Can someone please post a detailed step-by-step procedure. Given the circle with a radius a, what is the differential equation of the circle.
Find the differential equation of all circles of radius a
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3quoting you, "what is the differential equation of the circle"? – 2011-12-03
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0Please reopen, it seems pretty clear what is asked, even Wolfram Alpha can verify the usual differential equations for circles and the like: https://www.wolframalpha.com/input/?i=d+x%28t%29%2Fdt+%3D+y%28t%29+and+d+y%28t%29%2Fdt+%3D+-x%28t%29 which is found in the answer of @lhf . – 2016-10-24
3 Answers
We can express standard result/formula for curvature rectangular coordinates
$$ \frac{y''}{(1+y'^2)^{3/2}} =\frac{1}{a} $$
as the required differential equation.
EDIT 1:
In polar co-ordinates the ODE for curvature is
$$ \dfrac { r^2 + 2 r ^{'2 }- r r ^{"} }{ (r^2 +r'^2)^{3/2}}=\frac{1}{a} \tag{1}$$
The above can be derived from intrinsic/natural differential equation of a circle is
$$ \frac{d \phi }{ds} = \frac{d (\theta + \psi ) }{ds}=\frac{1}{a}\tag{2}$$
$$ =\dfrac{ \sin \psi}{r} + \frac{d}{ds} ( \tan ^{-1} \frac{r}{r'} ) \tag{3} $$
where $\phi$ is angle to x-axis, $ \psi$ is between arc and radius vector, $$ \tan \psi = \dfrac {r}{r^{'}} \tag{4} $$
Introducing above into (3) and differentiating, LHS is
$$ \frac{1}{\sqrt{ r^2 +r'^2 }} + \frac{r'^2 - r r ^{''}}{ r^2 +r'^2 } \frac{1}{\sqrt{ r^2 +r'^2 }} \tag{5=1} $$
Circle equation $$ (x-C_1)^2+(y-C_2)^2=a^2\quad (0) $$ Differentiate twice by $x$ $$ (x-C_1)+y'(y-C_2)=0\quad (1) $$ $$ 1+(y-C_2)y''+(y')^2=0\quad (2) $$ From $(2)$ we obtain $$ C_2=y+\frac{(y')^2+1}{y''} $$ Then substitute in $(1)$ and $(0)$ $$ (x-C_1)-y'\frac{(y')^2+1}{y''}=0\quad(3) $$ $$ (x-C_1)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2\quad(4) $$ From $(3)$ we obtain $$ x-C_1=y'\frac{(y')^2+1}{y''} $$ Then substitute in $(4)$ $$ \left(y'\frac{(y')^2+1}{y''}\right)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2 $$ After some simplifications we get $$ ((y')^2+1)^3=(ay'')^2 $$
From the implicit equation of the circle $(x-u)^2+(y-v)^2=a^2$, you get $$x'(x-u)+y'(y-v)=0$$ by implicit differentiation. Add the initial condition $$x(0)=u+a, \quad y(0)=v$$
You can write the differential equations as $$ x'=-y+v, \quad y' = x-u $$ which is especially nice for circles centered at the origin.
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0Thanks Norbert and lhf. Appreciate your answers. I am however surprised to see clean mathematical notations here. Do you use any software as such to paste these equations here? (Pardon me, I am using math.stackexchange for the first time. – 2011-12-03
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0@Nikhil, this site understands TeX. View the source of the web pages to see how it's done. See also http://math.stackexchange.com/editing-help. Welcome to MSE. – 2011-12-03
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0Thanks :-) BTW, I was also wondering if there is any kind of math application, which would give me sort of help to find the steps as in for example: Evaluate the integral x^2 (sqrt(a pow3 + x pow3)) dx. – 2011-12-13