Correction du TD 1 Analyse dimensionelle et ordres de grandeurs

15 →L
2→L2
→ML2T−3
→L
=LT −1v=d
t
=LT −2
=ML2T−2E=1
2mv2
=ML2T−3
=MLT −2F=ma
=LT −2F=P=mg
=M−1L3T−2F=Gm2
r2

l m g
mg
T=LαMβ(L.T −2)γ
β= 0, α +γ= 0,−2γ= 1
β= 0, γ =−1
2, α =1 2
ql
g
√2√n
n
2 2
2
2
?
6,02.1023 3.1024

6.109
10−15
?
l
21
3l1,26l
?240
3
4
7
4
1
5'0,25 3Vcell =Vhumain.2−40 '0,25.(103)−4'
0,25.10−12 3
Lcell =V
1
3
cell '10−410−5
10−4
Scheveux =2'10−2m2
Scell =L2
cell '10−8m2
Scheveux
Scell '106
d
dx
•(xn)0=nxn−1
•(1
xn)0=−1 xn+1 (1 xn)0= ( 1 f(x))0=−f0(x) f2(x)
(xα)0=αxα−1

•(√x)0=1
2√x
x1/2
√x=x1/2
(x1/2)2=x= (√x)2
•(ln(4x+ 3))0=4
4x+3
•(exp(−2x+ 3))0=−2 exp(−2x+ 3)
•(cos(4x−2))0=−4 sin(4x−2)
h→0
∆y
∆x
∆x
f0(x) = df
dx
df dx
dx
h
df ∆y
x
?
sin(x) cos(x)
(sin(x))0= lim
h→0
sin(x+h)−sin(x)
h

= lim
h→0
sin(x)(cos(h)−1) + sin(h) cos(x)
h
= lim
h→00 + cos(x)sin(h)
h= cos(x)
(cos(x))0= lim
h→0
cos(x+h)−cos(x)
h
= lim
h→0
cos(x)(cos(h)−1) −sin(x) sin(h)
h
= lim
h→00−sin(x)sin(h)
h=−sin(x)
limh→0sin(h)
h= 1 limh→0cos(h) =
1
d
dx
df
dx =d ex dx =exdf =exdx
dg
dt =d et+φ dt =et+φdg =et+φdt
dh
dt =d π3φ+4t dt =d(3φ+4t) dt
d(π3φ+4t)
d(3φ+4t)= 4π3φ+4tdh = 4π3φ+4tdt
πxex
e
(g◦f)0(x) = f0(x)(g0◦f)(x)
dk
dx =d(xyt+φ) dx = (yt +φ)xyt+φ−1dk dt =d(xyt+φ) dt =d(yt+φ) dt
d(xyt+φ)
d(yt+φ)=
yxyt+φdk = (yt +φ)xyt+φ−1dx +yxyt+φdt