PTF #AB 01 - Def¡nítion of a Limit

The intended heighl(or y -t,atue) of o function, f Q). (Remember thot the function

doesn'l octuolly hove to reoch thot height.)

Written: lim /(¡).t+c

Reod "the limit of f G) qs ,r qpprooches c

Methods for f inding o limit:t. Direct substitution2. Look ot the groph

Some reqsons why o limit would foil to exist:1. The f unclion opprooches o diff erent number f rom the left side thon f rom the right

side.2. The function increqses or decreoses without bound.

3. The function oscillotes between 2 fixed volues.

ø 1. Evoluot e tiry(zx' +l x)

= al)"r1 l.ù

lx+21limI*+-2 y])

3. Use the groph below to f ind thefollowing limits.

H(X)

2. Find the limit (if ¡t exists). v¿-9/\-k-

Y>'2- o

X

l¿n), i x< -LX>'LI a) lim rl(r) =

IèA

o'l

vq

x+ 'Lnü

b) lirnf{(x) = J.n e

o

a

PTF #Aù 02 - One-Sided Limíts

l1l /{xl meons the limit from the right.

lim /(x) meons the limil from the left

A curve hos o limit if ond only if lt+ f (x)=l\/(x) . (Left-hand linit = Right-hand

limít)

A curve is continuous on q closed intervol la,bl if it is continuous on the open

intervol (o,b) qnd lim f (x)=/(a) ond Jïlff"l=f þ). (Linitatfheendpoinfshasfo

mafch the function value af the endpoints.)

;

o

a

Use the groph to the right to f ind thefollowing limits, if they exist.

1. lim g(x) = '5

5. lim s(x)= -5

ó. ìim S(x) : - 5

7. 1gs',). -+

à,n e,8

9

lim¡+-3

lim.r+0

g(x) =

u+-3- 1g(x) -

2

3. lim s(.t) = +

,r1A st'l -- L

rr0

\¿

PTF #AB 03 - Horizontal Asymptotes & Limits at lnfiniÇ

Hori zontol Asymptotes:1. If fQ¡+c oS r-+*-,then y=c isohorizontol osymptote.

2. A horizontol osymptote describes the behqvior ot the for ends of the groph

3. It is helpful to think of on End Behovior Function thot will mimic the given

function (whot will dominote os the .x -values get lorge in both directions?)

Limits ot fnfinity:1. êraphically, o limit ot inf inity will level off ot o certoin volue on one or both ends.

?. Analytically,find on End Behovior Function to model the given function. Then use

direct substitution to "evoluqte" the limit.3. Short Cut:

Top Heovy: limit DNE

Bottom Heovy: limit = 0Eguol: limit = rotio of leoding coefficients

*Pleose be coreful with the shortcut. Some functions oct stronge and reguiresome extro thought. Also, wqtch our for limits of --, they con require extrathought.

Find the horizontol osymptotes ond

evoluote the limits.6. Tf the groph of y =+ hos o

9C+C

horizontol osymptote y- z ondoverticol osymptot? x=-3, then ai c =?ìiwt A -- Y 2 :. a-= 2.

2. lim 2x+5 -- D3xz +I

4

=æ0f

CrX=

åø , xX+C=o

:'(t= 77. For ¡>0, the horizontol line y =2 is

on osymptote for the groph of thefunction /. Which of the followingstotements must 6e lrue?

3

4. lim

lim

Ln¿-

(A)

(B)

(c)(D)

/(0f(xf(2limr+2

)=2)+2 for oll x>0

) is undefined

f(x)=*fQ)=2

5,i

PTF #AB 04 - Vertical Asymptotes & lnfinite Limits

Verticol Asymptotes:L. Tf fG¡-+*- os x-+c*,then x=c tsoverticol osymptote.2. ff o function hos o verticol osymptote, then it is not continuous.3. Vertical osymptotes occur where the denominator = 0, there is no common

foctor, ond the numerotor r 0.

fnfinite Limits:1. êraphically, qn inf inite limit increases/decreoses without bound qt o verticol

osymptote.2. Analytically, direct substitution yields o 0 in the denominotor only, with no

common foctor or indeterminate form.3. Numerically, substitute o decimol number opprooching the limit to see if the y-

vqlues ore qpprooching + or - infinity.

Find the verticol osymptotes ond intervolswhere the function is continuous.

Find lim/(¡) ond l+ff"l. y'A : ¡= |

-. _ x2 +5xy_

blx [x+ s

1. s(x)=+-r+5

Vfr : x'-'507S : (-,,,-5) r, cs,*)

(+-ùÞ*ÐItrz)

X 7 hole,'. f = -L

(- *,-)or,'ù(-L, *)

4

LTr | --T-- r

lìnl P¿x+ l+

-5 0

lirn.tlx+ t- æ

5. r(x)=# vA x-- I

r-) =

+

f)^

2

++-

lìrn 4¿*)X+ ¡+

lim&x)X-+ l-

t æ

t rc

PTF #AB 05 - The "Weir|' Limits

\/To work these problems you need to be oble to visuolize the grophs ond end behovior formost functions.

lim ¿' = "" lim ¿'=0 lim ¿-t = 0 lim ¿-'t = "",r-)æ x-)..€ )l--tæ x-)4

lim ln x--Ø.l-+æ

lim tanx--+rf 2-

lim tan-lE

.f-læ 2

lim ln¡-r-+0+

lim tanx -+x,-)El ¿

lim!=0.r)Ø x

-oo

,T=oê

lim tan-l1t

.r-â..€ 2lim 1=0

x-)4 x

.. Illm - = -oo

¡-+0- .x

.. 1llm -=oox-+0' x

\d,

EVoluote the limit of the "inside" functions f irst, ond then evoluote the "outside" functionqt thot number.

Evoluote the following limits.

1. limetl* eþl co o

(3 lim ln

,r-+æ J*v-t I

æ( x2 +2

-x'-5x-læ

2. lim tan-l.f-+æ

æ4.)-l(

oô¡-+0-

I an' @)fr

?-

t"i

o

z t_

PTF #AB Oó - Continuity at a Point

To prove o function is continuous of o point, c ,you must show the following three itemsore true:

1. /(c) exists2. tjg/(x) exists

equal)3. lry/{x)=f(c) (the function s value is equal to fhe linit af that x-value)

(fhe function has a y-value for the x-value in quesfion)(fhe function has a leff and ríght hand línít and they are

Stote how continuity ís destroyed of x-cfor eoch groph below.

T f - I - a - -r-lttrF-+-l--r-tllrL - t - J......r.-rlll!--!-J--'-rllrrtltr--i--r--r-tttt

i- - i - ì - -i-¿ttttr-f-r--r-3ttt¡ts - ¿ - ¡ - -r-2ttt,l.- .... l. .. L.. -.r... {rrt¡t

rFI

l-r

LI

-r-ì I I

lllr-ts-l-^t I

tttr- L - .¡ - -t- -llltt-Þ+-+-+(1 ,5)r r r- f - ¡ - r- -¡

e7.

lfnr ltx) Arl

X+l

r,v

7

6

5

4

tttt,.-.t-4--l-2ttttL - t - J - -r-1rttt

¡l-.itttt

- t - J - i - -tlttt-L-+-r--ltlrt

II

i - f - I - -t-

r-1-t--t=

f-T-l--l-

a-2-\-r-r--r-l

+1 7

i à å 4

tttt

rats

I

v

2

3

7).{2I

Þ+r7LtJ6

-l

I

..i

JJ

P, r"\-1- lim {¿'¡X+t

+l

+4ùrF

t

t..

I

-f-I

-!-I

!

a

+

.t.

