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Unformatted text preview: 2.1 Instantaneous rate of change and 2.2 The derivative function a? Cwmoi gassans) 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/‘éD : 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“winterval 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'ké
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
yAXIS 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 11122
[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.
 Fall '11
 Intro

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