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Find the equation of an osculating circle?
I need help with this question. I have already found k = 1/2, the radius 2, and N = -i -j but I am unsure how to locate the center of the circle
Find an equation for the circle of curvature of the curve r(t) = 2(ln t)i - [t+ (1/t)]j
e^-2 <= t <= e^2, at the point (0,-2) where t = 1
1 Answer
- Some BodyLv 73 years ago
The radius from the point of intersection to the center of the circle is perpendicular to the tangent line at the point.
r'(t) = (2/t)i - [1 - (1/t^2)]j
r'(1) = 2i
The direction perpendicular to this is, of course, ±j.
To find the sign, we use second derivative:
r"(t) = (-2/t^2)i - (2/t^3)j
r"(1) = -2i - 2j
The second derivative is negative, so the curve is concave down.
The center is at:
(0, -2 - r)
(0, -2 - 2)
(0, -4)
