rhe* v.tt 2.1 The Tangent and Velocity Problems Ex: When you jump off a swing, where do you go?
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1 2.1 The Tangent and Velocity Problems Ex: When you jump off a swing, where do you go? lf± # is.t *t, Ex: Can you approximate this line with another nearby? How would you get a better approximation? rhe* v.tt
2 Ex: A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after 5 minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. t (minutes) Heartbeats The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient s heart rate after 42 minutes using the secant line between the points with the given values of t. a) t 36 to t 42 m b) t 38 to t 42 m FIT any c) t 40 to t #f 4236 secant 530 ' beat 71.75k per minute min. m an 29* bpm d) t 42 to t 44 man What are your conclusions? l3 66 bpm Notice in this example that we calculated slopes, which gave us " # Change in total heartbeats So, the slope told us about the rate of change of the graph. Away ) change in time Heart rate
3 . Units: Position: Velocity: m meters ft feet cm centimeters ftp.ft/m..n.kmls ± Acceleration: ftlsr Average Velocity: Instantaneous Velocity: Ex: If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by y 10 t 1.86 t 2. Md F. s.si Img a) Find the average velocity over the given time intervals: i) [1, 2] ii) [1, 1.5] iii) [1, 1.1] m/s2 %, 9.8%2 Velocity calculated between two points. The velocity at a given point 1 : r 2 : s y CH y 10 (2) tf ftp * %. e ~ aatz.ly?t5ji2o6#5.3sm1s iv) [1, 1.01] u± I ie EE :O:* o 3 # to } at ) v) [1, 1.001] 4 AI % foot ioo At b) Estimate the instantaneous velocity when t 1. or 6.28 Mls 6.3 or 6.29
4 Given: 1) Any letter can change into an A. 2) Two Cs in a row can turn into a B. 3) Any word can be doubled. Any word can be # 4 octupled Ex: C C A # I C C # 3 Fatma ACAC ##}, cc# 3 A C # 1 ACAC # 1 A CAC # 3 o # o QED Ex: CAC CACIAC # 3 CABAC # 2 CAAAC CAAAC 0 # 1 Ex: C CC # 3 c C c C # 3 C CCC Cccc # 3 CCACACCC BACABC # 1 # 2
5 The Limit of a Function Ex: Graph this function: & ' ' ( &)* ' 2 1 " YH < 1) What is f(1)? 1 2) What is f(3)? 9 3) What is f(2)? Undefined 4) What would you guess f(2) should be? 5) Fill in this chart: 4 x f(x) x f(x)
6 The left side of the table tells us that lim &(') 4 0 ( 2. which is read as The lefthand limit of f(x) as x approaches 2 is equal to 4 or The limit of f(x) as x approaches 2 from the left is 4 The right side of the table tells us that lim &(') 4 0 ( 5 which is read as The righthand limit of f(x) as x approaches 2 is equal to 4 or The limit of f(x) as x approaches 2 from the right is 4 Note that x 6 7 means from the left and not necessarily negative numbers. The same is true of x 6 8 In this case, the lefthand limit and righthand limit are equal, so we can say that lim &(') 4 0 ( time limit The limit as approaches 2 from the left In general, 9:; >(<) L if and only if 9:; >(<) L and 9:; >(<) L < < 2 < 5 Example 7, page 92: ( 2,3 ) ( 5,2 ) G., ) ( 5,1) # a) lim 0 ( 2? ' 3 d) lim? ' 2 2 b) lim 0 ( 5? ' e) lim 5? ' I 2 Does c) lim? ' Not f) lim? ' 0 ( Exist DNE z g( 5) /
7 . Infinite Limits Ex: Find lim B 0 A 0 C if it exists. Approach 1: Table of values (Why can we use ±?) x ± 1 ± 0.5 ± 0.2 ± 0.1 ± 0.01 ± ' ( ,000 1,000,000 Approach 2: Sketch the graph M Hmmm it looks like f(x) keeps getting bigger and bigger as x approaches 0 from the left and the right. In this case we say: 1 lim 0 A ' ( Nole : Some books say this means the limit DNE Definition: Let f be a function defined on both sides of a, except possibly at a itself. Then lim & ' means that the values of f(x) can be made arbitrarily large (as large as we please) by taking x sufficiently close to a, but not equal to a. Note: Some texts phrase this as the limit does not exist. This can cause confusion, since that also refers to the case when the left and right hand limits do not agree.
