Curvature — Question 6

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Question 6

Find the osculating circle of the parabola y=x2y=x^2 at its vertex. State its center, radius, and equation, and explain which side of the tangent contains the center.

Original worksheet page 1: question and worked solution for 6-10-006
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Question 6 – Solution

Strategy Use curvature for the radius and the principal normal for the center direction.

See the diagram in the original worksheet below.

Curvature and normal From Question 2, κ(0)=2\kappa(0)=2, so ρ=1/κ=1/2\rho=1/\kappa=1/2. At the vertex, the unit tangent points right and the parabola bends upward, so 𝑵=⟨0,1⟩\mathbf N=\left\langle 0,1\right\rangle.

Center and equation C=(0,0)+12⟨0,1⟩=(0,12),C=(0,0)+\frac 12\left\langle 0,1\right\rangle=\left(0,\frac 12\right), x2+(y−12)2=14.\boxed{x^2+\left(y-\frac 12\right)^2=\frac 14}.

Geometry The center lies on the concave side of the curve, above the horizontal tangent. The circle has second-order contact with the parabola at the vertex.

Original worksheet page 2: question and worked solution for 6-10-006

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