1.

r J 18trt

t'r

4

- r - 1 - .ì- -lII'ILI

I

JJ

1s)

l-r -r--¡--1

1 -l

tll

II

.¡I

-r--f,-ì--l

I i31t2lr-1 i1

ù )'ne'

x+ä+

'- ()

4. Tf the function / is continuous ond if

rQ)=# when x*-2,

tx+Ð t*-z)

3x-7

-xa +3

x2 +9

then f (-2¡=)

x<-2

-2< x<2x>2

is ¿ not continuous?

f¿*) =

+(.ù= E-l5. Let h be defined by the following,

h(x) =

For whot volues of ¡Justify. \¡;r¡ (?x -1)= -17x++li rrn (:xr +z) ,'17x+-zr lim tt

X+ 2+

lim (x+ z-

x'*ù='l b

2 ù= l7+

no+ ulq Ð X=L6. For whot volue of the constont c is thefunction / continuous over oll reolsZ

r@) -cx*l x33cx' -l x >3

livYr (x-> 7-

lirn (¿x"-r

hc+t -- 4o'IZ =bO

¿x+7o+l

), 1o-

J-3

t""t

zr)

PTF #AB 07 - Average Rate of Change

The overoge rote of chonge of f (x) over the intervol [a,b] con be written os ony of the

following:,LydY¡.

^x dx

., f(Ð-f(a)b.

b-a3. Slope of the secont line through the points (o,f (")) ond (a, f (b)) .

* Average rqte of chonge is your good old slope formulo from Algebro f .

1. fn on experiment of populotion ofbocterio, find the ave?ogerote of chonge

from P to Q ond drow in lhe secqnt line.

(45f 340) ì

Q,

(23, 150)

# of doys

h?- Ç: {¿+5) - +1

4s - ¡-z^)340 - l50

LIS- ZV

.9Lo)ÈrJo-oo*

or g. brybtl

q5

Ition /r,

2. An eguotion to model thefree foll of oboll dropped f rom 30 f eet high is

f @) =30 -16x2. Whot is the ove?oge ra'le ofchonge for the first 3 minutes? Stote units,

.P,O. = ltt)- fttù Ív3 -O rnin

: -ll'Ì - 3ô{+3 rni tl

3. Use the toble below to

o) estimote /'(1870)b) interpreÌ the meoning of the volue

you found in port (c)

-{g Ærni 11

t (vr) 1850 18ó0 t870 1880

/(¡) (mitlions) 23.t 31.4 38.ó 50.2

\) 4(tsso)- ftrg .q+(8Ø o - tbbo

+(tsso)- lüf,Jù= l. lltÌ8go - tß-to 0k

4 l,r g¡o) - +(rßt,o) = .17

0lLLt

I,tt -

lBlo - lgt¿O

kte ú chango in mi

el uhe \car t910.

PTF #AB 08 - lnstantaneous Rate olChange

Theinstontoneousroteof chonge,orthederivotive,of f(x) of opointconbewrittenosony of the following:

1.f,(a)=mY.Thisfindsthevolueoftheslopeoftheton9entlineotthespecif ic point x--a.

2. Analyfically, find the difference guotient

.f ,(x) =¡^f (x)-f @) -n^f (x+h)-f (x)

= lim f (x+Lx)-f (x)

.+d x- a ¡;õ h ^r';õ MThis f inds 'lhe generic eguotion for the slope of the tongent line ot ony given point

on the curve.3. Graphically, it is the slope of the tongent line to the curve through the point

(o,f (")).

t. Set up the limit def inition of thederivotiveqt x=z for thefunctionf (x)=-x2 +2x?

4'(y)= -2¡+Lf'tù= -2(z)+7= -L

l'tr)= -L

2. Fill in the blonks:

The lim6(x + n)' -z(x + n\ +t - (o*' - 2x +7)

3. If / ¡s a diff erentiqble function, then

f '@) is given by which of the following?

r@+D- r(a)h

_-Erf. f@+h)- f(x) bx-a

h

(o) r on[(c) I cnd If onþonly(e) f, ff ond flf

fI onlyf ond fff

r)x(f (a)limx--ro

tr-TI.

h+0

f inds the

h

à¿r\v úiv u of

the function Lol,' - LL +1

\JTo f ind the eguotion of c tongent line to o function through o point, you need both o point

ond o slope:1. You moy hove to f ind the y -value of f he point on the groph by plugging in the given

x-value into the original equation.

2. Find the derivotive of / ond evoluota it qt the given point to get the slope of the

tongent line. (Most times you will plug in just the x-value, but sometimes you need

to plug in both the x -value and the y-value. The slope must be o number ond must

not contoin ony vorioble.)3. Use the point qnd the slope to write the eguotion in point-slope form:

! - !ruu, = ^(*- x,unr)

1. Let f 6e the function def ined by

f (x)=4x3 -5x+3. Find the eguotion of thetongent line to the groph of ¡ at the point

where x=-t lt-,) = 4¿_ù?5/_r)+ ?

{{-') - 'l+?*) = tz{-5/ '/-ù, tz -5-- 1

2, Tf the line tongent to the groph of thefunction / ot the point (1,7) posses through

(-2,-Z),then f'(1)=?

hn '2 -1=q v

-2

\"^/

o-

-)=I

,a

,"j

g-1 '-1(x+r)

)+ ' ¿,)

PTF #AB 09 - Tangent Line

3. Find the eguotion of lhe line tongent tothe groph of f G)= xn +Zxz st the point

where f '(x) = 1. You will need to use your

colculotor for this problem.

f '¿*) = tr?* 4 x* D. Llb oY O . L3-7

+(,Lrtà= o.llb

9- 0,1t5= t(X -o,LVb

PTF #AB 10 - Horizontal Tangent Lines

To f ind the point(s) where a function hcs o horízonfal fangent line.1. Find ¡ '(x) ond set it eguol Jo zero. (Remember thot q f raction is zero only if the

numerotor eguols zero.)2. Solve f or x.3. Substitute'lhe volue(s) for x into the originol function lo find the y -value of the

point of tongency.4. Not oll x-values will yield o y-value. ff you connot find o y-value, then thot point

gets thrown out.5. Write the eguotion of your tongent line. Remember thot since it is horizontol, ít will

hqve the eguotion ! = !,atue.

J

ç(

1. Find the point(s), if ony, where'rhefunction hos horizontol tongent lines.

q) f (x)= x3 +2x2 -L5x+14

¿'rJ ' ?y'* 4x -15 = a

(z*'r)(x + l)-' o

X-?2z7

-5- -9'Z,rr 'Ç1-Ð= 5a,)

3' (+) --

b) s(t) =2

It'-)

{

!' 5ö

-lo+' -b2L

L1I o

n0 l,\nt *avqurrt Unes

=-r%t

2. Let h be o function def ined for oll x * 0ond the derivotive of n is given by

u2 -',h'(x)-& ' for oll x*0. Find oll volues ofx

x for which the groph of n hos q horizontoltongent.

h'¿r) =

x Z=o

X

3. If o function / hos o derivotive

f '(x)=6-2sinx for o < x<2n, find the x-coordinotes of the points where'rhefunction hos horizontol tcngent lines.

('t*)' l, ' 7ç¡fi1, = o

sin x -- 12Lt

'tT zj

Y2 -L o

2-

L

ItX,,.. L

!

{= ?t Þ

w

1. Find o lineor opproximotion for f (2.1) if

¡6¡=)z Y:2x

f '¿*) -- -tLvX.

