Begin2.DVI

I7-7.
d~r
dt=~r

′, d~r

ds

ds

dt=

d~r

dt,

ds

dt=s

′= (~r′·~r′) 1 / 2

d~r

ds=ˆet=

~r′

s′,

d^2 s

dt^2 =s

′′= ~r′·~r′′

(~r′·~r′)^1 /^2

d^2 ~r

ds^2

=d

ˆet

ds

=κeˆn=s

′~r′′−~r′s′′

(s′)^3

, d~r

ds

×d

(^2) ~r
ds^2
=κeˆt׈en=κˆeb
I7-8. See derivatives from previous problem.
d^3 ~r
ds^3 =κ
dˆen
ds +
dκ
dsˆen=κ(τˆeb−κˆet) +
dκ
dsˆen
d^3 ~r
ds^3
=κτˆeb−κ^2 ˆet+dκ
ds
eˆn
d^2 ~r
ds^2
×d
(^3) ~r
ds^3
=κˆen×(κτˆeb−κ^2 ˆet+dκ
ds
ˆen) =κ^2 τˆet+κ^3 ˆeb
d~r
ds·(
d^2 ~r
ds^2 ×
d^3 ~r
ds^3 ) =
ˆet·(κ^2 τˆet+κ^3 ˆeb) =κ^2 τ
τ=
1
κ^2
d~r
ds·(
d^2 ~r
ds^2 ×
d^3 ~r
ds^3 )
In terms of a parametertone can write
d~r
dt
=d~r
ds
ds
dt
, d
(^2) ~r
dt^2
=d~r
ds
d^2 s
dt^2
+d
(^2) ~r
ds^2
(ds
dt
)^2
d^3 ~r
dt^3
=d~r
ds
d^3 s
dt^3
+d
(^2) ~r
ds^2
ds
dt
d^2 s
dt^2
+d
(^2) ~r
ds^2
2(ds
dt
)d
(^2) s
dt^2
+d
(^3) ~r
ds^3
(ds
dt
)^3
which can be express
~r′′=ˆet
d^2 s
dt^2 +κ
ˆen(ds
dt)
2
~r′′′=ˆet
d^3 s
dt^3 +κ
ˆends
dt
d^2 s
dt^2 +κ
ˆen 2 ds
dt
d^2 s
dt^2 + (
ds
dt)
(^3) [κτˆeb−κ (^2) ˆet+dκ
ds
ˆen]
~r′′×~r′′′=(stuff 1 )ˆeb+ (stuff 2 )ˆen+κ^2 τ(ds
dt
)^5 eˆt
~r′·(~r′′×~r′′′) =κ^2 τ(
ds
dt)
(^6) , but (ds
dt)
(^6) = (~r′·~r′) 3
andκ^2 can be obtained from problem 7-5 to obtain
τ=
~r′·(~r′′×~r′′′)
(~r′·~r′)(~r′′·~r′′)−(~r′·~r′′)^2
I7-9. (a) zero (b) zero
Solutions Chapter 7