8 . Ex: Find lim 0 0 K 2 07K, lim 0 0 K 5 07K 0 and lim 07K t.sn#siyf5oxlfxeeye.8.gn! Try 1 Tables Graphs Y 4 Not to scale y % e * t.gg#.oolinx.yao 4t 335,Io 7 s 3.9?,to 39 If?oIt ±
9 2.3 Calculating Limits Using the Limit Laws Limit Laws: Suppose that c is a constant and the limits lim &(') and lim?(') exist. Then 1) lim [ & ' +?(')] lim &(') + lim?(') 2) lim [ &(')?(')] lim &(') lim?(') 3) lim [P & ' ] P lim &(') 4) lim [ & '?(')] lim &(') lim?(') 5) lim[ R 0 ] W X TUV R(0) S(0) TUV W X S(0) if lim?(') 0 Ex: Use Law #4 to show that: : Slept Prove for nl lim [ & ' ]n [lim & ' ] n xjmaffh 'D [ lxigafh'd 't ' lxisma GAD 'ti mafg) fnjnafcx ) [lizfgd Step 2 : Assume for nk (6 Power Law) " ' E po0fnts True " be, Assume lxjna [ FHDK [ fszfcx ) ]k True step 3. Prove for nktl tiga [fc Dk " I [ lxignafcx ) ]k+ 1
10 Step I : Prove for n 0 or n I Step 2 : Assume for n K Step 3 : Prove for n Kt I ftp.#xdxwcxdffxbimaffadklismafcx)hgggit.fxiyafcxdkt '
11 Additional Laws: 7) lim ' c 8) lim ( a 9) lim ( Y + Y [ [ 10) lim ( + [ (As long as + is defined.) [ 11) lim (() [ lim (() (As long as [ lim (() is defined.) Ex: Evaluate these limits, justifying each step. 1) lim (2( ( 3( + 4) 2) lim (0C 7\08K (07\ lxig2x2 ljg3xtdjms4 2[ Laws 1+221law <w6 ;gxt 3jgx+ hjm4 3*3 +1*4 32*3 3 ' 3 ]t3( 3) + fight Low 2[ 3) 2 3( 3) +4 Law?
12 ninexlijafcxhfca I ) at D f (a) x xtgma.f( xk maf( )DNE lxignafcx ) # FG) sina.tk#xgma+fhs
13 lim (2)( 3) + 4): 1) lim (2) ( 3) + 4 ) lim (2) ( 3)) + lim 4 Law #1 2) lim (2) ( 3)) + lim 4 lim 2) ( lim (3)) + lim 4 Law #2 3) lim 2) ( lim (3)) + lim 4 lim 2) ( lim 3 lim ) + lim 4 Law #4 4) lim 2) ( lim 3 lim ) + lim 4 2 lim ) ( lim 3 lim ) + lim 4 Law #3 5) 2 lim ) ( lim 3 lim ) + lim 4 2 [lim )] ( lim 3 lim ) + lim 4 Law #6 6) 2 [lim )] ( lim 3 lim ) + lim 4 2(3) 2 3* Law #7, Law #8, Arithmetic 2) ( 3) + 4 lim 2) 3 1) lim (0C 7\08K (07\ TUV W a ((0C 7\08K) TUV W a ((07\) Law #5, plus lim (2) 3) 0 W a 2) TUV((0C 7\08K) TUV W a ((07\) B\ TUV W a ((07\) (Problem above) 3) 4) B\ B\ TUV W a ((07\) TUV (0 7TUV W a B\ B\ TUV (0 7TUV W a W a \ ( TUV W a 0 7TUV W a \ Law #2 W a \ Law #3 5) B\ B\ B\ ( TUV W a 0 7TUV W a \ ( \7\ \ Law #7, Law #8, Arithmetic
14 0 Ex: Find lim C 7K 0 7( 08( step I : Plugin 8 ' n' die " x t tiazaas 224 \8b Ex: lim C 7c b A b \ I 1 8 ooo / his thing tin h o 9+6 h+h 6h + hi T hc6t thing ( 6th ) 6 Ex: lim 0 B ) + 1 EF ) 1 G EF ) 1 *a eei iy OneSided Analysis Remember that: IJK N(L) L if and only if IJK N(L) L and IJK N(L) L L M L M 2 L M 5 Ex: Show that lim 0 A ) { if xso if < O
