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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.

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closed as unclear what you're asking by Rory Daulton, yoknapatawpha, Alex Provost, ComplexPhi, kamil09875 Jan 23 at 21:55

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

quoting you, "what is the differential equation of the circle"? – Ilya Dec 3 '11 at 18:40
Does this other account belong to you as well? If so, we can flag the moderators to merge these two accounts; it'll make it convenient for you to keep track of your questions. Thanks, – Srivatsan Dec 3 '11 at 23:45
Nope, that account does not belong to me Srivatsan. – Nikhil Mulley Dec 4 '11 at 5:12
up vote 1 down vote accepted

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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Thanks 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. – Nikhil Mulley Dec 3 '11 at 19:14
@Nikhil, this site understands TeX. View the source of the web pages to see how it's done. See also Welcome to MSE. – lhf Dec 3 '11 at 19:18
Thanks :-) 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. – Nikhil Mulley Dec 13 '11 at 12:46

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 $$ ((y')^2+1)^3=(ay'')^2 $$

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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.


In polar co-ordinates the ODE for curvature is

$$ \dfrac { r^2 + 2 r ^{'2 }- r r ^{"} }{ (r^2 +r'^2)^{3/2}} \tag{1}$$

The above can be derived from intrinsic/natural differential equation of a circle is

$$ \frac{d \phi }{ds} = \frac{d (\theta + \psi ) }{ds} \tag{2}$$

$$ =\dfrac{ \sin \psi}{r} + \frac{d}{ds} ( \tan ^{-1} \frac{r}{r'} ) \tag{3} $$

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,

$$ \frac{1}{\sqrt{ r^2 +r'^2 }} + \frac{r'^2 - r r ^{''}}{ r^2 +r'^2 } \frac{1}{\sqrt{ r^2 +r'^2 }} \tag{5=1} $$

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