PG. Clculus AB Bile (nd most importnt ook in the world) (Written nd compiled y Doug Grhm) Topic Limits Continuity 6 Derivtive y Definition 7 8 Derivtive Formuls Relted Rtes Properties of Derivtives Applictions of Derivtives Optimiztion Prolems 7 Integrls/Sustitution 7 Properties of Logrithms 8 Newton's method 8 Seprting vriles 8 Averge Vlue Continuity/Differentiility Prolem Rectiliner Motion Men Vlue Theorem Studying the grph of f ( ) Trigonometric Identities Growth, Doule-Life, nd Hlf-Life formuls Applictions of the Integrl ( Are, Volume, nd Sums) 6 7 Approimting Are ( Trpezoidl Rule, Left Endpoint, Right Endpoint, nd Midpoint) 8 st Fundmentl Theorem of Clculus 8 nd Fundmentl Theorem of Clculus 8 Integrl s n ccumultor Finding Derivtives nd Integrls given grph of f () Integrtion y Prts
LIMITS ** When evluting its, we re checking round the point tht we re pproching, NOT t the point. **Every time we find it, we need to check from the left nd the right hnd side (Only if there is BREAK t tht point). **Breking Points re points on the grph tht re undefined or where the grph is split into pieces. Breking Points : ) Asymptotes (when the denomintor equls ) ) Rdicls (when the rdicl equls ) ) Holes (when the numertor nd denomintor equls ) ) Piece-wise functions (the # where the grph is split) + = right hnd it f f ( ) = left hnd it **If left nd right hnd its DISAGREE, then the it Does Not Eist (DNE) t tht point. **If left nd right hnd its AGREE, then the it eists t tht point s tht vlue. **Even if you cn plug in the vlue, the it might not eist t tht point. It might not eist from the left or right side or the two sides will not gree. For emple: f = for for < f = DNE ecuse + = nd f f ( ) = Note : In generl when doing its, # = # = # = LIMITS LIMITS AT NON - BREAKING POINTS Very esy. Just plug in the # E# : + = E# : + 7 = = E# : + = HOLES IN THE GRAPH E# : + Fctor nd cncel or multiply y the conjugte nd cncel, then plug in # ( ) + = = + = 7 E# : + + = + + ( + + ) ( + + ) = + ( + ) + + = 6
RADICALS ( You must first check tht the it eists on the side(s) you re checking) If # mkes rdicl negtive, the it will not eist t tht #. When we check t the reking point (the # tht mkes the rdicl zero) there re two possile nswers: ) if the it works from the side tht you re checking. ) DNE if the it does not work from the side tht you re checking. E # : E # : E # : + = Since the it eists from the left t we cn plug in, then = = Since the it does not eist from the right t, then + + = DNE ecuse = DNE + = + = DNE ( oth sides don't gree ). + ASYMPTOTES (# ) (Since the point DNE we hve to check point tht is close on the side we re pproching) There re three possile nswers when checking t the reking point (the # tht mkes ottom = zero) ) If we get positive nswer the it pproches ) If we get negtive nswer the it pproches ) DNE If we get positive nswer on one side nd negtive nswer on the other side, then the it DNE E # : E # : E # : E # : = Check. which gives positive nswer so = 7 = Check 8. which gives negtive nswer nd 7. which gives positive nswer 8 + 8 so 8 7 + 8 = DNE ecuse the two sides do not gree. = Check. which gives negtive nswer so + + = ecuse ( ) ( ) = + + + = = oth sides gree ( ). TRIG. FUNCTIONS FACTS : E# : sin sin = sin tn = = sin tn cos cos = = tn = tn = = = E# : sin 8 = 8 = 8 E# : 6sin cos 6 sin cos = = 6 = 6 E# : π tn = =
