Calculus 111 Gylindrical Coordinates ‘As with two dimensional space the standard (x,,y,2) coordinate system is called the Cartesian coordinate system. In the last two sections of this chapter we'll be looking at some alternate ‘coordinate systems for three dimensional space. and Bere the same as with polar coordinates, Here is a sketch of a point in R*. (9.2) =(.82) The conversions fpr and y are the polar coordinates.\8o, if we have a poi mine cylindrical coordinates the Cartesian coordinates can ‘The third equation is just an acknowledgement that the 2-coordinate of a point in Cartesian and polar coordinates is the same. Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions. ©2007 Paul Dawkins 55 bttp:/tutorial.math.lamar.edu/terms.aspxCalculus It OR Let’s take a quick look at some surfaces in cylindrical coordinates. Example I Identify the surface for each of the following equations. (a r=5 (b) +27 =100 al (@z=r ‘Solution (a) In two dimensions we know that this isa circle of radius 5. Since we are now in three dimensions and there is no z in equation this means itis allowed to vary freely. So, for any given @ ‘= we will have a circle of radius 5 centered on the 2-axis. In other words, we will have a cylinder of radius 5 centered on the z-axis. (b) This equation will be easy to identify once we convert back to Cartesian coordinates. 4272100 Pt yte2? 2100 So, this is a sphere centered at the origin with radius 10, (c) Again, this one won't be too bad if we convert back to Cartesian. For reasons that will be apparent eventually, we'l first square both sides, then convert, [LErom the section of quadtic surfaces we know that this isthe equation of a cone. (©2007 Paul Dawkins 56 hntp:/utorial math Jamar.edu/terms.aspx.Calculus 111 In this section we iroduce spherical coordinates, Spherical coordinates can take a little Betting used to, Its probably easiest to start things off witha sketch, © (2) 2(2.8.9) a i». | J F | om ‘Spherical coordinates consist ofthe following three quantities. | First there isp This is the distance from the origin to the point and we will require p20. \ Ne eer ele Minicom | Next there is 8. This is the same angle that we saw in polar/eylindrical coordinates, Its the i angle between the positive x-axis and the Tine above (Which is also the same r asin | polarfeylindrical coordinates). There are no restrictions on 0. Finally there is . This is the angle between the positive z-axis and the line from the origin to tive point, We will require 03 97. ae @ In summary, isthe distance from the origin to the point, @ is the angle that we need to rotate down from ihe positive z-axis to get to the point and @ is how much we need to rotate around the ‘eaxis to get to the point. We should first derive some conversion formulas, Let's first start with a point in spherical coordinates and ask what the cylindrical coordinates ofthe point are. So, we know (, )) and want to find (r,0,z). Of course we really only need to find r and z since @ is the same in both coordinate systems. We will be able to do all of our work by looking atthe right triangle shown above in our sketch. With a little geometry we see that the angle between z and p is @ and so we can see that, (©2007 Paul Dawkins 37 ‘ttp-/tutorial.math Jamar eduterms.aspxCalculus Itt ZS poosp r=psing and these are exactly the formulas that we were looking for. So, given a point in spherical ‘coordinates the cylindrical coordinates of the point will be, r=psing 0=0 z= posg Note as well that, + Pag ? (cos* 9 +sin? p) = 2? Or, . Next, let's find the Cartesian coordinates of the same point. To do cylindrical cohversion formulas from the previous section. we'll start with the ‘Now all that we need to do is use the formulas from above for r and zto get, x= psingensd y= psingsin® -_ z= peosp Poege OP ont apeets ay Pee apeg ‘Also note that since we know that r? = x? +)? we get, ny preesth : Paxty tz ‘ Converting points from Cartesian or cylindrical coordinates into spherical coordinates is usually - = p=cos"(t)=2 ee p we = 63) 3 Notice that there are many possible values of ¢ that will give cos =+, however, we have restricted g to the range 0< @ <7 and so this is the only possible value in that range. So, the spherical coordinates of this point will are (24. ‘(Return to Problems} (©) Convert the point (~1,1,—V2) from Cartesian to spherical coordinates. ‘The first thing that we'll