Lecture Notes for 120 - UCLA Department of Mathematics

6.3. ACCELERATION 140
Exercises.
(1) Calculate
6.3. Acceleration
The goal here is to show that the tangential component of the acceleration of
acurveonaparametrizedsurfaceq (u, v) :U ! R 3 can be calculated intrinsically.
The curve is parametrized in U as (u (t) ,v(t)) and becomes a space curve q (t) =
q (u (t) ,v(t)) that lies on our parametrized surface.
The velocity is
˙q = dq
dt = dq
dt = @q
@u
du
dt + @q
@v
dv
dt = ⇥ @q
@u
@q
@v
⇤ apple du
dt
dv
dt
The acceleration can be calculated as if it were a space curve and we explored that
in Chapter 6. Using the velocity representation we just gave and separating the
tangential and normal components of the acceleration we obtain:
¨q = ¨q I + ¨q II
= ⇥ @q
@u
@q
@v
⇤
[I]
1 ⇥ @q
@u
@q
@v
⇤ t
¨q +(¨q · N) N
In Chapter 6 we worked with the normal component. Here we shall mostly focus
on the tangential part.
where
Theorem 6.3.1. The acceleration can be calculated as
2
3
¨q = ⇥ @q @q
@u @v
N ⇤ d 2 u
dt
+ u ( ˙q, ˙q)
4
2 d 2 v
dt
+ v ( ˙q, ˙q) 5
2
=
w ( ˙q, ˙q) =
✓ d 2 u
dt 2 +
X
w 1,w 2=u,v
II ( ˙q, ˙q)
◆ ✓ @q d u 2
( ˙q, ˙q)
@u + v
dt 2 +
w dw 1 dw 2
w 1w 2
= ⇥ du
dt dt
dt
◆ @q
v ( ˙q, ˙q) + NII ( ˙q, ˙q) ,
@v
dv
dt
⇤ apple w
uu
w
uv
w
vu
w
vv
apple du
dt
dv
dt
Proof. We start from the formula for the velocity and take derivatives. This
clearly requires us to be able to calculate derivatives of the tangent fields @q
@u , @q
@v .
Fortunately the Gauss formulas tell us how that is done. This leads us to the
acceleration as follows
¨q = d ✓ ⇥ @q
⇤ apple ◆
du
@q dt
dt
@u @v dv
dt
= ⇥ @q
@u
@q
@v
which after using the chain rule
⇤ " #
d 2 u
dt 2
+
d 2 v
dt 2
✓ d
dt
⇥ @q
@u
d
dt = du @
dt @u + dv @
dt @v
@q
@v
⇤ ◆apple du
dt
dv
dt