< f ( ) = + < E # : f E # : + PIECE - WISE FUNCTIONS The reking points re nd. = (Check in > ) E # : f + = 6 (Check in < ) E # : E # : f f These net three its re not t reking points, so we just plug in the numers. E #7 : 7 + = DNE (Both sides don't gree) E #6 : = E #8 : f f Check the powers of the numertor nd denomintor. = - = 8 E # : f ( ) = (Check in > ) f ( ) = (Check in < ) = (Both sides gree) f LIMITS THAT APPROACH INFINITY ) If the denomintor (ottom) is igger power the it =. ) If the numertor (top) is igger power the it = or -. ) If powers re the sme the it = E# : + = E# : + = = + = coefficient of the highest power of numertor coefficient of the highest power of denomintor E# : = E# : = E# : 6 E#6 : + 7 = E#7 : + 7 7 = = E#8 : = FINDING VERTICAL ASYMPTOTES AND HOLES A verticl symptote is the # tht mkes only the denomintor =. A hole occurs t the points tht mke the numertor nd denomintor = t the sme time. f ( ) = 8 f ( ) = + + f ( ) = 7 + = ( + ) ( ) ( ) ( + ) ( ) ( + 7) vert.sym. hole vert.sym. hole vert.sym. hole vert.sym. holes = 8 none none (, 8) none none = 7 (, ), f
Continuity Continuous functions hve no reks in them. Discontinuous functions hve reks in them (Asymptotes or Holes / Open Circles). ** To check for continuity t, you must check left hnd its s well s the vlue of the function t tht point f nd right hnd its f + f ( ). If ll three re equl then the function is continuous t. If f = f ( ) + f ( ) then the function is continuous t. If f is not equl to either one - sided it, then the function is not continuous (discontinuous) t. Continuous t Discontinuous t Derivtive t ll points f ( ) = h f(+h) f ( ) f + h h l Derivtive y Definition Derivtive t the point, f f = h f f + h h f() f(+h) f() l f(+h)=f() l +h +h Line l is secnt line h Line l is secnt line h Line l is tngent line f ( ) slope of secnt line l = f + h + h f ( ) f + h mens tht the distnce h is pproching nd h h the points get closer to ech other nd the two points ecome the sme point nd line l is now tngent line. The derivtive of function finds the slope of the tngent line!
E # : f = = Find f ( ) Use f ( ) = f + h h h h = h + 6h + h h f ( ) f + h h = h 6h + h h = h 6 + h = 6 E # : f = = Find f Use f = f + h h h h = h + 8h + h + h h f f + h h = h 8 + h + h = 8 tngent line to curve The derivtive finds the slope of the tngent line. The norml line is perpendiculr to the tngent line. E # : f = f norml line to curve = Find eqution of the tngent line nd norml line t =. + h h h = h + h + h h Eqution of Line (point - slope form) : y y = m = nd f = f Eqution of the tngent line : y = ( ) Eqution of the norml line : y = = h + h = ( slope of the tngent line) Questions from AP Test When you see these prolems, you need to tke derivtive of the given eqution. sin( + h) sin E# : = cos E# : = h h h h Eqution: sin Derivtive: cos Eqution: Derivtive: 6 ( + h) When you see these prolems, you need to tke derivtive of the given eqution nd plug in #. ( + h) E# : = 6 E# : h h h Eqution: Eqution: ( + h) = h Derivtive: Derivtive t = : 6 Derivtive: Derivtive t = :
Derivtive Formuls *Power Rule y = n y = n n E# : y = y = E# : y = y = y = y = *Product Rule y = f () g() y = f ()g() + g () f () E # : y = sin y = ( ) ( sin ) + ( cos ) *Quotient Rule y = f () g() y = f ()g() g () f () g() E # : y = sin y = cos sin = 6 *Chin Rule y = f () n OR y = f g( ) y = n( f ()) n f () y = f g( ) g = sin + cos cos sin E # : y = ( +) y = ( +) = 6( +) *Implicit Differentition- function in terms of s nd y s must write dy E# : y + y + = derivtive y + dy dy d + y d + = dy dy d + y d = y dy d + y *Trig. Functions d dy Now solve for d. = y everytime you tke deriv. of y dy y = d + y ( Tke the derivtive of the trig. function times the derivtive of the ngle) Function Derivtive Emple sin cos cos sin tn sec csc csc cot sec sec tn cot csc d d cos d sin( ) = cos( ) = cos( ) d = cos d csc d sin( ) = 8 cos( )sin d tn ( ) d = ( sec ) = sec ( ) = csc d d d cot d sec sin cot( ) = csc( )cot = sec sin = csc tn ( sin ) cos = csc 7