do here is find p. paletty ee = Jitit2 =2 Now we'll need to find gy. We can do this using the conversion for z, z=posp => = oncon'{ As with the last parts this will be the only possible g in the range allowed, Finally, let's find 8. To do this we can use the conversion for x ory. We will ise the conversion for yin this.case. sind = Now, we actually have more possible choices for @ but all of them will reduce down to one of the two angles above since they will just be one of these two angles with one or more complete rotations around the unit circle added on. We will however, need to de: Which one is the correct angle since only one will be. To do (© 2007 Paul Dawkins 59 bttp:/tutorial.math.lamar.edu/terms.aspx. J PeesbCalculus IIT this let's notice that, in two dimensions, the point with coordinates x=—T and y=I Ties in the second quadrant, This means that @ must be angle that will put the point into the second quadrant, Therefore, the second angle, @ = 2, must be the correct one. ‘The spherical coordinates of this point are then (222 4 Retw Now, let's take a look at some equations and identify the surfaces that they represent. ‘Example 2 Wdenlify the surface for each of the following equations. (a) p Solution (®) p=5 ‘There are a couple of ways to think about this one, First think about what this equation is saying. This equation says that, no matter what @ and are, the distance from the origin must be 5. So, we can rotate as much as we want away from the z-axis and around the z-axis, but we must always remain at a fixed distance from the origin. This is exactly what a sphere is. So, thi is a sphere of radius 5 centered at the origin, ‘The other way to think about itis to just convert to Cartesian coordinates. pas pi=25 wey tz? =25 ‘Sure enough a sphere of radius 5 centered at the origin. = turn to Problems} 7 Be eages Retuins ee ce eee ee the Steen pie axtaaye = b) p= lees any J epge In this case there isn’t an easy Way to convert to Cartesian coordinates so we'll just need to think about this one a little. This equation says that no matter how far away from the origin that we move and no matter how much we rotate around the z-axis the point must always be at an angle of $ from the 2-< ‘This is exactly what happens in a cone. All of the points on a cone are a fixed angle from the z- (© 2007 Paul Dawkins © bitp:/Autorial.math lamar edw/terms.2spx| | | | Calculus 111 axis. (Return to Problems) ‘axis. So, we have a cone whose points are all at an angle of © from the z (90-24 3 ‘As with the ast part we won't be able to easily convert to Cartesian coordinates here. In this case "No matter how far from the origin we get or how much we rotate down from the positive z-axis the points must always form an angle of 2% with the x-axis. Points in a vertical plane will do this. So, we have a vertical plane that forms an angle of 2 with the postive x-axis, Return to Problem (@) psing=2 In this case we can convert to Cartesian coordinates so let's do that. There are actually two ways to do this conversion. We will look at both since both will be used on occasion. Solution 1 Inthis solution method we will convert directly to Cartesian coordinates. To do this we will first need to square both sides of the equation. psn? g=4 ‘Now, for no apparent reason add pcos’ g to both sides. pisin? 9+ p* cos! 9 = 4+ pcos? g ? (sin? p+cos' 9) =4+ p? cos? piad+(peose)? ‘Now we can convert to Cartesian coordinates. Styled adez? xeys4 So, we have a cylinder of radius 2 centered on the z-axis. This solution method wasn't too bad, but it did require some not so obvious steps to complete. Solution 2 ‘This method is much shorter, but also involves something that you may not see the first time around. In this case instead of going straight to Cartesian coordinates we'll first convert to cylindrical coordinates, This won't always work, but in this case all we need to do is recognize that r = psing and we will get something we can recognize. Using this we get, (©2007 Paul Dawkins 6 butp:/tutorial.math.lamar.eduterms.aspxCalculus It psing=2 r=2 ‘Atthis point we know this isa cylinder (remember that we're in three dimensions and so this isn't acirelel). However, let's go ahead and finish the conversion process out. 2 red xeys4 So, as we saw in the last part of the previous example it will sometimes be easier to convert equations in spherical coordinates into cylindrical