-t2g

Y-, Lz)

9i

u5e

U,tpt

b.- =

Y

-lz\

It")Ç '¿x)

z3

x-

(r. t

,a

2. Evoluote J39 without a colculotor (use

I i neor opproxi motion).

t 'Fx Ð Y= 3lo

( .)

ù+L ^'/ +æ 75l.\J 8

l(z.l) ã t.bs

tt=¿--

x=zto a(tn)' l'L

J ltt (¿

l2- (x-bu)

zrfx

òù

5=,os

f4

t4-1,

o

Í;(*q- 7b) +bx (n-25

,t '11 x u, t-6

PTF #A811 - Linear Approximation

Stondord Lineor Approximotion: qn opproximote volue of o function ot o specified x-coordinote.

To find o lineor opproximotion:1. Write the eguotion of the tongent line of o" nice" x -velue close to the one you wont

?. Plug in your )r-volue into the tongent line ond solve for y .

3. Find o lineor opproximction for f (1.67) iff @)=sinx? v\be t : 1T

?-

QÚ)= T(LnI

f '¿*) = bþx

X TT

Lo

!-t-- o(x

3= I

| (t. at) : t

"r\/l2_./

aao

PTF #A812 - Deriratives of tnverse Functions

1. Find /'(¡).2. lv\oke sure thot you hove f igured out which volue is the ¡ ond y volues for eqch

function (/trl ond /-'(¡))3. Substitute'lhe x-volue for ¡ into /'(x).4. ïhe solution is- the volue you f ound in step #3

1. Tf s is the inverse function of r qnd

F(2)=3, f ind the volue of S'(3) for1

Y-

F(-r) =L+ x-¡.4

F '/x) "

L3x

F' tz) =

1+I l4)+' +t' 4

lt\,,T

?. Let f be the function def ined by

f (x)= x5 +2x-1. If g(¡) = ¡-'1x¡ ond

(1,2) ¡s on f , whot is the volue of s'Q)?

bJ +2-

1

l'¿r),( )+

3'(ù'-

l

1

Þ

:;È

(.ø

t/

Differentiobility meons thot you con find the slope of the tongent line ot thot point or

thot the derivotive exists ot thot point.

1. If o function is differentioble of x= c,'lhen it is continuous of x=c. (Remember

whot is meons to be continuous of o point.)

2. ft is possible for o function to be continuous of x=c ond not differentiqble QÌ x=c.

1. Let f be o function such thot

.. f (2+h\-f (2\ - meavlgr:--4-<;,-,ô I -J' î,tù--5Which of the following musl be true?

r'¡. / is continuous of x=2?,'/lt. f is diff erentiqble ot x=2?

\{J The derivotive of ¡ is

continuous at x=2?

f onþf qnd ff only

(e) II ond fff only

(b) II only(d) I ond lrr onþ

ip

2=7

PTF #AB 13 - Differentiability lmplies Continuity

2. Let f be o function def ined by

lzx-x' ,v<l-fQ)=\ , L

lx"+kx+p x>l'For whot volues of r ond p will f 6e

continuous ond diff erentioble of x = l?

liyvr (z*-'t!) = I

àl- l= l+k+?

tln $i kr+p) c t+ k* FX+ l*

I

livnX+l-

x<lx>l

o= Z+K

(t')=

-R= Y

L- z,l,

2x+ K

Q-zY) =

tc -- 2+R

/im (zx*dX+t+

l-- [+ K+F

/=K

PTF #A814 - Conditions that Destroy Differentiability

Remember for o function to be diff erentioble, the slopes on the right hond side must beeguol to the slopes on the left hond side. The?e o?e four conditions thot destroydtff eren'riobility:

1. Discontinuities in the groph. (Function is not continuous.)2. Corners in the groph. (Left ond right-hond derivotives ore not eguol.)3. Cusps in the groph. (The slopes opprooch r- oh either side of the point.)4. Verticol tongents in the groph. (The slopes opprooch t- oh either side of the point.)

x,Ð

1. The groph shown below hos o verticoltongent ot (2,0) ond horizontol tongents ot(1,-1) ond (3,1). For whot volues of x in theintervol (-2,+) is ¡ not di ioble2

a,+ f, = -2 ,t (enà p"in*+)

2 Let f be o function defined by

lx'+l x<0 -"f(x)={r\"'

l_xr+4 ¡>0â

o) Show thot / islis not continuous ot¡=o' I

lírn (xt+r) = I i yrr (-x2+.1X+O- X+ r)

+

1

)

t+ 4r' linn 4tÐ àvn u

K+o

:' f¿x) io NJ ðr cT 9

^i Y:o

b) Prove thot / islis notdiff erentioble ot x = 0.

eiutcc {1y.) iç hol- cTç

^t I=' ¡ *hen il uY)t)ù+be di++. +h ero.

¡l

Ð

PTF #AB 15 - lrnplÍcit Differentiation

1. Diff erentiote both sides with resp ect ,to x .

2. collect al ! terms on one side ond the others on the other side.ùc

3. Foctor out the Q.ùc

4. sofve fo, ! by dividing by whot's left in the porenthesis.ùe

Errors to wotch out for:

r r _ _ : :":":01 1i tl"j":':t,5 i :.î.1"1 :':'1 - r _ _ _ r _ r _ _3. If x2 + y' - 25 ,whot is the volue "f #ot the point (+,2)?

3x+ tlþ=oMÅi

- Ly- -x(4

7*5xâ= l{xb- tf rtI

Ðl**b - v 9v -1

34"3

b+6 a-55

2. Find the instontoneous rote of chonge ot(t,t) for x+2ry- y' =2.

¿{,â)

dr¡€-fÄX

à'

l'4++

t 3

r+ z¡j) n 2I- ,l*,.= o/-rf \\3/

à xtt,1)

-l -zT

t,,,)

1tþ_3

'3

-qI- Jlo3

LY-75

: -t- zb)z

1-zSj. uh-z

6-25

21

Ð 2(ù - 2(ù

PTF #A816 - Vertical Tangent Lines

To f ind the point(s) where o function hos o vertical fangent lina1. Find ¡'(x) ond set the denominotor eguol to zero. (Remember thot the slope of o

verticol line is undefined 'therefore must hove o ze?o on the bottom.)2. Solve for x.3. Substitutethe volue(s) for x into the originol function to find they -valueof the

point of tongency.4. Not all x-values will yield o y- value. ff you cqnnot find o y-value, then thot point

gets thrown out.

5. Write the eguotion of your tongent line. Remember thot since it is verticol, it willhove the eguotion x= xuotu".

d

,t,=Z-\

o) s(¡) = 3-{Æ

t *) --'L {"" ,7X) den=o 1o $tL w\ue' f'¿x)=rn

*r__o

1. Find the point(s), if ony, where thefunction hos verticol tongent lines. Then

write the eguotion for those tongent lines.

X=o

lz-

2. Consider the function defined byxy' - xty =6. Find the ¡-coordinote ofeoch point on the curve where the

lina is verticol. àen,Çtd=o

*('.j fJ) + 12r,) -

- x>Lt

o

At,iil - få*Lq

2-x u -xr uh À ruh¿n Ay!-y) = o

X(zf -X')= oX=¿r A. 2r-Xz=o

x-

{rÀ*

o

IL

x--

. ?lt$ -

L\4ù1

I -x" 'L-S¿L àen=o -to ,fi"/ vt\¿re. Subst. into ori h

eLual rrn3in al Ilo -D /A

"' f,t' o

L.?-

¡{,.t

v(*(

x5+

_x€

)

)

,_r,

-x5

x¿1r

Ç

),( (t

1-fL

I!t'Y.x"-

{'t*) =

9\Å, -4

Ð

PTF #A817 - Strategies for Finding Lirnits/L,Hospital,s Rule

Steps to evoluoting limits:1. Try direct substitution. (this will work unless you get on indeterminote onswer: 0/O)2. Try L'Hospitol's Rule (toke derivotive of top ond der¡votive of boTtom ond evoluote

ogoin.)