*Nturl Log E# : y = ln( +) y = y = ln( f ()) y = f () f () + = + E# : y = ln(sin ) y = cos = cot sin E# : y = log chnge of se y = ln ln y = ln = ln *Constnt Vrile y = f () y = f () f () ln ( steps: itself, derivtive of eponent, ln of se) E# : y = E# : y = E# : y = e y = ln y = ln y = e ln e = e *Vrile Vrile ( tke ln of oth sides) ln y = g( )ln f ( ) y = f () g() dy y d = g ( )ln f + f ( ) f ( ) g( ) dy d = f ( ( ) )g g E# : y = sin tke ln of oth sides then tke derivtive lny = sin ln dy y d = cos ln + sin *Vrile Vrile y = f () g() ( lternte wy) dy d = sin ln f + f ( ) f ( ) g( ) cos ln + sin Need to chnge y = f ( ) g( ) to y = e ln f ()g() y = e g()ln f () then tke derivtive. E# : y = sin sin ln y = e y = e sin ln cos ln + sin = sin cos ln + sin *Inverse Trig. Functions y = rcsin f ( ) y = rctn f ( ) y = rcsec f ( ) y = f ( ) f ( ) y = + f ( ) f ( ) y = f f ( ) E# : y = rcsin E# : y = rctn E# : y = rcsece y = y = 8 y = + 6 6 y = y = 6 8 + y = 6 e e f e e ( ) 8
Relted Rtes We tke derivtives with respect to t which llows us to find velocity. Here is how you tke derivtive with respect to t: derivtive of is d, derivtive of y is y dy, derivtive of r dr is r, derivtive of t is t = t dv V mens volume ; mens rte of chnge of volume (how fst the volume is chnging) dr r mens rdius ; mens rte of chnge of rdius (how fst the rdius is chnging) d is how fst is chnging; dy is how fst y is chnging Volume of sphere Surfce Are of sphere Are of circle Circumference of circle V = πr A = πr A = πr C = πr dv = πr dr da = 8πr dr da = πr dr dc = π dr Volume of cylinder Volume of cone V = πr h V = πr h use r h = r is not vrile in cylinder ecuse its' vlue is lwys the sme Due to similr tringles, the rtio of the rdius to the height is lwys the sme. Replce r or h depending on wht you re looking for. E # : The rdius of sphericl lloon is incresing t the rte of ft / min. How fst is the surfce re of the lloon chnging when the rdius is ft.? A = πr da dr = 8πr da da = 8π = 6π ft /min. The surfce re of the lloon is incresing t 6π ft /min. E # : Wter is poured into cylinder with rdius t the rte of in / s. How fst is the height of the wter chnging when the height is 6 in? V = πr h r = V = πh dv = π dh = π dh dh = π in / s The height is incresing t.7 in / s.
E # : Wter is leking out of cone with dimeter inches nd height inches t the rte of 7 in / s. How fst is the height of the wter chnging when the height is in? r h = r = h V = πr h V = π h dv = dh 8 πh 7 = 8 π dh The height is decresing t.8877 in / s h V = πh dh = 67 6π in / s E # : A 7 foot ldder is lening ginst the wll of house. The se of the ldder is pulled wy t ft. per second. ) How fst is the ldder sliding down the wll when the se of the ldder is 8 ft. from the wll? + y = 7 y = when = 8 d dy dy + y = 8 + = dy = 8 ft. per second. ) How fst is the re of the tringle formed chnging t this time? A = y da = d y + dy da = + 8 8 da = 6 ft. per second. θ 7 y c) How fst is the ngle etween the ottom of the ldder nd the floor chnging t this time? sinθ = y cosθ dθ 7 = 7 dy dθ = rdins per second E # : 8 7 dθ = 7 8 A person 6 ft. tll wlks directly wy from streetlight tht is feet ove the ground. The person is wlking wy from the light t constnt rte of feet per second. ) At wht rte, in feet per second, is the length of the shdow chnging? d = ( speed of the mn wlking ) dy Use similr tringles: + y = 6 y =? ( speed of the length of shdow ) y = 6 + 6y 7y = 6 y = 6 7 dy = 6 7 d dy = 6 7 dy = feet per second 7 ) At wht rte, in feet per second, is the tip of the shdow chnging? Tip of shdow is + y, so speed of tip is d + dy = + 7 = 6 7 feet per second 6 y
