Intersection of planes, curvature and osculating plane

1. Mar 5, 2012

trickycheese1

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?

2. Mar 5, 2012

HallsofIvy

Staff Emeritus
I assume you mean "intersection of the planes at (1, 0, -1)"

That was given as a "hint"? It isn't true. At (1, 0, -1), yes y= 0 at that point. It does not follow that y= 0 at any other point on the curve of intersection.
From x+ xy+ z^3= 0, $z= -x^{1/3}(1+ y)^{1/3}$ so the second equation , sin(xyz)= 0, becomes $sin(x^{4/3}y(1+ y)^{1/3})= 0$. Differentiate with respect to x and y.

3. Mar 5, 2012

trickycheese1

The hint wasn't from an instructor but an older student, seems it was incorrect.
If we differentiate with respect to x we get

d/(dx) sin(x^4/(3 y) (1+y)×1/3) = (4 x^3 (y+1) cos((x^4 (y+1))/(9 y)))/(9 y) = 0.

and if we differentiate with respect to y we get

d/(dy) sin(x^4/(3 y) (1+y)×1/3) = -(x^4 cos((x^4 (y+1))/(9 y)))/(9 y^2) = 0

I've only looked at examples where we work with position vectors of the form r(t) = x(t) * i + y(t) * j + z(t) * k, so I don't know what to do with the partial derivatives!