3. Try L'Hopitol's Rule ogoin (os mcny times os needed.)4. use foctoring ond conceling or rotionolizing the numerotor.

Find the following limits if they exist.

1. l,t$(2r'-xz +5)

= zLt\'-tÐ'* t5o

---r_-r_rr-_-r--r-r--r----rrr-----

?

lyn,

Å, a-

LO TT

.oo

xz +5x-63. lim

la

Y7

ttT(6

x)cosx5

'fT

s ¡J-¿\rol z)--

lo

Ð

.r-rl X-llim 2x+5X-rl ,._ :

4. lim I-a¡, (a+o))i-+o x' - a'

oö

Itlxt =X

li rvt

X+o

\itrnX+o

5. Hfft 3r<*>=3x3-5

zx7- 5 - Gs)¿

fiTT

L

Ia'to

9,xY"

I

6. lim¡-+0 Sin¡x+tarx q

D

l+ I

57

1-cosd2stn20

"(+;",{r)t5inÐ

lirnXro

7

*)5(ø

limr-+0

52-

8. lim0+0

lime -ro

ItmÐ -ro

4 :t"e cosë

- t.- (rrrz)

e +o CoSo

*.to

/tt¿¿ se

1

no à¿rl.9. tim?se"e = li, ê

ê*o

b 0

D

PTF #A818 - Related Rates

Set up the reloted rote problem by:

1. Drowing o diogrom ond lobsl.2. Reod the problem ond write "Find = ", "Whe?e = " , ond "6iven = " with the oppropriote

informotion.3. Wrife the Reloting Eguotion and if needed, substitute onother expression to get

dowtr to one vorioble.4. Find the derivotive of both sides of the eguotion with respect to r.5. Substitute the "6iven" ond "When" ond then solve for "Find".

'd

1. The top of a ?5-foot lodder is sliding

down o verticol woll of o constont rote of 3f eet pør minute. When the top of thølodder is7 f eet from the ground, whot is

the rote of chcnge of thE distonce between

the bottom of the ladder ond the woll?

|ittft^.'I,

v

Al' L*

L5

/xT+

-tr.rì

Àt

\tü h¿n '. y ='?

Given '. à: --vT+'

Wtk+."t EL' X?V

1z,

2+

tl

2uù. Dt^í-

+ ltE)=o

L25

T+6min

alrfr

-=,-c^ -Åhrr wrin - Tt

?. An inverted cone hos o height of 9 cmqnd o diometer of ó cm. ft is leoking woterot the rote of 1 cm3/min. Find the rote otwhich the woter level is when h=3

cm (' =I'r') \r= 1

V\,/h¿n, h=?¿n

).b1,?

hr4Y'V

V

V

q

á"(rr?.

'+L',\

t\n

2q h ff

2"1

#= Zth.bLt

Í(f)*/À

l:qqÍ

úLt,

Ð

PTF #AB 19 - position, Spee{, Velocity, Acceleration

1' Position Function: the function thot gives the position (relotive to the origin) of onobject os o function of time.2' Velocitr/ (rnstontoneousì: tells how fost something is 9oin9 of thot exqct instont

ond in which direction (how fcst position is chonging.)3. speed: tells how fost on object is going (not the direction.)4' Accelerotion: tells how guickly the object picks up or loses speed (how fost the

velocity is chonging.)

Position Function: s(r) or .r(r)Velocity Function: v(r) = s'1¡¡Speed Function: speed =lr(r)l

- _ _ -^:":":t:Tï'1': 1"-= 1'Ì: 3 r r - r r r _ _ r -1. A porticle moves olong the x-oxis so thotof time r (in seconds) its position isx(t) = tt - 6t' +9t +ll feet.

o) Whot is the velocity of the pcrticle ot

Ð irÈ)=J¡ri'=The ccceleration of I = 0?

d) Whot is the averdgevelocity on theintervol [t, t] ?

X(a)-x (r) tl- t 5).)

{'t3-l seb

x" lt) = v'tt) =&tü)= b&(oJ =

b) During whot

-â{+Seo

e) Whot is the ave?ageoccelerotion onthe intervol [r, o] ?

v(6)-vta) &Áro_ Is-o

(o -a Sttt vl5

S¿ÓLf) What is the totol distonce troveled

the

6

t

clefrom l=o to t=5?

) q++

) zD+t rÐ

'- rzL * 1v(o) =

v [+)=-T L' l, V

he

Le(+ (t,Ðvtt) < o

I

)

int

?oç nLù '.

lrln? a(t) :

toccelerotion positive? Negotive?

V'(t) . atr) = b+ -t/=ot'2-

.,. [t) +

L+

3

o

I

35

+L++

+ Ç+

-lL 2-ù

?;gh+ Gæ ) r) 7 ù

., 4+

28 ++ -à"+-¿^

PTF #AB20 - lnterrnediate Value TheorernIf these three conditions ore true for o function:

1. / is continuous on the closed intervol [a,b]

?. r@)+ Í(b)3. k is cny number batween f (a) ond Í(b)

Then there is of leost one numbør c in la,bl forwhich f @) -k.

f(b)

k hos be in

*As long os the function is continuous ond the endpoints

don't hove the some y-volue, then the function must tokeon eve?y y-volue between thos¿ of the endpoints.

f(o) you con find o c-volue in

hera thot will give you thot, k-volue:

1. Use the fntermediqt¿ Volue Theorem toshow thot /(¡) = xt +2x-l hos o zero in theintervol [o,r]. ]o

2. Let f @) be o continuous function on theintervol -23 xs2. Use the tqble of volues

below to determine which of the followingstotements must be true.Pb)= -l

&,)'2-I. /(x) tokes on the volue of 5

(A)(B)(c) only

sfI. A zero of /(¡) is belween -2 ond -1

Azeroof f @) is6

f onþIf on[

)

ilIf ond ffI, fI, ond ffl

)c -2 1 0 1 2

f (x) -4 t 6 3 -5

Ð

PTF #AB 21 - Mean Value Theorern & Rolle,s Theorem

lÂeon Volue Theorem:Whot you need: o function thot is continuous ond differentioble on o closed intervolWhotyouget:r,G)=rywherecisonx.volueinthegivenintervo|

Verbally it soys: The instontoneous raÌe of chcnge = average rote of chongaêraphically it soys: The tongent line ís porollel to th¿.".ont l¡n"

Rolle's Theorem (speciol cose of Meon volue Theorem):Whqt vou need: o function thot is continuous ond differentioble on q closed intervolAND the y -volues ot the endpoints to be eguolwhat you oet: r'@)=0 where c is on ¡-volue in the gtvenintervolVerbally it soys: The derivotive eguols ze?o somewhere in the intervolêraphically it soys: There is o horizontol tongent line (max or min)

t---II-I-_____I-

\

Ð

1. Let ¡ be the function given by

f @)= x3 -7x+6. Find the number cthot sotisfies the conclusion of the/\Âeon Volue Theorem for ¡ on [t,:].

I l"l -- 3x' -1itù-þ¿,1

õ-tz2

.€'r*) = b'3y,--1'' to

3v,'-- l?tt - l')/\¿ ,b

ne if Rolle's Theorem opplies. Ifc . Tf not, tell why.

