = x dx = x (igorig the costat of itegratio) ad have as our itegratig factor ρ = e R x. R The geeral solutio R the the differetial equatio will the be y = 1 ρ ρq(x) dx = e x e x e x x dx = e x e x dx = e x ( 1 + C). ex Now we eed to choose C so that this fits y() = 5: Puttig i for x ad 5 i for y we get 5 = 1 + C ad so C =. Thus the solutio is y = e x ( 1 +). ex (b) What will be the value of y(1)? Usig the result from (a), y(1) = e 1 ( 1 e +) = 1 e+e 1. This ca be rewritte i various ways, such as e +, or approximated usig a calculator as :949 :::. e Problem For each of the followig series, tell all you ca about its covergece. This should iclude whether it coverges or diverges, ad if appropriate whether it coverges absolutely or coditioally. Be sure to give reasos for your aswers! (a) 1X =1 ( 1) l() Sice this series has some egative ad some positive terms, it is appropriate to cosider absolute vs. coditioal covergece. If we take absolute values we have P l() : We ote that l() > 1 ad that P 1 diverges, so the series of absolute values diverges by compariso. The terms l() of the series do decrease with limit zero as!1, ad the sigs alterate, so by Leibiz' theorem the series does coverge. Hece it coverges coditioally. 1X 4 (b) cos As!1, 4 =1!, ad the cosie is a cotiuous fuctio, so the terms cos( 4 ) i this series coverge to cos() = 1. Hece the terms do ot go to zero so the series caot coverge. (The th term test".) 1X (c) 5 =1 This series coverges, ad there are may ways to show it does. The most direct may be to ote that it is a geometric series whose first term is, with ratio 5 r = 1, ad that a geometric 5 series with jrj < 1 coverges. 1X (d) ( 1) =1 p P Sice there are egative terms we first test for absolute covergece: p = P 1 p-series with p = > 1 ad so it coverges. Hece the series coverges absolutely. is a

2 Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t = ), how doesitprogress ad alog what path, ad where does it ed? (The path is oe of the coic sectios we have studied. You should idetify it as such.) (b) Give a equatio i rectagular (Cartesia) form relatig x ad y without the parameter t. (You may fid the idetity si t + cos t = 1 useful i derivig this equatio.) If we write x += si t ad y = cos t, we have (x +) + (y ) =1. Hece the path is a 9 ellipse. Its ceter is at ( ; ) ad it exteds horizotally from ( ; ) to ( 1; ) ad vertically from ( ; ) to ( ; ). Whe t = we have x = si() = ad y = + cos() = so we start at ( ; ), the top of the ellipse. As t icreases, si t icreases ad cos t decreases so x decreases ad y also decreases: We move couter-clockwise aroud the ellipse, begiig i a southwesterly" directio. Whe t reaches ß the values of x ad y have retured to their origial values, so we retur to the startig poit ( ; ). Problem 4 Fid the area iside oe leaf of the three-leafed rose r = si( ). To determie what rage of values cosititute oe leaf" we solve r =, si( ) =. We get that must be a multiple of ß. The first two such are =ad = ß, i.e. =ad = ß, ad I will calculate the area i that leaf. We get Z ß 1 (si ) d = 1 4 Z ß (1 cos ) d = 1 4» 1 si ß Problem 5 Use words to describe the followig sets of poits i space. = 1 ß 4 1 si ß + 1 si = ß 1 : (a) y = ad z jxj The restrictio y =implies the poits must be i the plae that cotais the x ad z axes. If we had z = jxj, that would correspod to a V" of two rays extedig out from the origi, tilted up 45 degrees, withi the plae y =. Sice we have z jxj all the poits i the y =plae above that V are also icluded. Hece the set is the sheet i the x-z plae comig dow to a poit at the origi ad bouded o the bottom by lies tilted up at 45 degrees. (b) 4» x + y + z» 9 Sice x + y + z is the square of the distace from the origi to the poit (x; y; z), this set cosists of the poits whose distace from the origi is betwee ad. Hece it is a hollow ball", cetered at the origi: The iside is the sphere of radius cetered at the origi, the outside is the sphere of radius cetered at the origi, ad the set cotais all poits o those spheres as well as the poits betwee them. (c) ß» ffi» ß ad»» ß. This is apparetly a set of spherical coordiates. The iequalities ß» ffi» ß restrict the poits to lyig o or below the floor", the x-y plae. The iequalities»» ß restrict the poits to lyig betwee the plae where y =ad the plae where x =. There is o restrictio o the distace ρ from the origi. So the set is oe eighth of space: It cosists of the poits which are southeast" of the origi, i terms of their x-y locatio, but at or below the floor level, i.e. where z».