Properties of Derivtives Derivtive is rte of chnge; it finds the chnge in y over the chnge in, dy, which is slope. d st derivtive m. nd min., incresing nd decresing, slope of the tngent line to the curve, nd velocity. nd derivtive inflection points, concvity, nd ccelertion. E # : Slope of the tngent line to the curve Given f ( ) = Find eqution of the tngent line nd norml line t =. = f ( ) = 6 = 8 f = f f Eqution of Line (point - slope form) : y y = m Eqution of the tngent line : y 8 = Eqution of the norml line : y 8 = incresing : slopes of tngent lines re positive f decresing : slopes of tngent lines re negtive f Properties of First Derivtive ( ) > ( ) < ( = ) f ( ) = mimum point : Slopes switch from positive to negtive. found y setting f minimum point : Slopes switch from negtive to positive. found y setting concve up : slopes of tngent lines re incresing. f concve down : slopes of tngent lines re decresing. f Properties of Second Derivtive ( ) > ( ) < ( = ) inflection points : points where the grph switches concvity. found y setting f slopes of tngent line switch from incresing to decresing or vice vers. incresing/concve down slopes re positive nd decresing incresing/concve up slopes re positive nd incresing decresing/concve up slopes re negtive nd incresing decresing/concve down slopes re negtive nd decresing M M M is Mimum; slopes switch from positive to negtive m m is Minimum; slopes switch from negtive to positive I is n inflection point; slopes switch from decresing to incresing I I m, I m M is Mimum; m is minimum; I is n inflection point
Appliction of Derivtives To find rel. m., rel. min., where the grph is incresing nd decresing, we set the first derivtive = E# : y = 6 + ) Plug #'s in ech intervl on the numer line into first derivtive. y = 6 6 6 f ( ) > mens grph is incresing on tht intervl. = 6( 6) f ( ) < mens grph is decresing on tht intervl. = 6( )( + ) f ( ) switches from + to, then point is reltive mimum. = nd = f ( ) switches from to +, then point is reltive minimum. ) To find the Y vlue of m. nd min. plug into originl eqution. + - f'() + - m min incresing decresing (, 6) (, 7), (, ) (, ) To find inflection points, where the grph is concve up nd concve down, we set the second derivtive = y = 6 = 6( ) = ) Plug #'s in ech intervl on the numer line into second derivtive. f ( ) > mens grph is concve up on tht intervl. f ( ) < mens grph is concve down on tht intervl. An inflection point occurs t the points where f ( ) switches from + to or from to +. ) To find the Y vlue of inf. pt. plug into originl eqution. f''() - +! inflection pt. concve down concve up,,, M - - - - - - - I -6-8 m
Optimiztion Prolems ) Drw nd lel picture. ) Write eqution sed on fct given nd write eqution for wht you need to mimize or minimize. ) Plug in fct eqution into the eqution you wnt to mimize or minimize. ) Tke derivtive nd set equl to zero. ) Find remining informtion. E# : An open o of mimum volume is to e mde from squre piece of mteril, 8 inches on side, y cutting equl squres from the corners nd turning up the sides. How much should you cut off from the corners? Wht is the mimum volume of your o? 8 8 8-8 8 8- V = (8 ) V = + = V = 7 + V = + 7 Cutting off mkes no sense (minimum o). = We need to cut off inches to hve o with = mimum volume. The mimum volume is V = )( =, = V = (8 ) V = in E# : A frmer plns to fence rectngulr psture djcent to river. The frmer hs 8 feet of fence in which to enclose the psture. Wht dimensions should e used so tht the enclosed re will e mimum? Wht is the mimum Are? P = y + A = y A = 8 y = A = 8 = y + A = ( 8 y) y y = A = 88 ft = 8 y A = 8y y = 8 E# : = A frmer plns to fence two equl rectngulr psture djcent to river. The frmer hs feet of fence in which to enclose the pstures. Wht dimensions should e used so tht the enclosed re will e mimum? Wht is the mimum Are? = + y A = y A = 6y = A = y = A = 6 y y y = A = ft 6 y = A = y y = 6 = Y Y RIVER RIVER Y Y Y
