Math 2200-01 (Calculus I) Spring 2020 Book 1 - fail for example ( - PowerPoint PPT Presentation

Math 2200-01 (Calculus I) Spring 2020 Book 1

- fail for example ( one input variable onion 27 Calculus I - variable calculus single - : y x , ( rates of change ) : diff calculus output variable Derivatives . ) . . - variable Calculus I i also Integral calculus single . . several input variables and/or several output variables multi variable Calculus III ie . : three output of an object at time t : one input t . position Kitt , gets , zits ) eg , variables xcts , gits , ziti . a function of position . Temperature in Tfx . g. z ) this Eg room as T ) ( three inputs output x. y , -2 ; one a function of position three outputs three inputs Wind velocity Ege , y , z x ; : as wind velocity . the components of Jan 28 are tangent line . lines to w.§ µ ( Tangent line secant wage is T.to#Ee 2¥ # - * y .

T a function of time t Temperature T as During the time interval , Et Et t ft . ti i.e . the temperature rises from . tz to Tr rate of change of temperature during T The over-age , . interval this time is in temperature , ← change - T Tz OT slope of the secant line from - t , ← time elapsed = - = on the graph - ta ot Hantz ) Lt . . T . ) to . . rate of change of temperature We want to understand instantaneous the rate of charge at time determine this consider t . To the first average . , smaller intervals ft . smaller time where and take , Ed te - t over we , and Hz gets closer closer to ti ) - Jan 29 . t.I.IT . Eg In my example , ti 3 - . We unite tiny = 2.2 degrees hour 4 2 The lift , 2.17 I is 2.2 I the 3. Yay 3. fo , limit of ✓ Tz - T . , is 2.2 . It - - the temperature at 3pm i - . is changing at astr approaches 3 ) rate of 2-31 2 a . 2. 2 degrees per hour .

second example using a polynomial function - fix ) = x ? A Find the y - of change at with rate of y respect to x x=z . line joining the points ( 2,47 and 13,91 t ÷ :÷÷÷÷÷÷ y ax slope curve has the tangent line . . :* :c . . . . . . . has slope # secant line the the curve on , i¥E±÷ .IE#.I = flx7-f# tgyg × The tangent line at Gay 3 2 . X - 2 is ffx ) -4=41 × -2 ) ie y . . - fczs f- ( x ) - Based on the y=4x - 4 . table of values X ← Az # H2 . I ? ' - ? ? ! - . . . . . the slope of the 339 Z i. e x line to the graph at G. 4) . is 4 tangent 3. s i. 5 . I 3 . .

a sufficiently . polynomial , then have nice formula a function has we If eg rules that provide , eliminating ways to evaluate limits definite algebraic the graph based or table of values guesswork on . the slope the tangent graph of y to the E , 4) Find of line - x ' at Eg - . . ) = ( x , x Their line from ' ) slope secant The to ( x , fix has 12,47 ) = K¥2 ¥- ( × t2¥z for xt2 = xtz = . . The slope of the target line is fig , lying ( xt 2) : :#÷ 2+2=4 = = . I ÷ Jan3 € ' o z - L functions satisfy lying ,flx)=4 of these Both

, "I¥z , lying fix ) lying = =3 Febio , : Friday 's quiz Compare Ess !m → ¥*= - fig - o = , 5- ÷ × 's } # defiant , y= Is fine # = A fig , - a = , ¥3 lying = o f → ¥3 = o

is continuous at Seattle him fix ) = Fla ) A function f if a . this requires that Explicitly ×→ a , . Ha ) exists be defined at in f must ie a ; , and limit ciii f at must have a ; a must agree and the values his ciii ) in cis . on the right , . for the function f Eg f discontinuous at 5 ; is • f Ms . - . . . . . • . not exist fix does # 31=1 , fig . , • f not continuous at 2 is . f- 127=3 lxiagzfcxl but these I = , do not agree two values ! Although Ling , f- is not is discontinuous fix ) =3 , f- at t • . defined at a . f- is except at 143,5 ( 0,7 ) continuous ie or * < 7 n . .

the cost of parking meter 254 for each at Eg is a . function of time . The minutes city cost 15 as a At each discontinuous at t , 45,60 30 , 15 is o - - . . - points of discontinuity , , - continuous of these Heft is c , not right . fiagaftti lie Ha ) ) - continuous but - - lim fits # flat ) ( ie . . f- → at y C- Feb about continuity ? - th ) 11 I why do we ,a# I - care continuous with f If is such that fits ) so and as cab f- Ca ) - o then there exists c , , ( Inbr Valse theorem ) flat = o there might be . The point not might Remarks be unique ie c : with this property . fa¥ more than one or fat c .

Consider fix ) does such a number exist ? Why - x' is E ? what - z - . a polynomial ( see By the Intermediate Sec 2.6 ) it F is continuous because is . Value theorem fa ) > o ) I ( since fro ) so there f # - , such that fk ) between O and exists 2 = o a . is only we 'll there such Later see one c as . FL , , L this value • We call Fc - 2 . two points which Another example : At this moment there are are surface having exactly same temperature Earth 's the the antipodes on the temperature . be the 1701,0 Etc 2T equator and Consider the let on , 0 longitude with respect to longitude ) equator ft lie O angle A is . . , ⑦ D in temperature between Lfo ) = difference Tl Ott ) - TCO ) = ⑦ = o f =p and its antipode ( at Ott ) longitude O . If flo ) - o " Tht ) < Tco ) fit ) > o then ie . . D- =3 such that There exists o cacti c Z , fcc ) = o = T eeth ) Tlc ) i.e . . .

= tiny . Hathy ca ) = lying µ *t A . tH¥a ' ' f % :O : :c :* . . .mx . I \ ' Febl4# SHHH - f ' ! . . six i - ha Es - - - gas = him . = fin "x ( im 941 gtz ) fins , I = i ×→ . I = = . g 't 3) = fine , = king . , = fig , - D - i C = . whereas Lingo ( finna # - i ) . ¥ = limo ¥ gto , = lying . "o = I does not exist = . g not differentiable at o . ' ¥9 ¥ not exist 1 × 1 ' g is co ) doses g - fi . if : gia , , - -0 . ( undefined if a = o ) - I .

EA 1¥ . position ( displacement ) s = time t = velocity = sit ) vet ) v - -