coordinates before converting into Cartesian coordinates. ‘Thi easier. ‘won't always be easier, but it can make some of the conversions quicker and “The last thing that we want to do inthis section is generalize the first three parts ofthe pr. example. > pza sphere of radius a centered at the ori e=a cone that makes an angle of or with the positive z—axis o-8 vertical plane that makes an angle of @ with the positive x—axis Partial Derivatives i : Introduction In Calculus | and in most of Calculus II we concentrated on functiges of one variable. In Calculus Ill we will extend our knowledge of calculus into funciiéns of two or more variables. Despite the fact that this chapter is about derivatives we will srt out the chapter with a section ‘on limits of functions of more than one variable. In the renpainder of this chapter we will be looking at differentiating functions of more than one varjable. As we will see, while there are differences with derivatives of functions of one v« ii bitpy/tutorial.math.lamar.edwterms.aspx. °a ih ! Rectangular Coordinale System: In rectangular Coordinates the Surfaces veprecenticl by eqpcttion of the form 2%, Y=4, and grh where %0, 4 and 3, are constantly | are constant sunfa te: 2% =Xe , Is plane parallel to gZ-plane. aged > is plane Parettel to x3- plane. | 23+% 2 16 plane paralfel te xy-plane. i e@, Calindvicad coordinates system} ) | In cylindrical Coordinates the Surfacer vepresenled Rt, , it Cyuations of te form Sh | i B=%e, =O, and 3rd where v, B, aad Rodeo consteubi, are constaxt surpaces. | Dex, 16 wight circular eglinder op radius % cenberedt | en geasiew at each pot Cr 6) on thi Cybinder ¥ has the | value v,, but & and % are unvertricted excapt for our blanket 6 assumption that 0f 662%. 7 be By (so half plone, altasheel along. an angle B,_with He petitive x-apis. At each point 0,3) ' én dNe curface, @ hou the valite 6, bub r and 3-are unrisbrelad except foriour blanket assump tion TO. 7 EER» We capeoe Jn), it a hergerlel PANE: At enok poil: (58; 35 3 hae He value J, bat v and & are unreitricttd Craft for our blanket ascumptions. a He Jeans and makingSpherical Coordindhe. 2ydlen: Ja sprovcal exord inal the surpaces veprezerted by y ofthe form 923, 62@ and P= R where LE adh are conta, are torttart Gucpacer. > Seke> this is the sphere ofcadiat be cerkered at engin. ce 2 te non-negallive - 2 G24, js hole i fr pee plane attached along z-aals malin an of 6 with the pecitive oc-axis. “a 7 Gates coreits og adi poids from which a bine « whe on'gin makes an angle # ¢, with positive aus. Deperdi on wheler OL 4E o Beger, tig wsill be a cone opening up or opening clown. (F = 2. ten the cone is flat, and dhe surface ° 2 1s the ~g- plane). Converting CeokDINATEs ? Conversion Formula (487)2> 3.) X=rcesO, Y=rSinG, Z=F C54) > 0583) = [oer ge , tano= % 5 BF (6:4) 268y) | Tepsing, =O > ZB-Peosp (685) 90,0.9) Ps Frage, Br E> banh= F (2,2,4¢) > GJ,Z) ¥=PSinpcosBs Y=PSinp sin®, gx Pea (54,390.06) Padeapage, tane=£ ,coop= 2 bOPALP) vg E aapRy [epht? Find the vedanoutay coordinates of the point | shose eylindrical coordinates are (48 W=(4, 5 Solukion ; 5-9. ———' we have bo find (44) Berl oo Be 2-3 o4oa X= ¥ C066 =4 cos © é 22 g=¥ sine = 4 sin 4d “9 je e Rectangular Coordinale oP given pont is (2, 2485-3). Erample21 Find the rectangular coordinater af tha. potrks iakose spherical coorduin (P, 0; p) are C4, B, z): As Solution? we have 1 find C4, 5): we, Heres Po4, OB, ge X= PEing CoP = Ysin& cos & Sxtecbeeele Je Ising he ees 2 PSINPSING — Y sin& Sin 4. oe Hsing soe <4. 8 = fe e aT COP» Hearse = Qn Reckangular coordinate of given point 's (1%, 1; are), Example 3? Fine) an equalion in Cylindricad) coordinater of the surface whose eqpallion in rectangular cobrdinciles (6 pe 4g avrg : Solukion; s¢=rcosb, fersin®, 7-5 a = ¥+_grdosPt sing. Example! Lind an equation in reelangular coordinates of tha Surface whee eqtalion. in c¥lindrical coorclinates i's re 4oose.Solitlion : Bra act, eer cose, gersing C= Yeose Mulkiplting both sides of equation by B= Uy cos O age = Ya ate ets =O 2 Y oc + 2) Salty > =o (x~ ty te y This ic’ aight circular circle parallel to pws. Example 5: Find an equalion of parabolord g=ahyg> in spherical coordinates. SolvSiOn? In spherical coordinates 2 = PpsinepeosB > gePsinpsing > grPcosd = ot Gan Peas p = Pr sin Geos O + Pbin¢ sin’ "2 pr sin (coset sin*e) = P*sinp vobuick pimplities to Psin*p= cos.