{NotebookFileName[],DateString[]}

style[
t_
]:=
Style[t,Bold,FontFamily->"Arial",FontSize->14]

$Assumptions={{x,y}∈Reals}

Mobius lesson

DSolve[{f'[x]==(1/2)Cos[(x/2)],f[π]==2},f[x],x]

Projectile thrown from hot-air balloon at altitude 2000 m, with initial velocity 20 m/s downward. Constant acceleration of -10 m/(s^2). How long does it take to reach the ground?

DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t]

y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]]

Solve[0==y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]],t]

Solve[0==y[t]/.DSolve[{y''[t]==-10,y'[0]==-20,y[0]==2000},y[t],t][[1]],t]//N

General antiderivative

g=UnitConvert[Quantity[1,"StandardAccelerationOfGravity"]]

gq=QuantityMagnitude[g]

Block[{tmin=0,tmax=1,cs=Range[-3,3],vs,legends},
vs=Table[
c-gq t,
{c,cs}
];
legends=Table[
"-g t + "<>ToString[c],
{c,cs}
];
Plot[Evaluate[vs],{t,tmin,tmax}
,PlotLegends->style/@legends
,AxesLabel->style/@{"t","v"}
]
]

Temperature gradient

Fourier’s Law: time rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area.

(OverVector[q])=-k∇T

(OverVector[q])  is the flux vector (W (m^(-2))), k the heat conductivity (W (m^(-1)) (K^(-1))), ∇T the temperature gradient (K (m^(-1))). One-D form

q=-k((dT)/(dr))

I can use that, maybe in the form Q=-k A((dT)/(dr)), where A is the area of the cylindrical surface across which diffusion is occurring, and Q is the heat transfer.

Graphics3D[{
FaceForm[Red,Blue],
Cylinder[]
}
]

Block[{rplate=4.5,hplate=1.0,rring=3.0,rvial=1.4,thickness=0.4,rθ=0 Degree,rQ=90 Degree},
Graphics3D[{
{
CapForm[None],White,
Tube[{{0,0,0},{0,0,thickness}},rring]
},
{
CapForm[None],White,
Tube[{{0,0,0},{0,0,hplate}},rvial]
},
{
FaceForm[Opacity[0.2],Opacity[0.2]],
Cylinder[{{0,0,0},{0,0,hplate}},rplate]
},
{
Black,Thickness[0.01],CapForm[None],
Line[{{0,0,thickness},{rring Sin[rθ],rring Cos[rθ],thickness}}],
Text[style["r"],(1/2)Total[{{rvial Sin[rθ],rvial Cos[rθ],thickness},{rring Sin[rθ],rring Cos[rθ],thickness}}],{0,-1}],
Text[style["A = 2π r d"],{-rring,0,thickness},{0,-1.2}],
Arrow[{{rplate Sin[rQ],rplate Cos[rQ],(thickness/2)},{rring Sin[rQ],rring Cos[rQ],(thickness/2)}}],
Text[style["Q = k  A T'(r)"],(1/2)Total[{{rring Sin[rQ],rring Cos[rQ],(thickness/2)},{rplate Sin[rQ],rplate Cos[rQ],(thickness/2)}}],{-1.2,0}],
Arrow[{{rring Sin[rQ],rring Cos[rQ],(thickness/2)},{rvial Sin[rQ],rvial Cos[rQ],(thickness/2)}}],
Text[style["vial"],{-rvial,0,thickness},{0,1.5}],
Text[style["petri plate"],{-rplate,0,hplate}],
Text[style["Q = heat flow is the same at all radii"],{rplate,0,0},{0,1}],
Text[style["d = depth of agar"],{rplate,0,0},{0,2.5}],
Text[style["A = area of surface across which heat flows at r"],{rplate,0,0},{0,4}]
}
}
,PlotRange->{rplate{-1,1},rplate{-1,1},0.9{0,hplate}}
,Ticks->Automatic
,ViewPoint->10{0.3,0,0.5}
,Lighting->{
{"Directional",Yellow,{{0,0,0},{1,0,1}}},
{"Directional",Red,{{0,0,0},{0,1,1}}}
}
,Boxed->False
]
]

Perfect!

Power rule

∫(x^(-1.001))\[DifferentialD]x

∫(x^(-0.999))\[DifferentialD]x

Other simple rules

∫((Sec[x])^2)\[DifferentialD]x

∫(1/((x^2)+1))\[DifferentialD]x

D[(E^(-(x^2))),x]

∫x (E^(-x^2))\[DifferentialD]x

D[Cos[ω x+ϕ],x]

∫Log[x]\[DifferentialD]x

Falling micro-black hole

Black hole calculator: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator .

The gravitational force due to a mass M is -((G M m)/(r^2)). The potential energy of a mass m at radius r is thus

$Assumptions={r>0,r0>0,r<r0,G>0,M>0,m>0}

M1=UnitConvert[Quantity[1,"EarthMass"]]

G1=UnitConvert[Quantity[1,"GravitationalConstant"]]

F1[
r_,
Mv_:M,
mv_:m,
Gv_:G
]:=
-((G Mv mv)/(r^2));
F1[r]

U1[
r_,
Mv_:M,
mv_:m,
Gv_:G
]:=Evaluate[
(∫_∞^r)((Gv Mv mv)/(r^2))\[DifferentialD]r
];
U1[r]

Escape velocity should be given by equating this to -(1/2)m (v^2).

Solve[0==(1/2)m (v^2)+U1[r],v]

Select[v/.Solve[0==(1/2)m (v^2)+U1[r],v], Simplify[Positive[#]]&]

Solve[0==(1/2)mv (v^2)+U1[r,Mv,mv,Gv],v]

v/.Solve[0==(1/2)mv (v^2)+U1[r,Mv,mv,Gv],v]

ve1[
r_,
Mv_:M,
mv_:m,
Gv_:G
]:=Evaluate[
v/.Solve[0==(1/2)mv (v^2)+U1[r,Mv,mv,Gv],v][[1]]
];
ve1[r]

ve1[r,M1,m,G1]

UnitConvert[Quantity[1,"EarthRadius"]]

ve1[UnitConvert[Quantity[1,"EarthRadius"]],M1,m,G1]

Yup, that’s consistent with the formula Wikipedia gives.

Now, suppose I have an object falling towards the Earth at escape velocity. Let us suppose that at time 0 it is at distance (Subscript[r, 0]), traveling towards the Earth with velocity (Subscript[v, e])((Subscript[r, 0])), where (Subscript[v, e]) is the escape velocity. Then it will be moving at velocity (Subscript[v, e])(r) at all times. That gives v as a function of r. I can’t integrate that and get anything useful. However, its inverse should be -((dt)/(dr)), and integrated with respect to r should give me t(r).

ve1[r]

$Assumptions

idi1=∫(1/(ve1[r]))\[DifferentialD]r//FullSimplify

idi1==

Solve[t0==idi1+C[1]/.r->r0,C[1]]//Simplify

idi1+C[1]/.Solve[t0==idi1+C[1]/.r->r0,C[1]][[1]]//FullSimplify

idi2=idi1+C[1]/.Solve[t0==idi1+C[1]/.r->r0,C[1]][[1]]//FullSimplify

idi2/.(Sqrt[G M f_.]):>(Sqrt[G M])(Sqrt[f])

idi2==(Sqrt[(1/(2GM))])(2/3)((Subscript[r, 0]^(3/2))-(r^(3/2)))//Simplify

t1[
r_,
Mv_:M,
mv_:m,
Gv_:G,
r0v_:r0
]:=Evaluate[
Assuming[
{r<r0v},
(∫_r0v^r)(-(1/(ve1[r,Mv,mv,Gv])))\[DifferentialD]r
]
];
t1[r]

au1=UnitConvert[Quantity[1,"AstronomicalUnit"]]

roort=UnitConvert[Quantity[2000,"AstronomicalUnit"]]

rsun=UnitConvert[Quantity[1,"SolarRadius"]]

Msun=UnitConvert[Quantity[1,"SolarMass"]]

t1[au1,Msun,m,G1,roort]

(Sqrt[((roort^3)/(G1 Msun))])

UnitConvert[t1[au1,Msun,m,G1,roort],"Years"]

UnitConvert[t1[0,Msun,m,G1,roort],"Days"]

r1[
t_,
Mv_:M,
mv_:m,
Gv_:G
]:=Evaluate[
r/.Solve[t==t1[r],r][[1]]
];
r1[t]

{
m D[r1[t],{t,2}],
F1[r1[t]],
m D[r1[t],{t,2}]==F1[r1[t]]
}

Hurray! It is indeed a solution of the DE!

Now, what if I had started at radius (Subscript[r, 0]) with velocity 0. Can I still solve it? Then I have at all times,

U1[r]+(1/2)m (v^2)==U1[r0]

Solve[U1[r]+(1/2)m (v^2)==U1[r0],v]

(1/v)/.Solve[U1[r]+(1/2)m (v^2)==U1[r0],v]

Select[(1/v)/.Solve[U1[r]+(1/2)m (v^2)==U1[r0],v],Simplify[Positive[#]]&]

v1[
t_,
Mv_:M,
mv_:m,
Gv_:G
]:=Evaluate[
Select[v/.Solve[U1[r]+(1/2)m (v^2)==U1[r0],v],Simplify[Positive[#]]&][[1]]
];
v1[t]

(∫_r0^r)(1/(v1[t]))\[DifferentialD]r

Nope, can’t do that in Monday’s lecture. Maybe when we get to the lecture on substitutions.