15 lim 1 1 : xia ' '' :o) 1 lixgfx Ktisnofxto or w±x Io + 0 jygolxl 0
16 e 0 Ex: Prove that lim 0 A 0 tin. tsmot does not exist. its mo 1 in CD o i Has txisno ' 1 1 Ex: If f(x) 2 ) +, ) < 2 ) +, ) 2, determine whether lim,()) exists. 0 ( figs f G) xhjz. tax 2 a F for 0 time fkk dig 2 cnn.fm Greatest Integer Function [[x]] The largest integer that is less than or equal to x. Ex: Same fig ffx ) D N E [[5]] 5 z [[2]] 3 %ni I g o_0 [[ 3]]? Eyes [ [ ] ] 3
17 Theorem: If f(x) < g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then lim,()) < lim 7()) Squeeze Theorem: If f(x) < g(x) < h(x) when x is near a (except possibly at a) and then lim,()) lim h()) L lim 7()) L Ex: Show that lim 0 A ) ( 9+: B 0 0. x 2 < ) ( 9+: B 0 < x2
18 Summary of Techniques So Far: A(?) 1) When in doubt, look at the graph. 2) If the function is defined at a, try substituting a for x. 3) If the function is not defined at a, try expanding and/or factoring to cancel out problem factors. 4) If you suspect that the limit might be infinite, look at a sign analysis/sign chart. 5) If you suspect that the limit might not exist, consider a onesided analysis. 6) If you can find g(x) > f(x) > h(x), use the Squeeze Theorem.
19 2.4 The Precise Definition of a Limit (DeltaEpsilon Proofs) What does x gets really close to a actually mean? The distance between x and a gets smaller and smaller. How do we express the distance between x and a? We use the Greek letter B (CDEFG) to express an upper bound for this distance: x a < B nm What does f(x) approaches a limit actually mean? The distance between f(x) and the limit, L, gets smaller and smaller. How do we express the distance between f(x) and L? We use the Greek letter H (epsilon) to express an upper bound for this distance: f(x) I < H ftp.ex In fact, if the limit actually exists, we should be able to get arbitrarily close to L as x gets closer to a. So we start with a value for H, and then solve to find a B that works. Translation: Step 1: Pick any value for H that you d like, or use one that someone else gives you. (It has to be positive, and usually it s between 0 and 1 often very close to 0.) Step 2: Set up the inequality L H < f(x) < L + H Step 3: Solve to find the interval of x values that will satisfy the inequality. Step 4: Use this interval to come up with an appropriate B.
