Mth Review Algebr Most of lgebr is pretty strightforwr. One thing tht comes up firly often in physics is the qurtic formul; tht is fining the solutions of the eqution: This hs solution b c + + = b b 4c - - = In prticulr if you re solving n eqution like = there re two solutions =. Powers n eponents come up lot. Some trivil formuls tht you shoul remember re = = / = n The following rules for combining eponentils re lso helpful: n m n m = + n n-m n = n m ( ) m = - n = Very often especilly when working with clculus we get epressions with e rise to vrious powers where e is the bse of the nturl logrithm. The three rules bove pply in prticulr to e : y y ee = e + e e -y = e n y ( e ) y One other useful fct to remember is tht the inverse function of e is the nturl logrithm ln so we hve e = y = ln y One smll item you might or might not hve lerne is hyperbolic functions which re efine by f -f f -f = ( e + e ) f = ( e - e ) cosh f sinh = e nm y n sinhf tnh f = coshf Your clcultor lmost certinly hs these functions built in whether you know it or not. The min ientity tht you nee to know concerning these functions is esily proven: cosh f- sinh f= These functions re closely relte to the trigonometric functions.
Comple Numbers In quntum mechnics we will commonly be working with comple numbers. The strting point for comple numbers is to efine the imginry number i s one of the squre roots of - so i = -. Then we cn efine n rbitrry comple number z s z = + bi where n b re rbitrry rel numbers. The number is clle the rel prt n b is clle the imginry prt of z. The comple conjugte of the comple number z which we will enote s z* is given by z* = bi. The mgnitue of comple number is given by z = + bi º + b In quntum mechnics we most often work with the squre of the mgnitue. It is esy to see tht this cn be rewritten s the number times its comple conjugte so zz* = + bi - bi = - b i = + b = z To fin the mgnitue squre of number therefore it is esiest to simply multiply by the comple conjugte which is just the sme epression with i chnge to i. For emple iq iq - iq iq - iq = = = = e e e e e Aing subtrcting n multiplying comple numbers is strightforwr for emple ( + bi) + ( c+ i) = ( + b) + ( c+ ) i ( + )( + ) = + + + = ( - ) + ( + ) bi c i c bci i bi c b bc i To ivie two comple numbers multiply the numertor n enomintor by the comple conjugte of the enomintor + bi + bi c- i c + bci -i - bi c + b bc- = = = + i c+ i c+ i c- i c + c + c + One importnt ientity tht will prove very useful is iq e = cosq+ isin q We will prove this when we get to it by Tylor epning the left n right sie of this epression. With the help of this ientity you cn esily show how the trigonometric n hyperbolic functions re relte to ech other nmely ( iq) = q sinh( iq) = isin q ( iq) cosh cos tnh = itn q.
Trigonometry In trigonometry we very often nee to fin the ngle n hypotenuse in right tringle from the legs or lterntively given the ngle n hypotenuse we nee to fin the legs. The following formuls c cn help fin these quickly: c = + b b tn q = or b sin q = c cosq = c The three trigonometric reltions re often memorize by the mnemonic SOH-CAH- TOA which mens sine is opposite over hypotenuse cosine is jcent over hypotenuse n tngent is opposite over jcent. From these we cn very esily prove the very useful ientities sin q cos q + = n tn q= sin q cosq For these three bsic trigonometric functions certin vlues come up often enough tht it is helpful to know their vlues. Rther thn memorizing ll of these it is esier to memorize the pttern for sine ( n for n = 3 4) n then memorize tht the cosine is the sme thing bckwrs n tngent is the rtio. In ition if you remember tht sine n tngent re o functions while cosine is even you cn get the vlues for the negtives of ech of these: sin (- ) =- sin cos( - ) = cos n tn ngle sin cos tn 3 p 3 6 3 45 p 4 6 p 3 3 3 9 p - =- tn In ition to the three stnr trigonometric functions bove there re three others: secq = cosq cscq = sin q n cosq cot q = = sin q tn q Some y you shoul memorize these s well but they on t come up s often. We lso - often encounter the inverse functions typiclly written s sin n so on. These cn be evlute with clcultor or for certin specil vlues by using the tble bove; for - p emple sin = 6. In trigonometry there re two stnr wys of mesuring ngles: in egrees or rins. There re 36 in circle n rins. Degrees re most commonly use when you re tlking bout physicl ngle. Rins re lwys use when you re working with clculus. Most clcultors cn be use to clculte using either of these units. Mke sure to check which one your clcultor is set for before you begin clcultion! b
