# osculating circle

### osculating circle

Find the osculating circle for the parabola defined by $$\vec{r}(t) = \ <t^2, t>$$ at $$t = 0$$.
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### Re: osculating circle

The "osculating circle" to a curve, at a given point, is the circle that passes through that point, has the same tangent line as the curve at that point and has the same curvature. From the first two conditions, the osculating circle to $$x= y^2$$, at (0,0), must have center on the x- axis- the center is at (R, 0) and the radius is R for some number, R. The curvature of a circle of radius R is $$\frac{1}{R}$$. What is the curvature of $$x= y^2$$ at (0, 0)?

HallsofIvy

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Joined: Sat Mar 02, 2019 9:45 am
Reputation: 119

### Re: osculating circle

We can write $$x= y^2$$ as $$x= t^2$$, $$y= t$$. The curvature when x and y are given as parametric equations x= x(t), y= y(t) is given by $$\kappa= \frac{|x'y''- y'x''|}{x'^2+ y'^2}$$. With $$x= t^2$$, $$x'= 2t$$ and $$x''= 2$$. With $$y= t$$, $$y'= 1$$ and $$y''= 0$$.

So here, $$\kappa(t)= \frac{|2t(0)- 1(2)|}{4t^2+ 1}= \frac{2}{4t^2+ 1}$$ and, at t= 0, $$\kappa(0)= 2$$. The radius of curvature is 1/2 and the "osculating circle" is $$\left(x- \frac{1}{4}\right)^2+ y^2= \frac{1}{4}$$.

HallsofIvy

Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 119

### Re: osculating circle

Thank you for the help. Your explanation is clear. Can you tell me why the book answer is $$(x - \frac{1}{2})^2 + y^2 = \frac{1}{4}$$?
Guest

### Re: osculating circle

Probably because it is the correct answer!

In my response I said "the radius is 1/2" and then wrote 1/4 where I should have had 1/2!
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