20 L < toragivn. type py xa value. y :L E, netted Ex: For f(x) x + 2, a 1, and L 3 find a B that corresponds to H 0.01 " Now it s possible that H 0.01 was a special case, so what we really want to do is to look at all possible values of H, not just specific ones. Ex: f(x) x + 2, a 1, and L 3, find a B that corresponds to all values of H we want to find J sit value. Ex: Prove that lim 3) 6 (for all values of H) 0 ( D E results 2 : Write out lygafcxkl Step We need to find /ka( or < S st 1 21<0. if.lx akd step } :D#h then. IFCH LKE Make the Step 1 Start with : look like, o <?x,? I%ff.is#IoIE43sa.iHaomot limfcx )L 3 Cxz, 14(0*2.99) x a < lxjm,fc ) 3 p of E, if / 1 +2 { < 1 11 E lfcx ) akd then lfcx ). 3/< E xt<e nyis5k ofstpl E STPZ LKE wengeedtfndo 1 11<8 gets <2 31 4<2 Let 1 21<5 o us there xts
21 1 51 I iiis y" k ± ah
22 0<1 0 Infinite Limits Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then lim ' ( means that for every positive number M there is a positive number, such that if 0 < x a <, then f(x) > M (Note that an infinite limit technically means the limit doesn t exist, just in a particular way.) Ex: Prove that lim (0 C 0 A We want to find or st. if 0 < 1 4<5 then f ( ) > M B Step I : We want f ( ) > M 2 > M Step 2 : We need to find of st. Step 3 : 0 / < Dea# 2 > M 1 1<5 < or ( P " 6) s mai*ij# a We should choose f Ff Fm Tm
23 1/2<4 If 0 O < XEO > * 2 < 2 If 1 1<2 0
24 2.5 Continuity A function f is continuous at a number a if lim ' ( '(+) This means three things: 1) f(a) is defined 2) lim ' ( exists (and is finite) 3) lim ' ( '(+) Ex: Draw examples of ways in which a function could be discontinuous. Removable discontinuity: Jump discontinuity: Continuous o from the left: lim ' ( '(+) 2 Continuous from the right: lim ' ( '(+) 5
25 Ex: Where are each of the following functions discontinuous? a) f(x) 0C 7 07( 07( fcxj.tk#ylyix+lxt2xoyfdiscb@xoc1fd:r b) f(x) B 0 C 4' ( 0 1 4' ( 0 c) f(x) 0 C 707( 07( 4' ( 2 1 4' ( 2 t.ee?s#i?ie***y d) f(x) [[x]] p 105 t.i.it#mxdisaaqz, or Frasers
26 . Theorem 7 The following are continuous at every value in their domains: 1) Polynomials 2) Rational Functions 3) Root Functions 4) Trig Functions 5) Inverse Trig Functions 6) Exponential Functions 7) Logarithmic Functions Ex: Evaluate lim 0 r sut 0 (8uvs 0 Steps: : A) Is the numerator continuous? SINX Yes ; B) Is the denominator continuous? Y Ztcosx C) Are there any points where the rational expression isn t defined? at is defined for b to Ztcosx 0 > If the function is continuous everywhere, then lim 0 r '(() f(@) find, F s In E Theorem 8 If f is continuous at b and lim A ( B, then lim '(A ( ) '(B), which means lim '(A ( ) '(lim A ( ) (See Example 8, Page 124)
27 Ex: Show that there is a root of the equation 4x 3 6x 2 + 3x 2 0 between 1 and 2. P ( x ) polynomial P ( D 4 ( ) ( i ) P (2) 4 (2) 3 6( 2) 2+3 (2) 2 EYE: EIY III. a li I the.it#teeieo x axis between * I I X 2.