= xo -2x2 for l-z,zl

Fi-) .+f -+x

?. Determiso, find

f (x)

C'Lç ,'àí{q. ,/

Ç(z)= 8= tFù,:'þllc'g thrn

A,gylies

')=4x (x o

0'z

(¿ X= O, l) I

o

¿NJ

3. Let f be o function thot isT5 differentiable on the intervol (r,ro). rf

f Q)= -5 ,"f (5) = 5, ond f Q) =-5 , whichof the following must be true? Choose ollthot opply. çlq) = Rù ?-o\t<tK !t6t-V¿ ) v / hos ot leost 2 zeros.lvT-

(-, of [r, al5

The groph

horizontolof ¡ hos of leqst one

tongent line. Fol\/çForsome c,21c15,thenf (c) =3 ÍvT u'a

5za

oñLY usôto V

lãx

-5

PTF #A822- Extrerna on an lnterr,al

Ertremo: the extreme volues,i.e.the obsolute moximums ond minimums

Extreme Volue Theorem: As long os / is continuous on o closed intervol, then / will

hove both on qbsolute moximum ond on obsolute minimum.

Findíng Extremo on o closed intervol:1. Find the criticol numbers of the function in the specified intervol.

2. Evoluote the function to find the y -volues of oll criticol numbers ond of eoch

endpoint.3. The smollest y -volue is the obsolute minimum ond the lorgest y -volue is the

obsolute moximum.

1. Find the obsolute extremo of eoch

function for the given intervql:

c. fQ)-x2+l on [t,z]

{'¿*ì -- 7x: o cñ i: nrl- in l,Y -O {_--- IL

+tù, Lhtz)-- 5

ubqt¡nax '

abs ,

¡¡¡in I

(

/L

2, ç)

l,L )

b. f @)=x-2cos(x) on lo,znl

{'¿*) = l+ L+ia* '- ogfnY

= LX 1¿*)

-Jt5. zq1

4. ovb

ahç nnin

il,tS r,¡tuy

ur 2rî - ;--- + ,28 V

t-

ínloô

4uab

1'L'(þ

o ,-z)(

( Ær 6'7q 1)

c. f (x)= )e +ezr -t on [o,l]

{',*l= l+ ZeL*=o-l2YAe=

2-N

I

Ie / .L

".' t/v ¿riir..rrl t,,ô: o * ë -l = l_f = O.

_ 3+ e,b-l = â+ tU= 1o 5,421

n¡ü

4¿")

{r+)

!^r¡aX

'Wli n

,o)t v, tos,.4

(o, o)

1. Find the criticol numbers.2. Set up test intervols on o number line.3. Find the sign of f '(x) (the derivarive) for eqch interval.4. Tr f 'çx¡ is positive then /(x) (the original funcfion\ is increosing.

Tf f '1x\ is negotive then /(x) (the original function) is decreosing.-_-__-_

rr-_r-r----r----rrr-r-----r

1. Find the intervols on which the function3.

f (x) - *t -i.*' is increosing ond

decreasing. Justify.

{ i*') = 3xt- 3X : o3*(t- r) = oX:or r +h)

û t

f, , *) b\cÇ'¿ù |, o

2. Let f be o function given by

f (x) - xn + x' -Z . On which intervols is /increosing? Justify.

'Ct¡ = .ty-'r ?¡ = 6àx(ax'-+t)=o

K=ô r x1- (=o

P¿*) +-

r,- I

K *---

I 7o

Ç¿u) ìr¡rsr Lo ,*) becauso-€'¿rl

> o

prF #AB 23 _ Finding lnffea sing/Decreasing tntervals

3. The derivotive, g' , of o function is

continuous ond hos two ze?os. Selectedvolues of I' ore given in the toble below.If the domoin of g is the set of oll reoll1grnþers,then s is decreosing on whichintervol(s)? fncreosing?

x -4 -3 -? -1 0 1 ? 3 4g'(x) 2 3 0 -3 -2 -1 0 3 2

I v\¿r . (- *, -21 v Lz,;)7,L)àrrr

ç

PTF #A824 - Relative Maximums and Minirnurns

First Derivotive Test:1. ff /'(x) chonges f rom + to -, then x is o relotive mox.

?. ff /'(x) chonges from - to +, then x is o relotive min.

Second Derivotive Test:1. If f " (x) is neg (tha function is ccd), then ¡ is q relotive mox.

2. TÍ f ' '(x) is pos (the function is ccu), then x is o relotive min.* x must be o criticol number*

To find the y-volue or the mox/min qnd to see if it is on obsolute mox/min:1. Toke the x -volues ond plug them bock in to the originol eguotion.2. Compore.

#

1. The function def ined by /(x) = x' -3)e'

for oll reol numbers hos o relotivemoximumot x =? Justify.

'¿r)= 4\øl = o4 (x -z-)--o

X--o,z{, + +

7- Y¿l.vnaY @ X' D

3. Whot is the minimum yolue off (x)= xlnx? Justify.

4'¿*) = x (d )* /^,x tr) . ol+-L,wX=Otl^" y = -l

f ',*) rvr in valu,re¿)

hn in

X-t

t¿J-e'(v

F

{ ('t

ohnay

)= (

V lc {,,Ð. ¿ ha,je9+/- +¿¿)

\rt o

-O,Vlôg2. Find the relotive moximum volue for

f @) = (x' -t)"'. Justify.

x'-r)ei + ej þ")nxú (*^þ2*-ùa,

ex x+))úx-r)=o +4. If / hos o criticol number of ¡ = 2 and

f " (x) = 3 , then whot con you conclude

obout f of x=2?(

o

t'l +-3,1 *L -{--L f "¿*l= 3 >o i- Ç¿x) cÒw

'o' là iç & vnìvrinrur,4

¿_fo X *fYlûx

"F 1_

rniYl

7)?

(oø

0,74

u\iX

Yel " ¡nnax valr¡ ¿

x) cÁ3+#

ÐPoints of Inflection¡ Points on the originol function wherethe concovity chonges.

1. Find where v" is zeroor undefined- theseore your possible points of inf lection(PPoIs)

2. Must test intervols to f ind the octuol POIs - they ore only where the secondder iva li ve chonges sign!

1. Write the equction of the line tongentto the curve ! = x3 +3x2 +2 of its pointof inflection.

5' = 3x^+ b\5" -- bx-rb = Q

b (x+l)=o¡ {,1*) +

-r-r-r- -----r--__-- -rr--r-r-------

x

Ð

Pot€¿-,) =

@ K=-r -t

f-r)ot >(-ù'*z-- -, +) tL3+

3(-t\'rbG,)= -3

-z(t,+t)

3' (-,) ;

4T. u-U

PTF #AB 25 - Points of tnflection

2. Given l"(x)=(x-3)(.r *I)' ,find thepoints of inflection of the grcph ofv=f@).

?o+,ìlole, PoI@

{"¿*)

K= 3,- I

-

chan

^9

t3

Pot oNL\l @ x= 7L 4" )tx qe 4

9i3

PTF #A826 - Finding ConcaveUp/Concave Down I ls

1. Find the PPOfs.

2. Set up test intervols on o number line.

3. Find the sign of .f ' '(x) (fhe second derivative)for eoch intervol.4. If f ' '(x) is positive then /(x) (the original function) is concove up (ccul.