E# : A crte, open t the top, hs verticl sides, squre ottom nd volume of ft. Wht dimensions give us minimum surfce re? Wht is the surfce re? V = y A = + y A = = A = + = y A = + A = = A = ft = y A = + = = y = y = Y E# : A rectngle is ounded y the -is nd the semicircle y = 8. Wht length nd wih should the rectngle hve so tht its re is mimum? y = 8 A = y A = 8 A = 8 A = 8 + ( 8 ) ( 8 ) ( 8 ) 8 A = A = 6 = ( 8 ) y = 8 6 = y = = ( Bowtie) ( ) A = A = 8 Y= 8- - Y
*Integrl of constnt Integrtion Formuls d = + C E# : d = + C E# : π d = π + C *Polynomils n d = n+ n + + C E# : + 8 d = + + C *Frctions ( Bring up denomintor, then tke integrl) E# : *Constnt Vrile E# : E# : d d = d = e d = *Trig Functions d = ln + C + C = + C ( steps : itself, divided y deriv. of eponent, divided y ln of se ) ln + C E# : d = e ln e + C = e + C E# : e d = ( Alwys divide y derivtive of the ngle) ln + C e ln e + C = e + C sin d = cos + C cos d = sin + C tn d = ln cos + C csc d = ln csc + cot + C sec d = ln sec + tn + C cot d = ln sin + C E# : cos d = sin + C E# : sin 6 d = cos6 + C 6 E# : tn d = ln cos + C E# : sec 7 d = ln sec 7 + tn 7 + C 7 E# : csc d = ln csc + cot + C E#6 : cot d = ln sin + C f ( ) *Nturl Log : d = ln f ( ) + C top is the derivtive of the ottom f E# : E# : d = ln + C E# : sin cos d = ln cos + C E# : d = ln + C d + = d = + ln ( +) + C *Integrl ( top is higher or sme power thn ottom) ( Must divide ottom eqution into top eqution). E# : + d Long Division = + + + d = + ln + + C E# : + d = + d = + ln + C
*Inverse Trig Functions d = rcsin + C d = + rctn + C d = rcsec + C Find vrile v nd constnt. The top MUST e the derivtive of the vrile v. E# : E# : E# : d = rcsin + C E# : d = v = = v = = d = 6 + d = rctn + C E# : d = +6 d = +6 v = = v = = = d = rcsec + C E#6 : 6 d = 6 v = = v = 7 = = 6 7 7 rcsec 7 + C rcsin + C rctn + C rctn + C d 6
*Sustitution When integrting we usully let u = the prt in the prenthesis, the prt under the rdicl, the denomintor, the eponent, or the ngle of the trig. function. E# : + d = E# : u = + du = d ( +) d = d ( u) du = u + C = ( +) + C d ( + ) d = + = 6 6 + = u du 6 = 6 u + C u = + du = 6 d = 8 + E# : + d = u u du = (u u )du = u u + C E# : u = + = u = ( + ) ( + ) + C du = d cos d = u = du = d E# : + d = u du = u cos d = cosu du = sinu + C = sin + C = = u = + You must switch everything from to u. Including the #'s. du = 6 d u = + = u = + = Properties of Logrithms + C logrithmic form eponentil form : y = ln e y = Log Lws : y = ln y = ln ln + ln y = ln y ln ln y = ln y ln 8 = ln = ln ln + ln = ln ln 7 ln = ln 7 Chnge of Bse Lw : y = log y = ln ln Fct : You cn t tke ln/log of negtive # or zero. We use logrithms to solve ny prolem tht hs vrile in the eponent. E# : e = ln e = ln ln e = ln = E# : ln = e ln = e = e 7 Memorize : ln e = ln = log = log = ln
*Newton s Method ( Used to pproimte the zero of the function) c f ( c) where c is the pproimtion for the zero. f ( c) E# : If Newton s method is used to pproimte the rel root of + =, then first pproimtion = would led to third pproimtion of = f () = + f () = + Plug in to find f () f () = or.7 = Plug in to find f / f / = 86 or.686 = DIFFERENTIAL EQUATIONS ( Seprting Vriles) (used when you re given the derivtive nd you need to find the originl eqution. We seprte the 's nd y's nd tke the integrl). E# : Find the generl solution given dy d = y dy d = y y dy = d y dy = d E# : Find the prticulr solution y = f () for E# given (, ) y = + C y = + C y = + C = 8 + C 7 = C y = + 7 y = + 7 E# : Find the prticulr solution y = f () given dy d dy dy = 6 d y = 6y nd (, ) = 6 d y ln y = + C y = e +C y = C e = C y = e If the rte of growth of something is proportionl to itself ( y = ky), then it is the growth formul. Proof : y = ky dy dy = ky y = k e ln y = e kt +C y = e kt e C y = C e kt dy = k y ln y = kt + C *Averge Vlue (use this when you re sked to find the verge of nything) f () d E# : Find the verge vlue of f () = from [, ] Avg. vlue = d = = 6 = 7 = 7 8