28 Common Mistakes 1) lim 0C A 0 ( 07( A DNE Hint: Always check to see if something will cancel bar " 's.ae#fxljn(x3)2 31 ( 2) lim 0 A 0 C ( A DNE Hint: If nothing will cancel, check the graph adf.lt#ksmoiao 3) lim 0C 7@08x 0 ( 07( liner liner 07( (07\) ( Hint: Know how formal your professor is a z tia" I z( 3 ) 231 Not necessary write # to I E EI I n Y O 2
29 ay i * y i i 3 for # 2 <j *'s t
30 2.6 Limits at Infinity Horizontal Asymptotes Look at the graph of f(x) B 0 g.to#xe*ror*o As x > +, f(x) > As x >, f(x) > As x > 0 +, f(x) > : As x > 0 f(x) > The line y 0 is called a horizontal asymptote of f(x), since lim +, 0 0 x Note: lim 0 x +, / can be read as: 1) The limit of f(x), as x approaches infinity, is L 2) The limit of f(x), as x becomes infinite, is L 3) The limit of f(x), as x increases without bound, is L Theorem 5 If r > 0 is a rational number, then 1 lim 0 x, y 0 If r > 0 is a rational number such that, y is defined for all x, then lim 0 7x 1, y 0
31 Ex: Evaluate lim \0 C 707( 0 C 8K08B * a (Using Theorem 5 above) s t *. F * (Note: Limit Laws only hold for finite limits, not infinite limits, so use the above technique to rewrite the following function.) Ex: Evaluate t.it#ot5pppoohgggpgftm@ (0 lim C 8B (Note:, (, 0 x \07@ sf IFi a F I
32 Ex: Evaluate lim 0 x (, ( + 1,) link Cat B) CA Ex: Evaluate lim 0 x (, (,) Hint: Use the conjugate. ) axftx ( ttf. fat B) tismalx ' x ) txftx ) linked?a5 AZB ' him 2+l#. xktjzxcx. Kat X ti a x Hint: Look at this as a product. D. A Ex: Evaluate lim 0 A 5 >?@A B 0 Hint: Let t 1/x, and then look at the graph of arctan lxignoarctantx As x ot Lettxt lim arctant tsoo, ttx a lim want /.bz?i2 t From the graph
33 Formal Limits Let f be a function defined on some interval (a, ). Then lim +, 0 x means that for every positive number M, there is a corresponding positive number N such that if x > N then f(x) > M Ex: Prove that lim 0 x C0 Step I : Given M, we want to find NS.t. if > N ) ex > M
34 ... We want " > r ex M ny.ie : t.ie#is /ne > 1mm I... > then Let, X > lnm N1nM if > NhM ) ex 's M is y.mx increased ' so it preserves order
35 Summary Visual Representations of Limits 1) Finite limits as x approaches a lim ' ( * means that for every + > 0, there is a number, > 0 such that if ( 1 <, then '(() * < + lxljmaf G) L : : : (Page 111) (Page 112)
36 2) Infinite limits as x approaches a lim ' ( means that for every 5 > 0, there is a number, > 0 such that if ( 1 <, then ' ( > 5 lxijna F G) a lim ' ( means that for every 6 < 0, there is a number, > 0 such that if ( 1 <, then ' ( < 6 lijnafcxt. a (Both images, Page 116)
37 3) Limits as x approaches infinity/gets large without bound: lim ' ( * means that for every + > 0, 0 y there is a number N such that if ( > 6 then ' ( * < + lxisnoof CD L lim ' ( * means that for every + > 0, 0 7y there is a number N such that if ( < 6 then ' ( * < + (Both images, Page 138) txignafcxk L
38 lim ' ( means that for every 5 > 0, 0 y there is a number N > 0 such that if ( > 6 then ' ( > 5 (Page 140) djgzfcx ) a
