every slope is a derivative. velocity = slope of the tangent line to a position vs. time graph...
TRANSCRIPT
Every slope is a derivative.
x
t
v
t
a
t
Velocity = slope of the tangent line to a position vs. time graph
Acceleration = slope of the velocity vs. time
graph
How then can we move up the stack of graphs?
dt
xdv
dt
vda
2
2
dt
xda
a
t
1t 2t
12 tata
12 tta
tav
a
ta
221 0
ttt A curve the under Area curve the under Area 21
tt
1221 00
tttt AA curve the under Area
v
t
1t 2t
221 0
ttt A curve the under Area
22
11
tavv
tavv
o
o
But
1v
2v
curve the under Area 21
tt
1221 00
tttt AA curve the under Area
1122 2
1
2
1vtvt
1122 2
1
2
1tattat
v
t
1t 2t
1v
2v
21
22 2
1
2
12
1
tatatt
curve the under Area
12 xx
21
2
1
tt
xtt
to from
curve the under Area
What if the graph is v
t
1t 2t1t 2t
v
t
1t 2t
Increase the # of rectangles
Fill the area with rectanglesv
t
1t 2t
be would to from
curve the under area
the calculate to method One
21 tt
221 0
ttt AreaArea 21
ttArea 1221 00
tttt AreaAreaArea
What if the graph isv
t
1t 2t
v
t
1t 2t1t 2t
v
t
1t 2t
rectangles of where
curve the under Area
#
lim
n
Arecn
.xy is slope the while
x,y is rectangle a of area the that Note
Increase the # of rectangles
Fill the area with rectangles
Definite Integral – the area under the curve between definite limits
v
t
1t 2t
2
1)()()( 12
tt tFtFdxxf
21)( ttxxf to from to respect with function the of integral The
Upper limit
Lower limit
Differentiation
Formula
Integration
Formula
Power Rule
aaxdx
d axdxa g
ege
1 nn nxxdx
dg
e
nge
n
n
xdxx
1
1
Let a, b, g, e, m, and n be constants.
axy ay
)(axdx
d
dx
dy g
ege dxadxy
nxdx
d
dx
dy
nxy nxy
ge
nge dxxdxy
1 nn anxaxdx
d
ge
nge
n dxxadxax
g
e
nge
n
n
axdxax
1
1
Differentiation
Formula
Integration
Formula
Let a, b, g, e, m, and n be constants.
naxy naxy
naxdx
d
dx
dy
ge
nge dxaxdxy
nxdx
da
dx
dy
Sum or Difference
)( mn bxaxdx
d
dx
dy
)()( mn bxdx
dax
dx
d
dx
dy
ge
mnge dxbxaxdxy )(
ge
ge
mn dxbxdxax
g
e
mn
m
bx
n
ax
11
1111 mn bmxanx
dx
dy
Differentiation
Formula
Integration
Formula
Let a, b, g, e, m, and n be constants.
mn bxaxy mn bxaxy
).1()4( 2 smt
smv
by givenvelocity a has object An
1 Example
? from ntdisplaceme sobject' the is Whata. sts 10
0 1 2 3 4
2
4
6
8
10
12
14
16
18
Method 1
graph vs. the under Area tvx
)4)(1(2
1)1)(1( s
msssmx
mx 3
smv
st
10 Ax
).1()4( 2 smt
smv
by givenvelocity a has object An
1 Example
? from ntdisplaceme sobject' the is Whata. sts 10
Method 2
10 dtvx
10 )14( dttx
10
114 0 dtdttx
1
0
1011
10
1
11
4
ttx
1022
ttx
0)0(21)1(2 22 x
mx 30 1 2 3 4
2
4
6
8
10
12
14
16
18
smv
st
).1()4( 2 smt
smv
by givenvelocity a has object An
1 Example
? from ntdisplaceme sobject' the is Whatb. sts 42
0 1 2 3 4
2
4
6
8
10
12
14
16
18
Method 1
graph vs. the under Area tvx
smss
ms
smss
msx
822
112
1642
114
mx 26
smv
st
40Ax x 2
040 AAx
).1()4( 2 smt
smv
by givenvelocity a has object An
1 Example
? from ntdisplaceme sobject' the is Whatb. sts 42
Method 2
42 dtvx
42 )14( dttx
4222
ttx
2)2(24)4(2 22 x
mx 260 1 2 3 4
2
4
6
8
10
12
14
16
18
smv
st
.
21(2by given is
onaccelerati whoseobject an for from )velocity in change the Calculate
2 Example
ta
stsv
1 2
21 dtav
21
2 )( dttv
2
1
12
12
t
v
2
1
3
3
1
tv
33 )1(3
1)2(
3
1 v
smv 3.2
2sma
st