*Continuity / Differentiility Prolem E# : f ( ) = At f ( ) =, < = 6, 6 = At f ( ) = f ( ) = f ( ) =. Therefore f () is continuous. + f () is continuous iff oth hlves of the function hve the sme nswer t the reking point. =, < f 6, At oth hlves of the derivtive = 6. Therefore the function is differentile. Since oth sides of f At f () = = 6 6 = 6 f () is differentile if nd only if the derivtive of oth hlves of the function hve the sme nswer t the reking point. nd f ( ) gree t, then f ( ) is continuous nd differentile t =. *Rectiliner Motion (Position, Velocity, Accelertion Prolems) -We designte position s ( t), y( t),or s( t). -The derivtive of position, ( t) = v( t) velocity. -The derivtive of velocity, v ( t) = ( t) ccelertion. -We often tlk out position, velocity, nd ccelertion when we re discussing prticles moving long the -is or y-is. -A prticle is t rest or is chnging direction when v t -A prticle is moving to the right or up when v t -To find the verge velocity of prticle =. > nd to the left or down when v( t) <. v(t) -To find the mimum or minimum ccelertion of prticle set t =, then check the vlues on numer line to see if nd how they switch signs. -Speed is the solute vlue of velocity, v t If v( t) nd ( t) gree If v( t) nd ( t) disgree -Distnce trveled is v( t) v( t) ( t) + + v t ( t) + +. speed is incresing. speed is decresing.
*Men-Vlue Theorem (Only pplies if the function is continuous nd differentile) f() Slope of tngent line = slope of line etween two points f (c) = f () f () According to the Men Vlue Theorem, there must e numer c etween nd tht the slope of the tngent line t c is the sme s the slope etween points (, f ()) nd (, f ()). The slope of secnt line from nd is the sme s slope of tngent line through c. f() f() f() *When given the grph of f (), it is like you re looking t # line. This is the grph of f (). Where f () = (-int) is where the possile m. nd min. re. Signs re sed on if the grph is ove incresing or elow the - is (decresing). c E : Grph of f incresing when f decresing when f m. occurs when f min. occurs when f ( ) from 6 6 ( ) > ( ) < ( ) switches from + to. ( ) switches from to +. The f ( ) is the slope of the tngent line of f ( ). concve up when slope is positive. concve down when slope is negtive. inf. pts. occur when slopes switch from + to or to +. f'() -6 - - + - + + -6-6 rel.min rel. m incresing decresing =, = (, ) (,6] [ 6, ) (, ) + - + - + f''() -6-6 inf. pts. concve up concve down =,,, ( 6, ) (, ) (, 6) (, ) (, ) 6 - - - - - - - - - Grph of f '() 6
*Doule - Life Formul y = C () t d *Hlf - Life Formul y = C *Growth Formul y = C e kt ( Comes from y = ky ) y = ending mount, C= intil mount, t = time, d = doule-life time, h = hlf-life time, k = growth constnt *Trigonometric Identities *Reciprocl Identities Identities *Doule Angle Formuls *Hlf Angle Formuls cosθ = secθ sinθ = cscθ *Pythgoren Identities sinθ cosθ = tnθ sin = sin cos cos sin = cosθ sinθ = cotθ cos = cos sin cos + cos = sin + cos = + tn = sec + cot = csc t h