Tf f ' '(x) is negotive then /(.r) (the original function) is concove down (ccdl.

s

1. Find the intervols on which

l@)=6(x2 +:)-' is concqve

down. Justify. ê¿r) =

2. Lel f be o function given by

,f (x) = 3x4 -!6x3 +24x2 +48. On whichintervols is / concove down? Justify

{'¿*)= t2x¿-4gxo+tgx

4"(*) = ?(o,AL- albx+,{ß =o

the function

up or concove

(x2+)f '¿*) -- (' 2x\T)

/ r-lLrl-+ x-(x'+ a)

=-lz*( x"+ l)o

{"l*), (xi )

-lz(v

,'u(zr'-gx+.1)=6'^ (4y-z )1x -z )*- t2-

+lz - (t l ("¿*f,= z L

L

Þtt',

*)+ +?3

(*i v)+'lz(x"+ a vz+3- 4x1

(x'+ a)

; - lz (- ?Xt+ 7)

[*a ru

+ ccLf "¿x)

G,ù(o

blc

- tz (-u*"*t)= o-2'2Y

¿_-rxX.X-

Cû u,ocr)

+)=oa4

J

+ +I -r r

(- *, -) u ¿t,*)b/cGr , r) bt. f ,,¿") < o

+ )

+ Lx)

n

"(r) > o#

PTF #AB 27 - "V-Substitution" Rule

1. Let u= inner function.2. Find au, then solve for ax.

3. Substitute u & du into the integrond (it should know fit one of the integrotionrules).

4. Tntegrote.5. Substitute the inner function back for u .

lþ-- gin 3É1

àq' ? c.or 3* ^*

x2 +3x+5\t l2r+tt-l¿xI \ V/\X,r!_ft 4. Intesrote I#-* 1y:- 1ø.n{

àq, = çctx

tanl(-

"fa r¡s À,^ Jr #*à*

J)r ç¿ txàx

l¿uÅo z frÇ=1

(f*3s+

¡

7

L

c,,

i11 +c-.

+ +()

2. Integrote fisin' 3xcos3x)dr E. rnresr at. !fta, = Jå, ( ++3

t C L,3 Jw dtþ¡ \^7

àu-=

l+¿xða* =f )- àw

ú-

f r+ð

2 -!^'l*l r c-Lr¿n= c'Dþ >*àx

3. fntegrote tet'*'dx =

rL'- 3xt I

dq, Zd*

7

+C)

ó. Using the substitution u=2x+l\'e2 t

-\x, J, (Jzt +t)dx is eguol to

(A) IIT,{"* (B) IË"tro"

(D) ffJ; a"

(E) I:J; o"

a

t

3 I

3 ð+u?.l.+

¿)I3

L4

3Y+LI )llì

trÀ"= À1. I

+()

PTF #AB 28 - Approximating Area

Finding o Lefl or Right Riemonn Sum or Tropezoidal Sum:

1. Divide the intervol into the oppropriote subintervols.2. Find th¿ y-volue of the function of each subintervol.

3. Use the formulo for a rectongle (un) or tropezoid (Iut^+tù)to find the oreo of

eoch individuol piece.

4. You must show work to eorn credit on thesel5. Alwoys justify o left or right Riemonn sum os on over or under opproximqtion using

the. fact thot the function is increosing or decreosing.Left Sum 5um

fhcreosingcurve

Þecreasingcurve

-I-III---Ir---I-1. Use oþÍf Riemonn Sum with 4 eguol 2. Volues of o continuous function f (ù are

t

subdivisioß to, Vt= { (t

below. Use o träpezoidol sum withubintervols of eguol lengthto

Y1-' +cpproximote ¡(x) d*( a

(r

L#.A^AI

L) rl

I o+l+{t q)ll un'

o. fs this opproximotion on over orunderestimote? Justify.

undtrcs* blu f¿*) = X' iztnCYUAgi¡1 ïVtr \=o to X= ,1.)o U+V ycc+4^^lee dye drtúrr¿o'r {h¿ C.lLf V8. ,

4,úr

d.{,l ù

,?)

ù

2.L{<dâx Ë IÇa) (u + 5'

LrLq) (5.t + {LGù (l'z+

+I 7 zl-,_na

(

Under opprox'. Over,opprox.

Olter opprox. Under opprox.

x I 1.3 1.6 1.9. 2.2

f (x) 6.0, 5.1 4.3 2.O 0.3

¡ 2,L

J, C*l ¿* 3 1'?u5

'Ð

Ð

o

PTF #AB 29 - Fundamental Theorem of Calculus

If / isocontinuousfunctionon [a,å] and r isonontiderivotive of ¡ onlo,bf ,then

li t¡<al* = F(b)- F(a)

Grcphicolly this meons the signed oreo bounded by x=ct, x=b , ! = l@) , ond the ¡-oxis.

1. Evoluote: f,þ; +6x)dx f:4. Evoluote: ,(t"') dx

X + 3x-lnu

{ L Ò5e 5

à

\nLe-:

r

f,'und,*

5e 5(Ð

L4

2. Evoluote:

tr- coþ Cos o

= -rfz aL1-

lotnlrin*¡a* '- 'Logy,

a

ln

4x

1.,L

¡ 2.+z) Llt-tJ,", I ?)

a

a6 he volues of tc for ,v

x3. Evoluote:

lL= -1y,

ùL-- -'1Åx

lir''* = ì1

.-4(l)Q2

which f *'a*-o?

*{x

,)äx

oß-

J K*tl,

=-a

t:"

0

aK

K

K

,-73t

?I

3

-)L K

t

\e'b

k?- tçliå^t4

,-4(r ++ ) +1 = o

lo75 e

l<= -3

f(: -7

12

PTF #AB 30 - Properties of Def¡nite tntegrals

1. If / is defined ot x=a,then f (/(r))¿r=o

2. If / is inregrobta onfa,bl, rhen Iif¡<.>la;-=-Ï(l@)ùr3. rf / ¡s inregroble, rhen

Il, (¡<.>)*=

.|" (¡trl)tu+ I!, (¡r.ù*

t

1. rf Í'(rt'l)*=then li {ft'l) dx=2

lo4tȿx=

ã

I

))ùt -7 ,

-rl

Y

J

"flo,

ItÀ¿*

"f

.-J +t')3

J,¡-* + ='?

il

'l

I

2. Which, if ony, of the following ore false?ltâ,r"VV

f, U <.>+ s(¡)) * = !' (f (x))dx+ ï kol)*r

rr. f"Uu>r<Ð)ù=ffffi*¡rrr. I!, þt <¡) * = ' I:

(r @)) ¿x

tr 4¿*cC

#,

Ð

PTF #A831 - Aver¿ge Value o( a Function

If / ¡s integroble onla,bf, then the overogevolue from the intervol isr l'( f (x\\dx

b-a Jo\J \e'lw

To find where this height occurs in the intervol:1. Set f @)=orlsw?F (averoge volue).

2. Solve for x.3. Checkto see if tha x-volue in the given intervol.

1. Find the ov"?age volue of l@)=sin¡ 3. The functionover lo,nl.

"+fi,nx = i¡ Cc"sx) ll

: -+ (-'- l)

r(t)-u**,(fi).r,r(#) ruis used to model the velo of o plone in

miles minute. Accordi to t,

whot is the plone for0<r<40?

flav¿ v¿l Wt¡

(

I2. Find the ove?aqe volue ofthe intervol [0, zf ,then f ind where this

volue occurs in the intervol.

y=f^Ifi s¡ {o-o f Stt) d tI

lnrô

o

q= ¡(ã+ I

du = ar-l ¡I¿u.f dx

0t IT

-_---

% *"t4

z-oLJ-z

J.z

ILq

Lol

!I

LI

3

I,l

u,

t2-

Ò

x"*

(t

(

¿ t^',

þlz-IL

ilr7t-r)- L

q,

t

Lô

(

0rv9

L1

L

ôl(

(

+

F

a

))1

v4lt,tC/ n r?+l

f,' t

5.1tt¡ rnifnin

eve

zb(Ð2

p 2 ,ctg1

uJh

ÇzAoß or ¡,tßl

1. For F(x)= Ii ffir,,find(a) F(z) --r:

¿_lv.'