The sum of ngles formul tens to come up lot in physics so let s ly it out for cosine n sine: cos y = cos cos y sin sin y sin y = sin cos y cos sin y I on t epect you to memorize these but I will use them occsionlly when I nee them. From these we cn esily prove the ouble ngle formuls s well: Vectors cos = cos - sin = cos - = - sin sin = sin cos Vectors come up so often in physics tht most physicists lern them in physics clsses not mth. A vector is quntity tht hs both mgnitue n irection like isplcement velocity force or ccelertion. A quntity with mgnitue but no irection is clle sclr. In this clss I will enote vectors by putting them in bol fce ( v ) though when I write it I normlly rw some sort of rrow over it ( v ). Sclrs will be generlly enote by mth itlic font ( s ). In three imensions vector hs three components: ( v vy vz) v = or v= v ˆi+ v ˆj + v k ˆ y z where î ĵ n ˆk re unit vectors in the - y- n z-irections respectively. The little roofs over the symbols re re ht n signify vector of unit length. The length of vector v cn be etermine using the 3D equivlent of the Pythgoren theorem. It is enote by v or just plin v n cn be compute using v º vº v + v + v. y z It is very common in two imensions to iscuss the length v n the ngle of vector v. The ngle is normlly mesure counterclockwise strting from the -is. With the help of the trigonometric formuls bove it is not too ifficult to convert from components to irections n vice vers. For emple suppose we were given the vector v=-ˆi- 3ˆj n ske to compute the mgnitue v n irection of this vector. The mgnitue woul be v = - + - 3 = 3 = 3.66 To fin the irection mke sketch of the vector s shown t right. The ngle t the origin cn be seen to be - = tn 3 = 56.3 The ngle however shoul strt from the +-is n is therefore 8 egrees more thn this for totl of 36.3. 3 y
Aing n subtrcting vectors is pretty strightforwr. To two vectors you simply mke little prllelogrm out of them by copying ech vector n v + w plcing its til on the he of the other vector s sketche t right. In components the vectors cn be e component by component tht is v+ w= + ˆ i+ + ˆ j+ + k ˆ. ( v w ) ( v w ) ( v w ) y y z z Subtrcting vectors in component nottion is similrly esy. Multiplying with vectors is bit more complicte. First of ll you cn multiply (or ivie) vector v by sclr s in strightforwr mnner. Geometriclly sv points in the sme irection s v but is s times longer. In components sv= sv ˆi+ sv ˆj + sv k ˆ y z Such multipliction is commuttive ( s º s r sv = rs v n istributive (( r+ s) v= rv+ sv n s( v+ w) = sv+ sw). Somewht trickier is vector multipliction. It turns out there re two wys to multiply two vectors clle the ot n cross prouct n it is importnt to keep them stright. The ot prouct is written vw (pronounce v ot w ) n prouces sclr quntity n is efine by w vw =vwcosq v v v ) ssocitive ( where is the ngle between the two vectors. In components it is esy to clculte: vw = vw + vw + vw. y y z z The other wy to multiply two vectors is clle cross prouct written v w (pronounce v cross w ) n prouces vector quntity. Like ll vectors it will hve mgnitue n irection. The mgnitue is given by v w =vwsin q The irection is efine to be perpeniculr to both v n w n chosen in ccornce with the right-hn rule. The right-hn rule works s follows: Put your right hn out stright but with your thumb pointe out n mke your fingers point in the irection of the vector v. Now twist your wrist so tht when you strt to curl your fingers your fingers will en up pointing in the irection w. At this point your thumb is pointing in the irection of v w. The only mbiguity occurs when v n w point in the sme irections (prllel) or ectly opposite irections (nti-prllel) but in this cse sin q = n v w =. In the picture bove the cross-prouct points out of the pper. In components the cross prouct cn be compute using the eterminnt: æ ˆ ˆ ˆ ö i j k v w= et v ˆ ˆ ˆ vy v z = vywz- vzwy i+ vzw- vwz j+ vwy-vyw k ç çèw wy wz ø It s messy but occsionlly it is useful. w v
When you combine ot or cross proucts with sclr multipliction or vector ition it is esy to show tht it is still ssocitive n istributive so tht we hve ( sv) w= s( v w) ( sv) w= s( v w) v+ w = v + w v+ w = v + w v+ w = v+ w v+ w = v+ w Also the ot prouct is commuttive. However the cross prouct is nti-commuttive: vw = wv but v w=- w v. Triple vector proucts involving three vectors combine with ot- n cross-proucts re messier n we generlly will not encounter them in this clss.