*Are Applictions of Integrls (Are,Volume, Sums) A = [top eqution-ottom eqution] d f() f() g() g() h() c c Are from to c Are from to c c A = [ f () g()] d + [g() f ()] d A = [ f () h()] d + [g() h()] d Find Are of y = sin from to π From to π, sin is on top of -is (y = ) sin π π c From π to π, -is (y = ) is on top of sin sin π π Are = sin d + sin d = cos + cos π π = + = π π *Volume V = A( ) d where A( ) is the Are of the cross section. Volume if cross section rotted is circle( A = π r ) Volume if cross section rotted is cylinder ( A = πrh) ( ottom function) V = π top function d V = π top function ottom function [ ] d rdius rdius height *Volume rotted out : Verticl cross section the - is the y - is ( g() ) [ ] π f () d π f () g() d the line y = k the line = c ( g() + k) π f () + k d π + c the line y = m the line = d ( m f ()) π m g() d π d [ f () g() ] [ f () g() ] d d y=m =-c y=-k f() g() =d
*Volume rotted out : ( Verticl cross section) the - is the y - is π d π d = 8π 8. = 8π. the line y = the line = π ( + ) + d π + d 7π = 7. = π.66 the line y = the line = 6 π d π 6 = 6π d 87.7 = 6π 7. y = y = = = 6 y = *Volume rotted out : ( Horizontl cross section) the - is the y - is π y y dy π y = 8π 8. = 8π. the line y = the line = π ( y + ) y dy π y + dy + 7π = 7. = π.66 the line y = the line = 6 π ( y) y dy π 6 = 6π 6 y 87.7 = 6π 7. dy dy For horizontl cross sections we must switch everything from to y. y to to y = = ± y
*Volume (Region is not rotted) V = A( ) is the Are of the cross section. d where A - Sometimes we will find the volume of regions tht hve different cross sections (not circle or cylinder). - These regions re not rotted ut come out t us. - We must first find the Are of the cross section, then tke it's integrl. E# : Let R e the region in the first qudrnt elow f nd ove g( ) from = to =. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the -is re : Squres( A = s ) f ( ) g( ) d f ( ) g( ) V = f ( ) g( ) Equilterl Δ's A = s f() g() f ( ) g( ) Semicircle A = πr V = ( f ( ) g( ) ) d V = ( f ( ) g( ) ) d f ( ) g( ) = dimeter V = f ( ) g Rectngle with h = A = h f ( ) g( ) f d π f ( ) g V = π ( f ( ) g( ) ) 8 d = rdius ( g( ) ) V = f ( ) g( ) Regulr Hegon A = p f ( g( ) )d ( g( ) ) V = f d f ( ) g( ) V = tn6 f ( ) g ( ) ( 6( f ( ) g( ) )) d V = ( f ( ) g( ) ) d tn6 f ( ) g( ) = pothem 6 f ( g( ) ) = perimeter
E# : Let R e the region in the first qudrnt under the grph of y = for. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the -is re : Squres A = s V = Equilterl Δ's A = s d = / / 6 R 7 8 d = ln = ln ln = ln = ln.8 Semicircle A = πr V = d = d = ln = ln. = dimeter V = π d = = rdius Rectngle with h = A = h V = Regulr Hegon A = p V = d = 6 d = = pothem 6 = perimeter Multipliers for other figures : π d = π 8 8 d = π 8 ln = π ln.8 d = d = ln = ln. d = d = ln = ln.7-6-( SL): -6-( LL): -6-( HYP) : 8 --( LEG): --( HYP): Regulr Octgon : tn 67. or tn π 8
Approimting Are We pproimte Are using rectngles (left, right, nd midpoint) nd trpezoids. *Riemnn Sums ) Left edge Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = + + + Totl Are=.7 8 ) Right edge Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = + + + Totl Are = 6.7 8 c) Midpoint Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = 7 6 + 6 + 6 + 6 6 Totl Are = 8.6 d d) Actul Are = + = + = =.667 *Trpezoidl Rule (used to pproimte re under curve, using trpezoids). Are f ( ) + f ( ) + f ( ) + f ( )... f ( n ) + f ( n ) n where n is the numer of sudivisions. [ ] - - - E# : f () = + Approimte the re under the curve from [, ] using the trpezoidl rule with sudivisions. A = f () + f + f () + f + f () = + + () + + 76 = = 76 6 = =.7 All you re doing is finding the re of the trpezoids nd dding them together! - 6