u-¡¡¿' à1ø =

(b) F'(3) - t+€ àt

l+ X-JJ-io

F'Ø) =

2. Evoluor ", ftf,,(e,

+t)at

Ii ë*r);tdxë̂t

-.à

- l'l4 1Ð

l+ ô- =

3x 5eô (Ð

3. Find

f/x)=(sec't)at

f^x

F?x)totL ¿t

1^z

)

O

- zf-to

PTF #A832 - 2nd Fundarnental Theorem of calculus

To find the derivotive of on integrol:

*l|trr,))ar]= r@).dx*Remember thot a must be o constont. If it is not, then you must use your propertiesof integrols to mqke it o constont.

p3x4. Given /(r) = Jo (+-zt)dt and

s(¡)= f (r,), find

(o) ,f '(-t) -E r*l'= ilq - zt=-ül'(-ù= 3("q+o)--

--rr----r---r--r-r---r--rrr-------r-

(b) s(x) in terms of on integrol

3 (rr) = lt"*) =

(c) g'(x)

$'(x) .t,

3eh [4- 4a"

(d) g'(o) : Zf fl - b;JÊ z (-z)= -l ro

(e) Write the eguotion for the tongent line

o)= { (e,)= f(,)àt , ¿|t-1.ÉJ:! 'lz-ft ?

! -) = -b(X-o) #

Ð

PTF #Aù 33 - Extensions of ffc

1. FTC as Accumulotion (fntegroÌe removes the rote!"):

o. Change in Populotion: f {r'fO) dt - P(b)- P(a) (gives totol populotion added

5eÌween tim¿ o ond b)

b. Change in Amount' f {n'<O)at = np¡-R(a) (gives totol amount added of

woter, sond, troffic, etc. between time o ond b)2. FTC os Finol Position ffntegrate to find the end!"):

time, b)

troffic, etc. of o given time, b)

1. A porticle moves olong the y -oxis so

thot v(r)=rsin(f2) for r>0. Giventhot s(Ð

is the position of the porticle ond thots(0) = 3, f ind s(2). V= t 2

, \ îL ^q--

*+¿HgLz)= 3 + \ v¿+)¿¿¿ö

2. A metol A metol of length I cm is heotedof one end. The function Z'(¡) =2x+3 gives

the temperoture, in oC ,of the wire ¡ cm

from the hected end. Find f (r'@¡)dx ond

indicote units of meosure. Exploin themeoning of the temperoture of the wire.ù

7?

ôLç¡nu.dw

fr't").t*-- Êan=(g-+2.ù_

gg.C)

IÒ

(o+ o)

Th¿ *cvnp of *lnc

%€ C ho*tcr ¿-T gth¿ wir¿ th4n qù

wirø isCYrr DYI tO cwt

çLù=.),gate?t3.8 z"l0f

b

PTF #A834 - Accurnulating Rates

Total=lnitiat Amt * !' **rAdded- lu

**r*emoved

foctory@)ui(w)=95 +0.1'wz -w bi

visuolize whot is hoppening in the situotion before you try to put the moth to work.

0 < rv <25 . They .o,@bicycles out ot o

da1àof s(w)l{eo 0sw<3 6ikes/week.\-z 195 3<w325

A cycles otp kes per for

1. How many bicycles ore in thefirslf weeks?

JL

öP(rr¡)¿w ¿ lgg , zu| bi k¿g

prcotv,4,¡þ'¡n 19+ Zuks

pLw)- qof wtd, (u,) - qu/ u"/

o

lo29

ootââe.vvv

lO,"!

Çä.,),lt - lI:10 - );o1o = -l' I'b1¿ö

N are hor¡s¿ âU

2. How mony bicycles ore in the worehouseion if

thp''bJfoctory need to stop production of ony timeduring the f irst 25 weeks? ff so, when?

9sx *rí*1 i{-qox-Lo=o

,.{

/ lEv

E( o- lnintrnu-ùyllo

',Ð kg

+ b0

K-¿-ç'ts

\l

t,&

PTF #A835 - Functions Defined by lntegrals

F(,r) = l)U r,>)¿,

derivqlive groph so onswer occordingly.

the number given (b).

Let f be o function def ined on the closed

intervol [o,z]. The groph of f , consisting of

four line segments, is shown below. Let g 6e

the function given by s(") = f' f e)dt.

ql /x) "' {tx)

3. Find the x -coordinote of eoch point ofinf lection of the groph of s on theintervol 0< x< 7. Justify your onswer.

4'r*l= lI\,= L,+t5

g" = {'cNo

*)+ 0t o¿-

4

3¡

I

C;O

7 IsV+, 4 l*4 4f. x=I ix=s" (*) ^l',;x) ís unñ a\ ,913

4. Let h(x) -

'2x+8

$t

?r+,

(

-tGraph r¡¡ = 3

l. Find g(3) , g'(3), ond g "(3) .

I, rrro,-!.'. Find olt

ta¿¿t= irù(++à-_

criticol volues for h(x) ond clqssify themos q minimum, moximum or neither. h'tx)t ¿h'¿Ð = +/*) - ?zx = o +¡) -T= u

{z*)= ?x ¡loNFg(z) =

g't+)'=

3" (+) -

b{¿a -vu(ù=oX2. Find

the intervol o < x<3 .

3- .{)

3-o 3

onI

?Y= oX

h't*)Lô

)7hA ,\-^ 1

g(o)= [þtat= 'Sl+u>Lu''1

@

lò

2__L

roÌe of c of

/ \ 2/ \ / 7

/ z \D \

?

y-_? h¿x) lnasanÀ chancq+

x'o , h[x) Iu,ç a

PTF #AB 36 - Solvin g Differential Equations

1. Seporote the voriobles (usuolly worth 1 point on o free response guestion).

2. fntegrote both sides, putting "C" on the side with the dependent vorioble (found on

the botlom of the differentiol). (If there is no "C", you lose oll points for this porton a f ree response guestion.)

3. Tf 'there is on initiql condition,gel to o point where it is eosy lo substitute in theinitiol condition ond then solve for "C".

4. Use the "C" you found ond then continue to solve for f (x) (if needed.)

1. Fínd o solution y = f (x)to the

diff erentiol eguotio n 9={ ,oti, fyingu)e e'

z(.a(

,Â

Ç

I

2âúIIt')

lr,.t urq

J_=

t-i,J

t-Lj

2

U--2vdui:^

'2_

Y\

J' x^/ X

{+ü e (r¡.

Ia./.

_(,^

Jn

!tI/7tL-

4

4

=

I

'?lç,; (-

^ 2u)

(L:

tu(,':

.: C,

'r,L¡2u

,,1,',lO r

2. rf

x7+c-i.n ( zx7r¿.)

*=tr'ond if ! =-! when x=r ,

I

I

then when x=2 , y =?.ZÀi,(

ttl2Xlo

.J- - z+ú-l-l :(t

-'- irr - I

- l -'zx___t _

I '2-v

(.r ,\

1,1:,)I

I

I'-'il

I t¡))

3. Find y = f @) by solving lhe diff erentiol

eguotion *= r'(e -zx) with initiolax

Icondition f(3)=

.")

(u - r*)ax

---- bx - \n + (,

-i : tg-¿l r L,-lV '= (-

',2-x +bx -t7L

X -bx'tt,

L1

t-A.)

I

4:

Y

.,'2.x -bx.t t )ù(

bx - y,"- l?

