(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The equations sin(xyz) = 0 and x + xy + z^3 = 0 define planes in R^3. Find the osculating plane and the curvature of the intersection of the curves at (1, 0, -1)

2. Relevant equations

Osculating plane of a curve = {f + s*f' + t*f'' : s, r are reals}

Curvature = ||T'|| where T is the unit tangent vector

3. The attempt at a solution

I guess my biggest doubt here is determining the position vector I want to be working with. Since we're looking at the point (1, 0, -1), then sin(zyx) = 0 implies that y=0. (I got this hint but I don't really understand it). So now we got the intersection x + z^3 = 0, and if we parametrize x(t) = t^3, z(t) = -t and y(t) = 0 then we get r(t) = t^3 * i - t * k where (i, j, k) is the standard basis for R^3. Now we differentiate and take lengths in turns of r to get the vectors we want to work with to get the curvature and osculating plane. Is this the correct method?

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# Homework Help: Intersection of planes, curvature and osculating plane

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