Differentition Most of physics is written in terms of ifferentil equtions n it is importnt to be ble to tke erivtives of even complicte functions quickly. Fortuntely it turns out tht few rules llow you to tke the erivtive of ny function f() no mtter how complicte. The erivtive of function f() is enote either by f ( ) º f ( ) Higher erivtives cn be enote by similr nottion 3 f º f f º f etc. 3 It is importnt to know the erivtives of few simple functions from which you cn buil up the erivtives of rbitrrily complicte functions. The bsic functions you nee to know to get erivtives of more complicte ones re ( n n re rbitrry rel numbers): n n = n - = e = e ln =. You shoul memorize ll four of these. In ition the following erivtives of trigonometric n inverse trigonometric functions come up lot: sin = cos cos =-sin tn = sec - - - sin = cos =- tn = - - + You shoul memorize t lest the erivtive of sine n cosine. To tke the erivtive of more complicte functions you nee rules for tking the erivtive of sums ifferences proucts n quotients of functions s well s for functions of functions: ( f g) = f g fg æ f ö fg fg = f g+ fg - = ç çèg ø g éf g ù f g g ë û = Keep in min tht in mthemtics only the vrible will be enote by letter like but in physics there will generlly be lot of constnts enote by letters s well. As n illustrtion let s fin the erivtive: é Asin( k + f) ù æ Asin( k + f) ö Asin( k + f) Ne = N e = Ne Asin k + f ê ë úû çè ø Asin( k + f) =- NAe cos( k + f) ( k + f) Asin( k + f) =- NAke cos k + f ( )
Integrtion There re in fct two ifferent types of integrtion clle efinite n inefinite integrtion. If f () is n rbitrry function of then the inefinite integrl F () (sometimes clle n nti-erivtive) is efine to be tht function F whose erivtive is f tht is to sy F = f. It is enote by putting no limits on the integrtion symbol. F = f F ( ) = f ( ). Becuse the erivtive of constnt is zero the inefinite integrl F is efine only up to constnt n hence in proper formlism the nswer to n inefinite integrtion shoul lwys look like F + C where C is n unspecifie constnt of integrtion. Often this constnt cn be ignore. Mke sure you keep the ifferentil in your integrtion; if you ever chnge vribles this fctor cn be importnt! A efinite integrl hs limits of integrtion = n b n is efine s the re uner the curve f() strting from the point = to the point = b. The funmentl theorem of clculus reltes the efinite integrl to the nti-erivtive nmely b b f ( ) = F º Fb -F where F f =. Becuse the efinite integrl involves the ifference of F between the two enpoints the constnt of integrtion C lwys cncels out n is therefore irrelevnt in efinite integrl. In contrst to ifferentition there re no simple rules to perform integrtion. Generlly you o your best to mnipulte your integrtion into reltively simple form then you either immeitely recognize the integrl or you look it up in n integrl tble (or better yet lern how to use Mple to o integrtion for you). Mny integrls cnnot be written in simple close form in which cse moern computers cn numericlly clculte the result often to high ccurcy for most relistic problems. A few rules tht llow you to fin inefinite integrls will help you. If is n rbitrry constnt n f () n g() re functions whose integrls re F() n G() respectively then it is not hr to show tht f ( + ) = F( + ) = = f f F ù = = é ë f g û f g F G Note lso tht it is commonly possible to o integrtion by substitution nmely let be = g y n then substitute this in. However note ny function of new vrible y so tht the ifferentil trnsforms to ( ) becomes = g y = g y y so the new integrl = ( ) ( ) = ( ) f f g y g y f g y g y y.
