Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
1  
Nope, that account does not belong to me Srivatsan. –  Nikhil Mulley Dec 4 '11 at 5:12
add comment

2 Answers

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.

share|improve this answer
    
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 math.stackexchange.com/editing-help. 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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.