39 . 2.7 Derivatives and Rates of Change Let s look at secant lines again: For a given curve y f(x), we re interested in what the tangent line at P (a, f(a) ) looks like. So we look at another point, Q (x, f(x) ), to the right of P, and investigate what the secant line through P and Q looks like: A delta ay may* ax Since we now understand a little about limits, we can look at what happens as we move Q closer and closer to P by letting x approach a. The slope of PQ ~;ÄÅ É R 0 7R(J) ~ÇY 0 07J so, as we nudge Q closer to P, we get
40 12 1) R 0 7R(J) m lim 07J This is the slope of the tangent line at P. (Provided the limit exists.) Ex: Find an equation for the tangent line to the parabola y x 2 at the point P(1, 1). 4 limipgyttsmope m tiny m2, ftp.ia#x2.fla1tisxtig' ' tig* lxigcx+d2 ( 1,1 ) If we look at the original idea of points P(a, f(a) ) and Q(x, f(x) ), but, instead, consider the x values as x and x + h, the new graph looks like this: y ( x tenant
41 l m 3 ( l, 2) Point slope : y y, m ( viable t#, ) y 2 3 ( ) y z y#i< Slope Intercept I to, D
42 Ex: (#14 in the book) If a rock is thrown upward on the planet Mars with a velocity of 10m/s, its height (in meters) after t seconds is given by H 10t 1.86 t 2. a) Find the velocity of the rock after one second. t H Average Velocity t.is OsOn H m b) Find the velocity of the rock when t a. to H 0 st# 8.14 % ta HlOa 1.86 a Average Velocity ( 10 a 1.86 a To ' ) al a a Mls ' c) When will the rock hit the surface? jh&ekigs i I: o±a I# ' tdo to t or # s d) With what velocity will the rock hit the surface? It should be 10 m/s Velocity T (F)
43 To get a good approximation for the instantaneous Velocity, pick pts close to ta is calc. the average velocity. t a t ath HH. lot 1.86T ' H (a) 10 a 1.86 a2 H ( at D 10 ( ath ) 10 a + Koh ( ath ) ' ( dtzahth ') Hcath ) Average H (a) Velocity ( ath ) a
44 10h3.72aht.SI I h X( A # 1.86k ) <7 A. velocity between a and Instantaneous I (, njmo Average Velocity ) velocity ath ntgm ( ) A
45 . 2.8 The Derivative as a Function! ($) lim b A! $ + h!($) h * a rise run Ex: If f(x) 2x 2 4x + 3, find f'(x). ftp.nliggfcxth#hlignoeathf4cx+h)+d [ ] L ;no#4 htzh *4_I#*8 tin 4 42*44 h O, Liz H4xt# ) nlism (4 +244) Notation: n x f'(x) y' ÖÉ Ö0 ÖR Ö0 Ö Ö0 f(x) D f(x) D x f(x) first derivative f''(x) ÖC É Ö f'(x) second derivative Ö0 C Ö0
46 A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a, b) or (a, ) or (, a) or (, ) if it is differentiable at every number in the interval. Ex: Where is f(x) x differentiable?.ie#ii*, Some ways a function can fail to be differentiable: Corner/kink: Yet Discontinuous: n :& Vertical Tangent Line (with or without a corner):
47 .. Ex: If f(x) 4x 3 2x 2 + 3x 12, find f'(x), f''(x), f'''(x) and f (4) (x) Ffx ) him h O fcxth ) f#.in#xthf2lythiddf_4x22xhx.id h O, L ;z[4cx 'xh2thy ' xht 4h L ;g % ) qfijno ( ) X(Rx4Rxh+4h2 # lim, 12 2hH2xh44h34xhth2+h 0. h fig
48 . f f " ( D Lsgfkthhtftxj nlingo hipo[ *4+3[ ] #. 4h h+ke#4ht3x#xx high tin hso ( ) Lingo ( )24 4
49 f " '( ) lim f"c th"cx ) hso h qnhjmo[ ][24 44 Ko h him 2 Liz 2424 f " )( ) 2k ;mf"gth#(j 1 he Koen tie Lingo 00
50 . f ( ) f ' G) f " ( ) x 24 4 f " G) f ' " ( x ) o
51 Pascal 's A 111( xth 1,,2, ) ' +h ( +h) x72xh+h ' ) I 3 3 I ( x+h)?xi3x'ht3xh2+h3 \, in, I 4 6 Y I (xthjhx 't h3+h4
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