# Thread: Find the Osculating Circle

1. ## Find the Osculating Circle

Cheers everyone.

I have to find the osculating circle of r(t) = <2cos(t),3sin(t)> at t = $pi$/2.

I find r'(t)=<-2sin(t),3cos(t)> and |r'(t)|=(4sin^2(t)+9cos^2(t))^(1/2). I converted that into |r'(t)| = (4 + 5cos^2(t))^(1/2). I am stuck on taking the derivative of the Unit Tangent vector to find the Principal Normal Unit vector in order to find the curvature. The derivative is getting horrible. Am I on the right track?

2. Originally Posted by palmaas
Cheers everyone.

I have to find the osculating circle of r(t) = <2cos(t),3sin(t)> at t = $pi$/2.

I find r'(t)=<-2sin(t),3cos(t)> and |r'(t)|=(4sin^2(t)+9cos^2(t))^(1/2). I converted that into |r'(t)| = (4 + 5cos^2(t))^(1/2). I am stuck on taking the derivative of the Unit Tangent vector to find the Principal Normal Unit vector in order to find the curvature. The derivative is getting horrible. Am I on the right track?
I'm not sure what method you are using here. The curve r(t) = (2cos(t),3sin(t)) is an ellipse centred at the origin, with its axes parallel to the coordinate axes. When $t = \pi/2$, r(t) is the point (0,3) on the y-axis, so the centre of curvature will also be on the y-axis. To locate it, you just need to know the radius of curvature at that point.

With $x = 2\cos t$ and $y = 3\sin t$, the curvature is given by $\kappa = \frac{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}{(\dot{x}^2 + \dot{y}^2)^{3/2}} = \frac{6\sin^2t+6\cos^2t}{(4\sin^2t+9\cos^2t)^{3/2}}$. Evaluate that at $t = \pi/2$, then the radius of curvature is $1/\kappa$, and that will tell you where the centre of curvature is.