3 (d) r = z The distace r outward from the z axis is to be equal to z. That is distace outward i ay directio, sice there is o restrictio o. If we look at a cross-sectio obtaied by cuttig dow alog the z-axis, i.e. i ay vertical plae that goes through the origi, we will see two parabolas. They are r = z goig outward away from the z-axis i both directios i our plae. Sice that is true for ay such plae, the set is the surface obtaied by rotatig these parabolas aroud the z-axis. It is sort of a ifiitely large pulley", with a groove goig aroud the z-axis. Problem Let ~v = ~ i + ~ j ~ k ad ~u =8 ~ i +4 ~ j 1 ~ k. (a) Fid the (scalar) compoet of ~u i the directio of ~v. We ca compute this as ~u ~v j~vj = p = p 8. (b) Fid the (vector) projectio of ~u o ~v. We could use the result from (a) ad divide by the legth of ~v ad multiply the resultig umber oto ~v. Or we ca start from scratch with the formula ~u ~v ~v = 8 ~v ~v ~v = 14~ i+ 8~ j 14 ~ k. (c) Write ~u as a sum of two vectors, oe parallel to ~v ad oe orthogoal to ~v. The vector we use parallel to ~v is Proj ~v ~u which we computed i (b) to be 14( 1~ i+ ~ j 1 ~ k). If we call this ~u 1, we ow let ~u = ~u ~u 1 be the other vector: This gives us ~u = (8 14 )~ i +(4 8 )~ j +( ~ k) = 1 ~ i 1 ~ j ~ k. Problem 7 Let A = (1; ; 1), B = (; ; ), ad C (Cartesia) coordiates. = (1; ; 1) be three poits i space i ordiary rectagular (a) What is the area of the triagle whose vertices are A, B, ad C? We eed two vectors lyig alog sides of the triagle: I will use! AB = ~ i + ~ j + ~ k ad! AC = ~ i + ~ j + ~ k. The area we wat is the 1 of the size of the cross product of the!! vectors. Computig, AB AC = ~ i ~ j + ~ k. The legth of this vector is p 1+4+1= p so the area of the triagle is p. (b) Fid a vector of uit legth perpedicular to the plae cotaiig these three poits. Oe vector perpedicular to the plae is the cross product we just computed, so we fid a uit vector i that directio. We have already computed the legth of that vector ad we get for the desired vector p 1 ( ~ i ~ j + ~ k) = p 1 ~ i + p ~ j + p 1 ~ k. (c) Fid a equatio for the plae cotaiig these three poits. We could use either the cross product computed i (a) or the uit vector from (b) as a vector perpedicular to the plae. Avoidig the square roots, I will use ~ i ~ j + ~ k. The plae goes through A, B, ad C so we ca use ay of these as a poit o the plae." If we use A we get for the plae we get (x 1)( 1) + (y )( ) + (z + 1)(1) = or x y + z = 1 4 1=. This is a little prettier as x +y z =. We ca check that each of the three poits A, B, ad C satisfies this equatio. (d) Fid equatios for the lie through B perpedicular to the plae cotaiig the three poits. We ca use the same vector as the directio for this lie. We wat the the lie through (; ; ) i the directio of ~ i ~ j + ~ k. I parametric form this is x = t, y = t, z = t. I symmetric or Cartesia form it is x = y = z. 1 1

4 Problem 8 Cosider the graph of 9x 1y x 18y 4 = : (a) For some traslated coordiates x ad y, the ceter (or vertex if this is a parabola) will be at the origi of the coordiate system. Write the equatio i x ad y. We complete the squares ad get 9(x ) 1(y +4) = 144. If we let x = x ad y = y +4, ad divide by 144, we get (x ) ) 1 (y 9 =1: (b) Is this graph a circle, a ellipse, a parabola, or a hyperbola? Sice we have it i stadard form ad the (x ) ad (y ) terms have opposite sigs, it is a hyperbola. (c) If it is a circle, ellipse, or hyperbola, where is its ceter? If it is a parabola, where is