208 
ym—aVsta(tag V3). 
This gives us 
| À 31 
"TA Qa? 4 3aig(kar V3) Ne le 
The case a =O has been discussed in 9; we therefore put 
a=0. When p=0,r=a:(t + 2a’), ie. a value between r=a 
and r= 3e. When p approaches to 7: av 3 (a value which, from (30), 
exceeds a) 7 should approach to zero, according to (31). But first 
r must become equal to «a, viz. when gp becomes 
2 V2—6a? 
Ek, rn: WIE Un 
3a 
and for this, according to (27), an infinite time is required as then 
z—2. So the motion is as follows: p changes from — ¢, to ¥,; 
corresponding to + = a. The greatest value of r is reached at the 
moment when gy — 0, viz. 
when p=— p, (as well as when y= ~y,) 7 becomes a. If ap- 
proches to zero, p‚ increases indefinitely and the motion approaches 
more and more to that which has been discussed in 9. 
1 Phe. case e= 6; = EPs 
Put e, =e, =a’, e, = — 2a’, then (23) passes into 
) dz. (32) 
iG = == — REE a cel 
(za) Vata’ 
As 2 > — 2a’, we may put z=— 2a’-+ 7’. Then we get 
2 dy 
dp = — = 
y—d a’ 
Now, if z >a’, and therefore y° >3 a’, we get 
y =aV3 cotgh (Hap V3) 
and 
a 33 
cars 1— 2a? + 3a’ cotgh? Gap Bend (RES 
If, on the contrary, z< a? and consequently 7° < 30’, 
y—aVYdstgh (tag V3), 
and so 
a 5 
Se 4— 2a? + 34° th’ (Lag ER rr Cn 
z cannot pass @ and must moreover lie between — + and 2. 
