+1 862 207 3288 3. Let I! =flx) and v = g(x) be two planar curves, such that in an interval [(7,1)] there holds 11 ≤ v
and at a particular point xo ∈ (a,b) there holds 11 = v. Show that the (scalar) curvatures ∣⇂↗≀∣ and k1,
considered as positive when the curvature vector is pointing upward, satisfy k,,(xo)
Hint: Observe that the two curves must be tangent at the indicated contact point. Make a
rotation preserving the sense of “upward”, and such that both curves become horizontal at xo, and
observe that the scalar curvature at any point of a curve is invariant under rotation. Show that the
curvature of a curve y = F (x) at a point where F ’(x) = 0 is just F ’ ’(x).
4. Definition: 11 = 0(1/1‘) means ∣⊔∣∣ is bounded as r → 0. II = o(l/r) means lrlll → 0 as r → 0.
Show that if ot + y < 75/2 and if K > 0 then every solution 11(x, y) of divT II = K11 in a wedge domain
of opening 2a and making contact angle 7 on the boundary satisfies 1! = 0(1/1‘), but there is no
solution satisfying 1/ = o(l/r). Hint: Follow the procedures in Notes 7 and in Notes 8.
Thus, every such solution must tend to infinity at some set ofpoints approaching the vertex, at
least as fast as the inverse of the distance of those points from the vertex, but there is no such set
of points at which any solution can approachl infinity faster.
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