{NotebookFileName[],DateString[]}

The pendulum

DSolve[{θ''[t]==-(g/L)Sin[θ[t]]},θ,t]

Approximate thermal expansion coefficient of steel.

steelTEC=13×10^-6

gstandard=N[UnitConvert[Quantity[1,"StandardAccelerationOfGravity"]]]

Solve[Quantity[1/2,"Hertz"]==(1/(2π))(Sqrt[(gstandard/L)]),L][[1]]

UnitConvert[L/.Solve[Quantity[1,"Hertz"]==(1/(2π))(Sqrt[(gstandard/L)]),L][[1]],"Inches"]

Gravity as a function of latitude. From these pages: http://curious.astro.cornell.edu/our-solar-system/the-earth/42-our-solar-system/the-earth/gravity/94-does-your-weight-change-between-the-poles-and-the-equator-intermediate and https://van.physics.illinois.edu/qa/listing.php?id=186.

gpoles=Quantity[9.83,(("Meters")/(("Seconds")^2))]

gequator=Quantity[9.78,(("Meters")/(("Seconds")^2))]

gpoles/gequator

latWaterloo=43.4643 Degree

latDallas=32.7767 Degree

gφ[
φ_
]:=
gpoles+(gequator-gpoles)((Cos[φ])^2)

gφ/@{0,latDallas,latWaterloo,(π/2)}

((gφ[latDallas]-gφ[latWaterloo])/(gφ[latWaterloo]))

UnitConvert[Quantity[(1/2)((gφ[latDallas]-gφ[latWaterloo])/(gφ[latWaterloo])),"Days"],"Seconds"]

Abs[((gφ[latDallas]-gφ[latWaterloo])/(gφ[latWaterloo]))]>Abs[steelTEC*10]

steelTEC*10

steelTEC*10//N

Solve[Quantity[(1/2),"Hertz"]==(1/(2π))(Sqrt[(gstandard/L)]),L]

Lstandard=L/.Solve[Quantity[(1/2),"Hertz"]==(1/(2π))(Sqrt[(gstandard/L)]),L][[1]]

Solve[(1/2)==(1/(2π))(Sqrt[(g/L)]),L]

T1[L_,
gv_:g
]:=2π(Sqrt[L/gv]);
T1[L]

$Assumptions={g>0,L>0}

D[T1[L],L]//Simplify

Tp=D[T1[L,gstandard],L]/.L->Lstandard

ΔL=10*steelTEC*Lstandard

Tp*ΔL

((UnitConvert[Quantity[1,"Days"],"Seconds"])/(Quantity[2,"Seconds"]))

((UnitConvert[Quantity[1,"Days"],"Seconds"])/(Quantity[2,"Seconds"]))*Tp*ΔL

Quantity[1,"Days"]*(1/2)*steelTEC*10

UnitConvert[Quantity[1,"Days"]*(1/2)*steelTEC*10,"Seconds"]//N

Logarithmic differentation

D[(x^x),x]

D[(x^(x^x)),x]

D[Log[(x^(x^x))],x]

Log[(x^x)]//FunctionExpand

D[Log[(x^x)],x]

Assuming[
x∈Reals,
D[Log[Abs[x]],x]//FullSimplify
]

Assuming[
x∈Reals,
D[Log[Abs[x]],x]==(1/x)//FullSimplify
]

f1[
x_
]:=
((((x^5)+(x^2)))^9)(Sqrt[Cos[x]])(x^x)

D[f1[x],x]//Simplify

D[Log[f1[x]],x]//Simplify

Cube root

r1[
x_
]:=
(x^(1/3))

D[r1[x],x]/.x->1728

(1729.03^(1/3))

N[((1729.03`20)^(1/3)),20]

N[((1.03`20)/432),20]

N[(1/1728)(1/3)12,20]

N[((1.03`20)/1728)4,50]

(4/1728)

dfdx0=.

Block[{x0=1728.0,x,xmin=1000.0,xmax=2450.0,f=(x|->(x^(1/3))),dfdx,dfdx0,tangent},
dfdx=D[f[x],x];
dfdx0=dfdx/.x->x0;
tangent=x|->f[x0]+dfdx0(x-x0);
Plot[{
f[x],
tangent[x]
},{x,xmin,xmax}
,PlotLegends->{(x^(1/3)),(x0^(1/3))+dfdx0(x-x0)}
,Epilog->{
PointSize[Large],
Point[{{x0,f[x0]}}]
}
]
]

(512^(1/3))

(1/432)(512-1728)+12//N

(1/432)((10^6)-1728)+12//N