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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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    Please 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 3

1

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} $

7

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

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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    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.2011-12-13