1810.
We need some assumptions:
 flat universe
 W_{m} ~ 0.3
 mass of galaxy M_{g} ~ 10^{12}M_{sun}
 present horizon radius d ~ 2/H_{0}
Let M(t) = ^{4}/_{3}pd^{3}(t) r_{m}(t) be the mass contained within the horizon radius d. Using (18.47), the constancy of r_{m}a^{3} and the present value r_{m}(t_{0}) = W_{m}r_{crit} = 3W_{m}H_{0}^{2}/8p we find
 dM
dt

= 4pr_{m}(t) d^{2}(t) so at present  dM
dt

~ 6W_{m} ~ 1.8. 

The required time Dt is therefore
Dt =  M_{g}
dM/dt

~ 
10^{12} M_{sun}×(1.5 km/M_{sun})
1.8 ×(3×10^{5} km/s)
 ~ 10^{6} s ~ 1 month. 

1820
The Friedmann equation (18.77) becomes (with â(t) = a(t)/a_{0})
 æ è
 dâ
dt

ö ø 
2

= W_{v} â^{2} + 
W_{m}
ä



This can be solved by separation of variables (the substitution y = â^{3/2} helps),
â(t) =  æ è 
W_{m}
W_{v}
 ö ø
 ^{1}/_{3}

sinh^{2/3}(^{3}/_{2} W_{v}^{½}
H_{0}t). 

The present moment occurs when â(t_{0}) = 1, namely
t_{0} =  æ è
 2
3H_{0} W_{v}^{½}
 ö ø
 sinh^{1}  é ë
 æ è
 w_{v} W_{m}
 ö ø
 ^{}1/_{2}

ù û  . 

1824
a) Look for the maximum of U_{eff} of (18.78): this occurs when (W_{m}/â^{2}) + 2W_{v}â = 0 or r_{m} = 2 r_{v} = L/4p.
b) From the Friedman equation (18.77) we have U_{eff}(â) = ^{1}/_{2} W_{c} or
^{8p}/_{3} (r_{m} +r_{vac} ) a^{2} = 1 hence a = L^{½}
so the volume is V = 2p^{2} a^{3} = 2p^{2} L^{3/2}.
c) The plot of U_{eff} in Fig 18.9 shows that a small change in r_{m} will either cause the universe to expand to infinite volume, or to collapse to a singularity.
1827
a) From (18.63) and (18.77) we have
r(a) =  
3
4pa^{2}

U_{eff}(a) 

The first law of thermo (18.20) gives
p(a) =  
d(ra^{3})
d(a^{3})

= 
1
4pa^{2}


d[aU_{eff}(a)]
da



That's your parametric representation.
b) The potential should be like (18.78) down to a universe radius when k_{B}T ~ 10 Mev; for smaller a it should rise beyond the value it has a a_{max} in order to cause a bounce.
c) From the above equations,
r+ 3p = 
3
4pa^{2}


dU_{eff}
da



So r+ 3p must be negative in the 'smaller a" region of b).
1830
a) We need r(c) and z(c) such that the geometry to be embedded can be written
dS^{2} = dc^{2} + sinh^{2}cdf^{2} = dr^{2} + r^{2} df^{2} + dz^{2}
Evidently r(c) = sinh(c), but then

æ è

d\z
dc

ö ø

2

= 1  cosh^{2}c < 0 

does not work.
b) Exchange sinh and cosh in the above, and the equation does have a solution namely
r(u) = cosh u z(u) = 
ó õ

u
0

(1  sinh^{2}u¢)^{1/2} du¢ 

which defines an embedded surface but only up to u = sinh^{1}(1) = 0.88.