98. Simultaneous change of both independent and dependent variables.
x
=
ρ
cos
θ
{\displaystyle x=\rho \cos \theta }
y
=
ρ
sin
θ
,
{\displaystyle y=\rho \sin \theta ,}
the equation
f
(
x
,
y
)
=
0
{\displaystyle f(x,y)=0}
becomes by substitution an equation between ρ and θ , defining ρ as a function of θ . Hence ρ, x, y are all functions of θ .
Illustrative Example 1. (42 ), §103 ,
(A )
R
=
[
1
+
(
d
y
d
x
)
2
]
3
2
d
2
y
d
x
2
,
{\displaystyle R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}},}
into polar coördinates.
Solution. Since in (A ) and (B ) , §97 , t is any variable on which x and y depend, we may in this case let
t
=
θ
{\displaystyle t=\theta }
(B )
d
y
d
x
=
d
y
d
θ
d
x
d
θ
{\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}}
(C )
d
2
y
d
x
2
=
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
(
d
x
d
θ
)
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{\left({\frac {dx}{d\theta }}\right)^{3}}}}
Substituting (B ) and (C ) in (A ) , we get
R
=
[
(
d
x
d
θ
)
2
+
(
d
y
d
θ
)
2
(
d
x
d
θ
)
2
]
3
2
÷
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
(
d
x
d
θ
)
3
{\displaystyle R=\left[{\frac {\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}}{\left({\frac {dx}{d\theta }}\right)^{2}}}\right]^{\frac {3}{2}}\div {\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{\left({\frac {dx}{d\theta }}\right)^{3}}}}
(D )
R
=
[
(
d
x
d
θ
)
2
+
(
d
y
d
θ
)
2
]
3
2
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
{\displaystyle R={\frac {\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}{{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}}}
But since
x
=
ρ
cos
θ
{\displaystyle x=\rho \cos \theta }
y
=
ρ
sin
θ
{\displaystyle y=\rho \sin \theta }
d
x
d
θ
=
−
ρ
sin
θ
+
cos
θ
d
ρ
d
θ
{\displaystyle {\frac {dx}{d\theta }}=-\rho \sin \theta +\cos \theta {\frac {d\rho }{d\theta }}}
d
y
d
θ
=
ρ
cos
θ
+
sin
θ
d
ρ
d
θ
;
{\displaystyle {\frac {dy}{d\theta }}=\rho \cos \theta +\sin \theta {\frac {d\rho }{d\theta }};}
d
2
x
d
θ
2
=
−
ρ
cos
θ
−
2
cos
θ
d
ρ
d
θ
+
cos
θ
d
2
ρ
d
θ
2
{\displaystyle {\frac {d^{2}x}{d\theta ^{2}}}=-\rho \cos \theta -2\cos \theta {\frac {d\rho }{d\theta }}+\cos \theta {\frac {d^{2}\rho }{d\theta ^{2}}}}
d
2
y
d
θ
2
=
−
ρ
sin
θ
+
2
cos
θ
d
ρ
d
θ
+
sin
θ
d
2
ρ
d
θ
2
.
{\displaystyle {\frac {d^{2}y}{d\theta ^{2}}}=-\rho \sin \theta +2\cos \theta {\frac {d\rho }{d\theta }}+\sin \theta {\frac {d^{2}\rho }{d\theta ^{2}}}.}
Substituting these in (D ) and reducing,
R
=
[
ρ
2
+
(
d
ρ
d
θ
)
2
]
3
2
ρ
2
2
(
d
ρ
d
θ
)
2
−
ρ
d
2
ρ
d
θ
2
{\displaystyle R={\frac {\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}{\rho ^{2}2\left({\frac {d\rho }{d\theta }}\right)^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}}}}
Ans.
![{\displaystyle R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97530f670d9115e1f0052796113113dd374dd0a4)
![{\displaystyle R=\left[{\frac {\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}}{\left({\frac {dx}{d\theta }}\right)^{2}}}\right]^{\frac {3}{2}}\div {\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{\left({\frac {dx}{d\theta }}\right)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bd5304779f0168d342b016d446086fc815edbeb)
![{\displaystyle R={\frac {\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}{{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76a801bb57d4f3c0c3029afcb98cc0310eded690)
![{\displaystyle R={\frac {\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}{\rho ^{2}2\left({\frac {d\rho }{d\theta }}\right)^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f328bff9bf791158f6ca042374357fbfa6e269ff)
