Lecture 10 Notes (MATH M-119; Brief Survey of Calculus I, Staff)

Lecture 10 Notes (MATH M-119; Brief Survey of Calculus I, Staff)

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Unformatted text preview: 2.1 Instantaneous rate of change and 2.2 The derivative function a? Cwmoi gas-sans) From chapter 1, there were 2 rates of change:/ \efii LL, Mt \ ‘C .. at” 2 s 3; Average rate of change: an 0‘2 SW1 “Wt “ A“ J? (me Percent rate of change: t L f0 * ’) 5mm W/W We mentioned that the derivative fimction is a slope and so is the average rate of change, but there is a difference: The derivative is the slope at 1 point, and the average rate of change uses 2. The lines that show these slopes are the tangent and the secant line respectively. The only problem with 1 point is that our slepe formula requires 2. The good news is that if the 2 points are REALLY close, they look and act like 1 point. A “any” fiflb/‘é-D : A“°-m"’°‘: 3‘ i \ t / ((3: D) I (-SII gm Let f (x) = x2. What is the average rate of change om x=0 and x=3? Use a very small L“ flu“ I“winter-val to estimate the instantaneous rate of change at ,7 % m do! A '_.___°°°“°‘ “31:: fl: «3‘3" A)? wt X20 m: ems—c .oo'l ' (mm mm 1-: not} Lop) (mg Ile'“) COP) Leolvma D V XI?! ' :: (“g/f9,” 8mm Mmfiosuauaq‘5'm09Q jz'q‘fi“ m a) nee—Elfin“?- 1q11 33" 1‘73“) 45' SM 1 Lg (,3 q) (23‘ l _ to *- 5.?fi‘ifii'k-é 1W? 1 1 (6 “m 41 fl 5. as “km 4.. - 0° _ A 1; ' *- f" A¥g ‘00 m_ Wz.q1 -3 m W “l flaw—Lt Note that in each case, the instantaneous rate of change is just the slope of the tangent line. \(Euooui 15- MR‘S ‘1 Estimating instantaneous rate of change from a table. 1 W‘“5 m -—- Ex. (4) Find the avera e veloci over the interval 0 S t S 0.8 , and estimate the velocity at t=0.2 of a car whose position, S, is given by e o owmg table: t(sec) 0 0.2 0.4 0.6 0.8 1 S (f?) 0 0.5 1.8 3.8 6.5 9.6 (o 03 = W‘- q ($35.5) AJC' F b nulan 0919541 “"1" it?” 2 MATE- e: Glee-'1- fizz I ‘ i U (oldilq’lfi C.Z.5)Cq‘8 For the graph given below, estimate the instantaneous rate of change (=slope of the tangent line=derivative) at each “major” point Use the table to fill it} the following table X .... height slope . M?” {nut mar W W3 W MW“! W at“? Foams “’ 9”“ HBO “:5 WW y-AXIS Use the table to fill in the following table 3..X..=f(x) m=f'(X) {Away-"5‘ Wflwhw Is [3 gwfm ‘33.“ Icy-(n 2.3 Interpretations of the derivative. Let’s look at the two types of linear rate of change: Average rate of change Instantaneous rate of change Thus there are several different ways to notate the derivative: f'(x) f'(3)=-2 fl Q = _2L/-—-.:'~D!\ air dx 1:3 Slope of the tangent line at x=3 is 1112-2 [61+- 0 Ex (13) A mutual fund is currently valued at $80 (per share) and its value per share is increasing at a {egg of $0.50 per day. Let V(t) be the value of the share 2‘ days from now. A) Express the information above in terms off and f’. . \ ( B) Assuming the growth rate stays constant, estimate and interpret f0 0). 4:.) “r Cie- 5 ‘” aiufl. \f U50 1’ i i \L j’c‘ita J tum t Cop sup? \[ L03 ‘3 41 L3 Local Linear approximation What we just did in the prior example of taking the current linear trend and continuing it is called local linear approximation. What did we do as far as a formula? Ifdelta x is close to 0 then, we know from doing instant rate of change problems that: 52 as A—y or solving for delta y, we get Ay ~ d—y - Ax = f'(x) - Ax . This gives us _/" 0‘55 Ax mum- was. 0'36 {Wecw 6m f(x+Ar)=y+Ay m f(x)+f'(x)Ax ‘- / CHI-“5' t UK} cL. Fur-:1 5 [run/2. ’ L": "- (no -.: Y (‘00 "- m Y ' -— M - The bigger Ax is, the sure we are about our estimate. in (X F‘) m: (24:01) :f'oo. (“I 45’ I meant UNE. 124;“) Him ' M ...
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This note was uploaded on 12/21/2011 for the course BUS 100 taught by Professor Intro during the Fall '11 term at UCSB.

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Lecture 10 Notes (MATH M-119; Brief Survey of Calculus I, Staff)

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