# Thread: Curverture of a Osculating Circle at a Point

1. ## Curverture of a Osculating Circle at a Point

I need help with this curverture problem. I am having troubling setting up the equations. For the curverture k, i calculated using the cross product of magnitude of the r'(t)xr"(t) and then divided by the magnitude r'(t) cubed. I took the reciprocal of that and my answer for Radius = 16^2/sqrt(257). I am not too sure of the numbers. I plugged in t=pi/2 before doing the cross product and before the cubing.

I am stuck on how to find the center and setting up the equation. I have 2 submissions left! Help would be appreciated.

2. ## Re: Curverture of a Osculating Circle at a Point

Assuming that 16^2/sqrt(257) is the radius then the center of curvature lies on the line through (0, 16) perpendicular to the curve and distance 16^2/sqrt(257) from that point. I presume you have already found, or can find, the tangent line to the curve at that point so finding the equation of the perpendicular should be easy.

But are you required to do that? Your curve is given by r= <x, y>= <16 cos(t), 16 sin(t)>. So x= 16 cos(t) and y= 16 sin(t). x^2+ y^2= 16^2 cos^2(t)+ 16 sin^2(t)= 16(sin^2(t)+ cos^2(t))= 16. So the curve IS a circle! It should be clear what the "osculating circle" to a circle is!

3. ## Re: Curverture of a Osculating Circle at a Point

Sorry, I am still not understanding this concept very well, and I am just trying to follow the formula. So, is the C(t) equation a vector of a circle in terms of t?

4. ## Re: Curverture of a Osculating Circle at a Point

That's what the problem said, isn't it? "Find a parameterization of the osculating circle". I would think that, if you are dealing with "osculating circles", "cross products", and "derivatives" you certainly should have learned that "$\displaystyle x^2+ y^2= 16^2$" is the equation of a circle! So, again, what is the osculating circle of a point on a circle?

5. ## Re: Curverture of a Osculating Circle at a Point

Okay. i think i get it. So the circle they give us is parametrized by x = 16cos(t) and y=16sin(t)

so the equation of the osculating circle is the whole whole itself, which is c(t) = (16cos(t))^2 +(16sin(t))^2-16^2?

6. ## Re: Curverture of a Osculating Circle at a Point

Oh, dear. Did you not notice that (16sin(t))^2+ (16cos(t))^2- 16^2 is identically 0? c(t) must be, again, a pair of parametric equations. You were given that parametric equations of this figure, a circle, are <16 cos(t), 16 sin(t)> and, as you now say the osculating circle at any point on a circle is that circle. So parametric equations for the osculating circle are parametric equations for the circle!

7. ## Re: Curverture of a Osculating Circle at a Point

Sorry, i probably made no sense! I'm just starting to make sense of the vector stuff. Anyways, is the curverture c(t) of the convulating circle always the equation of the circle for any circle?

The answer is just <16cos(t),16sin(t)> ? The same they gave us?

8. ## Re: Curverture of a Osculating Circle at a Point

NO! The curvature of any circle, at any point is 1 over the radius of the circle.

9. ## Re: Curverture of a Osculating Circle at a Point

Okay, gotcha. I'm confusing the terms! C(t) is different from curverture. And in this case, we don't really need to find the radius of the circle because we aren't finding the curvature. The c(t) is just <16cost,16sint>