# Thread: Osculating circle at given points

1. ## Osculating circle at given points

Find the osculating circle at the given points.

$r(t) = $ at $t=0$

I really have no idea where to even start this one. My professor has not gone over anything about osculating circles

If i could get any tips on how to do this one that would be great

Thank you!

2. Originally Posted by mybrohshi5
Find the osculating circle at the given points.

$r(t) = $ at $t=0$

I really have no idea where to even start this one. My professor has not gone over anything about osculating circles

If i could get any tips on how to do this one that would be great

Thank you!
To start have a look here:

Osculating Circle -- from Wolfram MathWorld

or here:

Osculating circle - Wikipedia, the free encyclopedia

3. The "osculating circle" to the graph of a fuction, f, at a given point, is the circle that passes thorugh that point, whose center lies on the normal line to that graph at that point, and whose radius is 1 over the curvature.

So the first thing you want to do is to identify the normal vector to the curve and find the curvature at that point.

By the way, you might find it intersting to look up "osculate" in a dictionary.

4. Ok i found the normal vector using $N(t) = \frac{T'(t)}{||T'(t)||}$ to the curve (it is very messy so i am not going to type it up).

The curvature k is just $k(t) = \frac{||T'(t)||}{||r'(t)||}$

Not sure where to go from here...

5. so, this is what i have came up with....

r(0) = <0,0> so the circle is at the origin
N(0) = <0,1>
k(0) = $\frac{2}{\sqrt(5)}$
p = radius = $\frac{\sqrt(5)}{2}$

i know the equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$

i am just not sure how to finish it up now

6. To find the center of the osculating circle, you need a line to the center. You have a direction (N(t)) and you have the magnitude (the radius). Make a vector line out of it and find the point @ t = 1.

for r^2, you can simply put 5/4.