its vertex? (Be sure to use (x; y) coordiates for this ad the followig aswers!) The ceter is where x =ad y =, so x =ad y = 4. The ceter is (; 4). (d) (i) If it is a circle, what is the radius? What are the coordiates of two poits o the circle? (ii) If it is a ellipse, where are the eds of the major axis (the two poits furthest from the ceter)? Where are the foci? What is the eccetricity? (iii) If it is a parabola, where is the focus? Give a equatio for the directrix. (iv) If it is a hyperbola, where are the foci? Where does it cross the x or y axes (the traslated axes you foud i (a)). (You ca idetify these poits usig either coordiates but make clear which you are usig.) Give equatios (i x ad y) for the asymptotes. Usig the otatio we have used i class ad i the text, a =4ad b =. Hece c = p a + b =5. Thus the foci are 5 uits from the ceter, alog the x axis ad so horizotally displaced from the ceter. That puts them at ( ; 4) ad (7; 4). The curve does ot cross the y axis sice settig x =gives (y ) = 9. It crosses the x axis at x = ±a = ±4. These crossigs are at (±4; ) i x -y coordiates, or ( ; 4) ad (; 4) i x-y coordiates. The asymptotes are y = ± 4 x : Usig x = x ad y = y +4 we get y +4= 4 (x ) ad y +4= (x ). Those ca be simplified to 4 y = 4 x 11 ad y = 4 x 5. Problem 9 Some iitial terms of the Maclauri series for e x are used to approximate f(x) =e x, for :1 <x<:1. How may terms are eeded i order to guaratee that the approximatio is withi ±:1 of the actual value of f(x) for all x i that rage? Write out the polyomial showig those terms. Your aswer should both tell the umber of terms eeded (or the degree of the highest power term eeded, but be sure to idicate which!) ad the polyomial you would use as your approximatio. (You may use the fact that je x j ad je x j are both less tha for :1 <x<:1.) The Maclauri series for e x starts out 1 x + x x + x4 :::. The derivatives of are 4 e x alterately e x ad e x. Hece the remaider term R (x; ) (i Lagrage form) is ±e c x +1. (+1)! Thus if we use the terms of the series through ± x we are guarateed the approximatio at! x is withi j ±e c x +1 j of the correct value, for some c betwee ad x. We eed to choose (+1)! so that, for ay x betwee ±:1, that is guarateed to be at most :1. Sice :1 <x<:1 ad c is betwee ad x, we kow :1 < c < :1 also. Usig the hit, je c j <. Sice jx +1 j icreases as we move away from, the largest that part ca be is (:1) +1. Hece the

5 remaider term satisfies jr (x; )j» (:1)+1. We compute this for a few values of : Whe (+1)! =we get :1 = : ::: which is ot withi the specified tolerace of :1. Whe =we get :1 = :1 < :1 so =is good eough. 4 1 Summarizig, we use the terms through x we use the polyomial 1 x + x x. Problem 1, the first four terms icludig the costat, so Fid all solutios of y +y =4x +: (a) Fid all solutios of y +y =: The characteristic equatio is r +r =, with roots r = ad r =, so the solutios to y +y =are y h = C 1 e x + C e x = C 1 e x + C. (b) Fid a solutio of y +y =4x +: Sice the fuctio 4x + is a polyomial, ad is a root of the characteristic equatio, we use a polyomial Ex + Fx for our particular solutio y p to y +y =4x +. The y =Ex + F ad y =E. Puttig these i the equatio we get E +4Ex +F =4x +. p p Rearragig we have (4E)x +(E +F )=4x +, so 4E =4ad E +F =. These give us E =1ad F = 1. Hece our solutio is y p = x + x. (c) Write a expressio for all solutios of y +y =4x +: We combie the fuctios y h ad y p ad get C 1 e x + C + x + x to the full equatio. represetig all solutios