I'm presented with finding the equation of the osculating circle at the local minimum of $\mathbf f(x) = 3x^3-9x^2+5x-1 $.
Finding the local minimum wasn't that hard; I take the first derivative of $\mathbf f$, finding the critical values $\frac {5}{3}$ and $\frac{1}{3}$. Taking the second derivative, I can determine that $\frac {5}{3}$ is the local minimum, with a corresponding y-value of $-\frac{34}{9}$.
I then get stuck here: I don't know how to find the osculating circle. I've learned that the osculating circle has a radius $\frac {1}{\kappa}$, but the equation for curvature that I've learned is $\kappa = \frac{||\mathbf T'(t)||}{||\mathbf r'(t)||}$. This would require reparameterizing the function into something like $(t,3t^3-9t^2+5t-1)$, right? I tried implementing this, but my unit tangent vector becomes a mess with rather tedious derivations to be solved.
There has to be an easier way! Can someone provide some guidance here?
Best Answer
For a two-dimensional parametric curve $(x,y)$, the signed curvature can be explicitly obtained as $$k=\frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}$$ Here $x,y$ are understood to be functions of $t$. For your cubic curve this works out as $k=12$, so the radius of the osculating circle is $\frac1{12}$.
Now note that the tangency is at a local minimum, so the osculating circle should be directly above it; the vector from the osculating centre to the point of tangency points straight down. So the centre of the osculating circle is at $(5/3,-34/9)+(0,1/12)=(-5/3,-133/36)$. From this you should be able to work out the circle's equation.