The " Kissing " Circle A.K.A the Osculating Circle , ANNOTATED

Question

Find the parametric equations of the osculating circle to the curve resulting as the intersection of the plane that the vector valued function r(t) = < cost , sint, sint > lies and the cylinder x² + y² =1 at t = π/4.

Center of osculating circle :

Radius of osculating circle :

Answer 1

Hi Asked,

As usual, "Asked" questions are those for which a tutor couldn't figure out the student's problem set question, I suspect....

Anyway, visualization of problems is important. What is the plane of the figure referred to? Since it starts at x=1, (y,z) = (0,0) (for t=0) and must also include x=-1 (y,z)=(0,0) (for t=pi), and also includes point x=0, (y,z)=(1,1) (for t=pi/2), that is a plane that includes the x-axis, and (y,z)=(1,1) -- in short, at a 45 degree angle into the (y,z) plane.

OK, the stated cylinder is up the z-axis, with a radius of 1.

You should be able to visualize it now, a cylinder slicing a plane. That generates an ellipse; transform that into the (x,y) plane and solve by your usual geometric means. Hmm?

-- Cheers, --Mr. d.

Answer 2

IN A NUTSHELL

x = cos t , y = sin t , z= sin t. ⇒ x² +y² = 1 and y = z . Therefore the curve r(t) = < cost , sint , sint >

is the intersection of the right cylinder x² +y² = 1 and the plane y = z .

T(t) = ( 1 / √(1+cos²t)) < -sint, cost, cost > and T( π/4 ) = ( 1/√3)< -1,1,1> . A fast way of finding N ( π/4 ) is

to get the normal vector is to observe that the normal vector of the plane y = z is < 0, -1 , 1> hence

B= 1/√2 < 0,-1,1> Therefore since N = B x T then N = (1/√6) < -2, -1,-1>

The curvature at t = π/4 can be easily calculated , and κ(π/4) = 4√3/9.

Therefore the center of the osculating circle to the curve resulting as the intersection of the plane that the

vector valued function r(t) = < cost , sint, sint > lies and the cylinder x² + y² =1 at t = π/4 is at the tip of the vector



(√2/2) <1,1,1> + (3√3)/4 (1/√6) N = (√2/2) <1,1,1> + (3√3)/4 (1/√6)<-2,-1,-1> = < - √2/4, √2/8, √2/8 >.

And finally the parametric equation of the circle

C(t) = < - √2/4, √2/8, √2/8 >.+ (3√3)/4 ( N (π/4) cost + T (π/4) sint)

Answer 3

This is a pretty complicated problem that cannot be explained in text only. I solved this problem on Zoom and recorded a video. Here is the link:

https://us02web.zoom.us/rec/share/m-Tpe58x2oS-n-aYVV0R90dTcKWAyN2RELH7vmRrSg1Z6t5AknxElIW-8X7eu9Lw._VQNLBplNRsbw_ow?startTime=1629818058000

Not sure if WyzAnt allows posting of external links. Please let me know if this works. If not we will figure out something else.