PTF #AB 37 - Slope Fields

1. Substitute ordered poirs into the derivotive to compute slope volues of those points.2. Construct short line segments on the dots to opproximote the slope volues.3. For o porticulor solution, sketch in the curve using the initiql condition ond guided by

the tongent lines.

1. Consider the differentiol eguotiond'

- *'l v-l).t \r tux

c¡. On the oxes provided, sketch o slope

field for the given differentioleguotion at the twelve pointsindicoted.

c. Find the porticular solution y= f (x)

to the given diff erentiol eguotionwith the initiol condition .f (0) = 3 .

DI

Xt.{ x.J

,;)-tn' Iq

.Cr- U

J-r- t,,J -, I =

t'b= 3X+U

t)'2tr +(_,

/

/l+

/

/l

Ð_

¡

L

þnL-t-

b. Describe oll points in the xy-plonefor which tha slopes one positive.

B-

U -1"-

tiY,

/t

t

,

(,

i'*o

xq+

þ(-'l ) >0

>lX+o U-l>ô

e\ (Lv'o

uuh ¿rr Xf o ,rwrt 1> I

9\rt ptøàt:te

PTF #A838 - Exponential Crowth & Decay

x

* k is colled the constont of voriotion ond must be found in eqch problem by using theinitiol conditions.

dt

t. ff ø= ky ond r is q non-zero constont,dt

then y could be

2. The number of bocterio in q culture is

growing qt q rote of 3000¿2'/5 per unit oftime r . AÌ t = 0, the number of bacteriapresent wos 7,500. find the numberpresentot r=5.(B) 2e* (C) eo'+3(A) 2ekry

(D) kty+5 (E) Lw,2'

I+-2

4,þ )

2¿+

L

(0,(s,

u

de

15@

?)

C üKt

booù ø

C

Zç

dL

(v\ ôiG)

r,J .: a

+-=-L Þ

9Lrdu

= drc

tl=

I

2-bt4 +L

)r1

7t

('r\"t'b0ùo

Vooo / E\(;)

( 0Teot - tzoo(t)å + c7ÇoD=7500 +u

c

1,LL

= 7b,$Ls

\¿(,5)= vooo

Vlùn'rooez d 55,4[1 ,

55,4tJ "

lzo4z"l

PTF #AB 39 - Particle Motion 5u mmary

A porticle moves olong lhe x-oxis withvelocity of time r > 0 given by v(r) = -I+ et-. .

Attime t=o,s=2. þrù

L'2-

v(¿)=-l+¿u-

3. Find oll vqlues of t for which theporticle chonges direction. Justifyyour cnswa7., _t

11{t)= -trl o1. Find ø(3) , y(3) ond s(3) t-t

q(þ)= - ¿'Aþ) = - e-'

(-'

\\) tr l+a ")*L=l

L-ìl= 0

¡¿ì +

ya o clto¡9 c\it'¿ c)-,",t alblL V¿¿) o dnÀ clnctnqrg

9t n9nd the displocement ond totol

distonce of the porficle over the timeintervol o<r<3.

Þirp\, l"Lr)¿r =lE

+,

9 (a) =

-..)

e(o) t t-+nlH"ve+Jw

9 P,) = 'zFr'2¿

Jo Gtn ø

U..FLt- ')¿t

¿)u=--l

-t

r your onswertncr V¿

| +e,-?-l-

ot time t =3? Give a ùì./,reoson fo

rS lpeelv&) ctt =

6lrrr¡9t læ qavne 4tt) i aLt)

')n

tQ í¡OY sin cu Y[1)i ,apce

ú\t?-)

l"ßr )6Y 1"8-t'{

-"9(d üY_.ab6

x -.115

v(a).a,L1) '-

-,8tnl- ,lò6j

LLYe t4fl¡, hr3rrHn'9

PTF #AB 40 - Area Between 2 Cuwes

If / ond g aîe continuouson [a,b] ond s(r)</(x) bounded bytheverticol lines x--a ond

x=b, then the qreo between'îhe curves is found by

e= to çf {r)- s(x))dx

To f ind the qreq of c region:L. Sketch or drow the grophs.

2. Determinewhether you need dx or dy (going verticolly or horizontolly)

3. Find the limils from the boundories, oxes or intersections.4. Set up the integrol by Top-Bottom if ax or Right-Left if dy.

5. Tnlegrate ond avoluote the integrol.

,l (1,4

1. Find the oreo ofgucdront thot is

!=x3 +8 ond

)

n the f irstthe grophs

xu= lrx

7'a= L.Åx

.Lclu , AX

2e-L

A= f [rx rs)- /rt*! a,{)

= lu'6-x7 )¿x

-g

I12l= tx-4L

+tI

)-tz+

A.fo #,,1Y

2.

zÇoe" Àn

I(¡

- (o) :

2. The qreo of the region bounded by thelines .r =o , x=2 ond y=0 qnd the curve

y = e,12 is

-. z

e=O

X--'¿

A

x ,L;¿tn .{ (,

0

ó

3. Find the oreq of R, th¿ region in the firstguodront enclosed by the grophs off (x)=1+sin(2x) ond s(x) =e't'.(calculafor)

,"13 Ø

[a¡ - s(-,-] ¿*-føó Þ

r..,

"421

PTF #AB 41- Volurnes of Slabs (Cross Sections)

Volume = I^ rea

Volume of Slobs (Cross Sectionsl:

certoin shope.

The y-oxís)

Eguiloterot Triongle i A=f IRectonglet A=s(height)

fsos. Rt. Tri (on hyp.) , o=Ir'

Semicircle: A=!sz8

Sguore: A= s2

fsos. Rt. Tri (on leg): O=I,

L Let n 6e'¡he region ín the f irst

guodront under y -+ for +< x19.,lx

n¡n¿ tn"@ of thesolid whose bose

?. Find the volume of the sold whose bose isenclosed by *' + y' =1 ond whose cross

sections token perpendiculor to the bose

r,__ {T-_çA -- i* $rs)-

is the reg ion R qnd whosecut by plones r tosluore9.

to

x-ox a?e

x

rtru|= g*9= T-b

,g l'J ¿r

"'81I

-- itr(r -x) !=V11' )n (t-x),,Ix

=

Atx)¿lv

f .',[,-r),]x

the circte (x+t)2 +yz =9 ond whose

cross-sections hove orec formulo givenby

r

V

1

1

I

t,J"'^'

ITIz

3. Find the volume of o hose bose is

ta,F*(É -o)'¿*

+dY -2x.)

VA(x

Id= ¡rX

¡u= r ly

) = sin(zx

Atx)/x1

+

Ly, rly-

1zkl-L* q - -L- 1 X

2.

)

tr

ÍÍ, Lo9-Try

X I-+ I

semicircles

'ff

Ll-( tr ù

4

PTF #AB 42 - Volumes of RotatÍons (DÍscs & Washers)

Volume of Disks:

line of rototion.eh

Volume of Wosher:

ogoinst the line of rototion.el

)]f (bl -- ALt

o,"'\^

r(

1. Find the volume of the solid generoted

v--

3. Find the volume of the solid generotedby revolvin! x=.F with y=3 cnd

t the y -oxi\x=0 'L

: = l+r/J K- I =t+, R+- Y

X-l

) ) Lvì

" (løx -32-u: - 1z+ !--rr tnl

\

n l"'(o-oÞx

-,b^l,nl )

L2. the region enclosed

line x= 3 , ond the curve y =.Æ ia-f'@oboutvolume of the

T

the ¡ -oxis. Whot is thesolid gene roted? dX

Gø)-'ï (u, ,a")I:= "þ, Ð É,___!

dy

È3 diEl'

V

7

,l o'"ix

" ('t rt\r H *'t)xdY-

g TT