Integrl Tbles The first four integrls on the left sie n perhps the integrls of cosine n sine shoul be memorize. For more complicte integrls I recommen tht you use Mple to ssist you with the integrl or you cn look it up in the following tble. Note tht ll inefinite integrls hve n implie + C which cn be ignore whenever you re performing efinite integrl. In ll epressions below it is ssume tht b n n re rel non-zero constnts. = n+ n = n¹ n + = ln e = e æ ö e = ç - e ç çè ø æ ö ç 3 e = - + e ç çè ø sin =- cos cos = sin tn =- ln cos sin sin = - 4 sin cos = + 4 tn tn = - ( ) ( + ) ( - ) = ln + + + æ ö = ç > - çè ø æ ö - sin tn - = > + çè ø - = ln > - + = 3/ = + + =- - > - = ln + + =- 3/ + =- > - 3/ æ ö b- = b> > b ç b è ø ( + ) + b ( -) ( + ) sin cos sin cos = sin + cos =- + - sin
Prtil Derivtives n Multiple Integrtions In mthemtics it is common to work with only one vrible which we typiclly cll but in physics it is common to hve t lest three imensions ( y n z) n sometimes four (incluing time t). Hence quntities re commonly functions of severl vribles t once we might write such function s f ( yz ) for emple. When ifferentiting it is then common tht we nee to tke erivtives in more thn one irection n in such cses we nee nottion of prtil erivtives. A prtil erivtive is just like n orinry erivtive ecept we tret every vrible ecept the one we re ifferentiting with respect to s constnt. For emple the prtil erivtive with respect to is written s ( + )- ( ) f hyz f yz f ( y z) º lim h h Just s with orinry erivtives we rrely use this efinition inste just using our orinry rules for ifferentiting. For emple suppose we h the function f yz = Ay- z n we were ske to tke vrious prtil erivtives of it. When tking the erivtive for emple we woul tret y n z s constnts n hence the first term woul yiel erivtive of Ay while the secon term woul yiel nothing since it is constnt. So the three prtil erivtives of this woul be f = Ay f y= A n f z=- Az It is lso very common tht in physics we nee to perform multiple-imensionl integrtion (these will lwys be efinite integrls). In such circumstnces the integrtion shoul be worke from the insie out; tht is you first nee to o the innermost integrl then work your wy out to the outermost integrl. One of the hrest prts of oing such n integrtion is setting it up in the first plce since epening on the shpe of the region you re integrting over it my be very ifficult to figure out the limits of integrtion. Often the limits of the inner integrtion will epen on the vlue of the vrible in the outer integrtion. When oing the innermost integrtions ll of the outer vribles cn be trete s constnts. Let s o n emple to see how this works. Suppose we re fce with y( + + y ). We strt by performing the y integrl (since tht is insie). We tret n both s constnts n hence we hve y= 3 4 3 y + + y = y + y + 3 y = 3 + y=. We then cn esily finish the -integrtion. 4 3 4 3 3 4 4 4 5 4 y( + + y ) = ( 3 + ) = ( 3 + 3 ) = 3 + 3 = 3.
Mny multi-imensionl integrls cn be simplifie gretly when there re symmetries involve. For emple suppose you hve to perform n integrtion over isk of rius R but the integrl is such tht the integrl epens only on the istnce r from the center of the isk. The integrtion cn be performe by iviing the isk into thin nnuli bsiclly slightly thickene circle of rius r n R r thickness r. The thin nnulus cn be thought of s rectngle of length r n with r tht hs been bent into circle n therefore hs re rr. Hence the re ifferentil A cn be replce by rr. For emple to clculte the integrl of r + over isk we woul hve R R A prr p p + + = = r + = R + - r r This metho cn be use to clculte volume integrls for cyliners cones n spheres s well. However for spheres the most common sitution is one where you must perform n integrtion over sphericl volume n the integrl epens gin only on the istnce r from the center of the sphere. In this cse the most efficient wy to o the computtion is in terms of thin sphericl shells hving rius of r n thickness r. The volume of this thin shell is the re of the shell (4 r ) times the thickness so we replce the volume element V with 4 r r. For emple if you re tol tht the totl chrge ensity of sphere of rius R is given by (r) = Ar then the totl chrge of the sphere is R 4p 4p Q= rv = ( Ar ) r r= Ar = AR 5 5 5 R 5 4p