meanings of the derivatives
DESCRIPTION
Meanings of the Derivatives. I. The Derivative at the Point as the Slope of the Tangent to the Graph of the Function at the Point. The Tangent to a Curve Example: The tangent to the graph of the function f(x) = (x-2) 2 + 3 at the point x = 2 is the line y = 3. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/1.jpg)
Meanings of the Derivatives
![Page 2: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/2.jpg)
I. The Derivative at the Point as the Slope of the Tangent to the Graph
of the Function at the Point
![Page 3: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/3.jpg)
The Tangent to a Curve
Example: The tangent to the graph of the function f(x) = (x-2)2 + 3 at the point x = 2 is the line y = 3
![Page 4: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/4.jpg)
The Derivatives as the Slope of the Tangent
.)(
)()('
))(,())(,(&))(,())(,(,
00
0
000
000
lim0
xfxt
xftfapproachesitsThat
xatngenttatheapproachestftandxfxjoininglinetheofslopethesoxatngenttatheapproachestftandxfxjoininglinethextAs
xt
![Page 5: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/5.jpg)
Examples (1)For each of the following functions, find the equation of the tangent and the normal at the given point
014)1(45:.25)1(.1
:014
')1(45
:)5,1())1(,1(4
4113)(
:1
!13)(
:123)(
)1(
xyxythatShowfthatShow
QuestionsyxsThat
xyislinethisofequationThe
ftoinpthethroughandslopewitlinestraighttheisngenttarequiredThis
xf
toequalisxatftongenttatheofslopeThe
thatShowx
xf
Solutionxatxxxf
Example
![Page 6: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/6.jpg)
0)56(210)2
(53:.2
3)2(.1
:0)56(210
'
)2
(53
:
)3,2())
2(,
2(4
5)1(5)0(3)(
:2
sin5cos3)(:
2cos5sin3)(
)2(
yxxythatShow
fthatShow
Questionsyx
sThat
xy
islinethisofequationThe
ftoinpthethroughandslopewithlinestraighttheisngenttaThis
xf
toequalisxatftongenttatheofslopeThe
xxxfSolution
xatxxxf
Example
![Page 7: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/7.jpg)
3:)3,2()3,2())2(,2(
2:0)2(2
:,.0)2(2)(
:int
3)2()(:
)3(2
yistoinpthethroughlinehorizontalaofequationThe
ftoinptheatngenttahorizontalahasfThisx
getWex
letThusislinehorizontalaslopeThexxf
SolutionhorizontalisftongenttathewhichatpotheFind
xxfLet
Example
![Page 8: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/8.jpg)
36.:int132)(
)4(23
isngenttatheofslopeTheawhichatpoThexxxf
Example
![Page 9: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/9.jpg)
08036:')3(3628:)28,3(
04536:')2(3627:)27,2(
)28,3())2(,2()27,2())2(,2(:36
32:
3666
:,36int
66)(
:int36int
132)(
2
2
23
xysThatxyisatngenttatheofequationThe
xysThatxyisatngenttatheofequationThe
fandfareisngenttatheofslopethewhichattsinpoThe
xorxgetWe
xx
letThusequationthissatisfywillisngenttatheofslopethewhichatxpoThe
xxxf
isxpoanyatderivativetheofvaluethegivingformulaTheisngenttatheofslopethewhichatpoThe
xxxf
![Page 10: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/10.jpg)
II. The Derivative at the Point as the Instantaneous Rae of Change
at the Point
![Page 11: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/11.jpg)
The Derivatives as the Instantaneous Rate of Change
![Page 12: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/12.jpg)
Find:1.a. A formula for v(t)b. The velocity at t=2 and at t=5c. The instances at which the particle is at rest( stops temporarily before changing direction). When it is moving forward/backward?
2.a. A formula for a(t)b. The acceleration at t=2 and at t=3c. The instances at which the particle experiences no acceleration (not speeding). When it is speeding up/slowing down?
lyrespectiveparticletheofonacceleratitheandvelocitythebetaandtvLetttimeoffunctionaasparticleaofpositionthebe
ttttsLet
Example
),()(
425
31)(
)1(
23
![Page 13: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/13.jpg)
1.a. A formula for v(t)v(t) = s'(t) = t2 – 5t + 4
b. The velocity at t=2 and at t=5v(2) = -2v(5) = 4
c. The instances at which the particle is at rest( stops temporarily before changing direction). When it is moving forward/backward?
c.1. Let: v(t) = 0= → t2 – 5t + 4 0 → ( t – 1 )( t – 4 ) = 0 → t = 1 Or t = 4The particle becomes at rest at t = 1 and at t = 4
c.2.The particle is moving forward when: v(t) > 0v(t) > 0 → t2 – 5t + 4 > 0 → ( t – 1 )( t – 4 ) > 0 → t > 4 Or t < 1
c.3.The particle is moving backward when: v(t) < 0v(t) < 0 → t2 – 5t + 4 < 0 → ( t – 1 )( t – 4 ) < 0 → 1 < t < 4
![Page 14: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/14.jpg)
2.a. A formula for a(t)a(t) = v'(t) = 2t – 5
b. The acceleration at t=2 and at t=3a(2) = -1a(3) = 1
c. The instances at which the experiences no acceleration (not speeding up or slowing down). When it is speeding up/slowing down?
c.1. Let: a(t) = 0= → 2t – 5 = 0 → t = 5/2 The particle experiences no acceleration at t = 5/2
c.2.The particle is speeding up when: a(t) > 0a(t) > 0 → 2t – 5 > 0 → t > 5/2
c.3.The particle is slowing down when: a(t) < 0a(t) < 0 → 2t – 5 < 0 → t < 5/2
![Page 15: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/15.jpg)
Example (2)Let s(t) = t3 -6t2 + 9t be the position of a moving particle in meter as a function of time t in seconds1. Find the total distance covered by the particle in the first five seconds2. Graph S(t), v(t) and a(t)
Solution:v(t) = 3t2 - 12t + 9 = 3(t2 - 4t + 3) = 3(t-1)(t-3)v(t) = 0 if t=1 or t =3 →The particle stops temporarily at t=1 and again at t=3v(t) > 0 if t > 1 or t > 3 →The particle moves in one direction (the positive direction) from before t=1 and after t=3v(t) < 0 if 1 <t < 3 →The particle moves in the opposite direction (the negative direction) from between t=1 and t=3
![Page 16: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/16.jpg)
s(0) = (0)3 - 6(0)2 + 9(0) = 0
s(1) = (1)3 - 6(1)2 + 9(1) = 1- 6 + 9 = 4
s(3) = (3)3 – 6(3)2 + 9(3) = 27 – 54 +27 = 0
s(5) = (5)3 - 6(5)2 + 9(5) = 125 – 150 + 45 = 20
The total distance traveled in 5 seconds
= |s(5)-s(3)|+ |s(3)-s(1)|+ |s(1)-s(0)|
= |20-0|+ |0-4|+ |4-0| = 20 + 4 + 4 = 28
![Page 17: Meanings of the Derivatives](https://reader036.vdocuments.us/reader036/viewer/2022081420/56815de1550346895dcc0726/html5/thumbnails/17.jpg)