How do I find the equation of an osculating circle when I'm given the parabola? This is a question given out by my calculus professor, and I'm completely stumped as to how I need to go about solving it.
Let the parabola $y=x^2$ be parameterized by $r(t)=ti+t^2j$. Find the equation of the osculating circle for the parabola at $t=1$ by performing the following steps.
a) Find the formula for $\kappa(t)$, the curvature of the parabola and compute for $\kappa(1)$
b) The radius of the osculating circle we want is $\rho={1\over \kappa(1)}$. Find the center of the osculating circle by computing the unit normal $N(1)$ and calculating the sum $C=r(1)+\rho N(1)$.
c) Use the center and the radius of the osculating circle to write the equation of the circle in standard form.
To begin with I'm not sure how to find the formula for the curvature of the parabola, and even from there I don't know what to do. Where do I begin?
 A: As the comment from Rory Daulton said, you can look up the formulas for calculating curvature here. These same formulas must be in your class notes or textbook, too, or else your teacher wouldn't be asking you this question.
There are two choices for the curvature formula: one if you choose to think of the curve in parametric form $\mathbf{r}(t) = \big(x(t), y(y)\big)=(t,t^2)$, and a different one if you choose to think of the curve as the graph of $y=x^2$.
Anyway, use whichever formula you want to calculate curvature $\kappa$ at $t=1$.
Then, the radius of the osculating circle is $\rho = 1/\kappa$.
Find the tangent at $t=1$. It's in the direction $(1,2)$. So the unit tangent is $(1/\sqrt5, 2/\sqrt5)$.
Rotate 90 degrees to get the unit normal, so this is $(-2/\sqrt5, 1/\sqrt5)$.
To get to the center of the osculating circle, travel a distance $\rho$ along this normal vector from the point $(1,1)$.
Now you know the center and radius of the osculating circle, so you can write down its equation.