*Approimting Are when given only ( no eqution given) To estimte the re of plot of lnd, surveyor tkes severl mesurements. The mesurements re tken every feet for the ft. long plot of lnd, where y represents the distnce cross the lnd t ech ft. increment. 6 7 y 8 6 7 6 6 6 6 7 67 ) Estimte using Trpezoidl Rule ) Estimte using Midpoint sudivisions A 8 f + f ( ) +...+ f ( ) + f A [ ] A [ 6 + 6 + 6 + 7 ] A 8 +6 + + + +8 + +8 + 67 A 78. A 78 f ( ) + f ( ) + f ( 7) + f ( ) c) Estimte Avg. vlue using Trpezoidl Rule d) Wht re you finding in prt c? 6.7 The verge distnce cross the lnd. Avg.Vlue 78. e) Estimte using Left Endpoint f) Estimte using Right Endpoint A f + f ( ) + f +...+ f ( ) 8 A 8 A 8 + 6 + 7 + 6 + 6 + 6 + 6+ 7 f ( ) + f + f ( )...+ f [ ] A [ 6 + 7 + 6 + 6 + 6 + 6+ 7 + 67] A 778 A 7 *Approimting Are when given only ( no eqution given) Unequl sudivisions: You must find ech Are seprtely. y ) Estimte using Trpezoids A = ( + )h A ( + ) + ( +) + ( + ) ) Estimte using Left Endpoint ( A = wih left height) A + + Trpezoids shown c) Estimte using Right Endpoint A = wih right height A + + 7
st Fundmentl Theorem of Clculus Just plug in the top # minus the ottom #. E# : d = = 8 = 8 E# : π π sin d = cos π π = nd Fundmentl Theorem of Clculus (When tking the derivtive of n integrl) = + Plug in the vrile on top times its derivtive minus plug in the vrile on ottom times its derivtive. E# : E# : d d d d d d f t = f t = = E# : t + = + + = + + Integrl s n ccumultor A definite integrl finds the chnge in the eqution ove it. The integrl of velocity from to is the chnge in position (distnce trvelled) from to. The integrl of ccelertion from to is the chnge in velocity from time to time. The integrl of f () is the chnge in f (). d d t = 6 = 7 E# : If f = then find f. f ' Since f () d = f f it finds the chnge in f from to. Since the re under f from to = this will help find f. f = f + f () d = + = 7 E# : If f = then: f Integrls going left re negtive. Integrls going right re positive. = + f ( ) d f - = + f ( ) d E# : Given v v( ) = v + ( t) -. = +. =. f = = 6 f 8 = 8 nd ( t) = sin t + - - find v. 8 - - - - = + f ( ) d = + f ( ) d 8 Grph of f'() π 6 = + = 7 8 = + π = 6 + π = 8 +.6 = 7.766 (The integrl of ccelertion finds the chnge in velocity)
Finding Derivtives nd Integrls given grph of f() - - - 6 Grph of f() 7 8 The derivtive is the slope of ech line. = f ( 7) = DNE = DNE f ( 8) = = f ( ) = DNE = f (.) = f f f f The integrl finds the totl re etween f() nd the - is. f d = 7 + = 8 f d = f d = 7 + + = f d = = 8 ( f ( ) + ) d = 68 f ( ) d = 8 - - - 6 Grph of f() 8 6 7 8 8 7 6 8 7 6 7-6 7 8 velocity of runner in meters per second - 6 7 8 velocity of runner in meters per second E# : Find the velocity of the runner t t = nd t = 7 seconds. v = = v( 7) = Find the ccelertion of the runner t t = nd t = 7 seconds Since v ( t) = ( t), you find ccelertion y finding the derivtive (slope) of velocity. = ( 7) = Find the distnce trvelled y the runner from t = nd t = seconds E# : Given () = find () nd (). () = + v(t) = + = 6 () = + v(t) = + 8 = Distnce trvelled = v t v t = 8
*Integrtion y Prts (used when tking n integrl of product nd the products hve nothing to do with ech other) Alwys pick the function whose derivtive goes wy to e u. There re two specil cses. Cse : When ln is in the prolem it must e u. Cse : When neither eqution goes wy, either eqution cn e u (the eqution we pick s u must e u oth times) nd we perform int. y prts twice nd dd to other side. f ( ) g ( )d = f ( )g g( ) f ( ) du more simply u dv = uv v du E# : e d = e e d Cse : E# : ln d = = e e + C = u = dv = e d u = ln dv = d ln d ln + C du = d v = e du = v = *Tulr method E# : cos d = sin + cos sin + C Deriv. Integrl + cos sin + cos sin *Specil cse E# : Deriv. d = Integrl + ln ( ln ) ln ( ln ) + C ( neither function's derivtive goes wy so we use integrtion y prts twice nd dd integrl to the other side) st time u = e du = e d nd time u = e du = e d dv = sin v = cos dv = cos v = sin E# : e sin d = e + e cos d = e cos + e sin e sin d e sin d = e cos + e sin e sin d = e cos + e sin + C