Question refers to limits and continuity. 2 (a): Show that as x approaches 0+, t

Question

Question refers to limits and continuity.
2 (a): Show that as x approaches 0+, the secant circles approach a limiting position given by the circle that passes through the origin and is centered at (0,1/2). As shown in the accompanying figure, this circle is the osculating circle to the graph of y=x^2 At the origin.

(b): Show that if we replace the curve y=x^2 by the curve y=f (x), where f is an even function, then the formula for C (x) becomes:
C (x)=1/2 [f (0)+f (x)+(x^2)/(f (x)-f (0))]

[Here we assume that f (x)f (0) for positive values of x close to 0.] if lim (x->0+ C (x)=Lf (0), then we define the osculating circle to the curve y=f (x) at (0,f (0)) to be the unique circle through (0,f (0)) with center (0,L). If C (x) does not have a finite limit different from f (0) as x approaches 0+, then we say that the curve has no osculating circle at (0,f (0)).

Explanation / Answer

you can look up the formulas for calculating curvature https://en.wikipedia.org/wiki/Curvature#Local_expressions

There are two choices for the curvature formula: one if you choose to think of the curve in parametric form r(t)=(x(t),y(y))=(t,t2)r(t)=(x(t),y(y))=(t,t2), and a different one if you choose to think of the curve as the graph of y=x2y=x2.

Anyway, use whichever formula you want to calculate curvature at t=1t=1.

Then, the radius of the osculating circle is =1/=1/.

Find the tangent at t=1t=1. It's in the direction (1,2)(1,2). So the unit tangent is (1/5–,2/5–)(1/5,2/5).

Rotate 90 degrees to get the unit normal, so this is (2/5–,1/5–)(2/5,1/5).

To get to the center of the osculating circle, travel a distance along this normal vector from the point (1,1)(1,1).

Now you know the center and radius of the osculating circle, so you can write down its equation.

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