this is the problem.

Let surface X(u,v)=(u,v,u^2+v^2). Then a point p=(1,1,2) in on X.
principal vector at (1,1,2) is (1,1)&(1,-1). And principal curvature at (1,1) is 2/27. And pricipal curvature at (1,-1) is 2/3.

Let a curve a(t)=(t,1,t^2+1) on X. Then find normal curvature kn at (1,1,2) on a(t)⊆X .

I have a question.
1.Why the principal vector is not 3-dimension but 2- dimension?
If the principal vector above are not correct, how can I find the pricipal vector?
In fact, the suface is very smooth like surface of revolution but not suface of revolution. So I think that kn can't be applied latitude&longitude.( If the surface is revolution, I wanted Euler's thm using latitude&longitude as principal vector for finding kn.)
How can I find the principal vector? If you can explain with some pictiure, I really appreciate..

2.And how can I find the kn?

3.In fact, I have a solution. But I think the solution is a litte strange.
Because, it says, the direction of tangent vector of a'(t) is (1,0)& the angle between a'(t)& principal vector is π/4 (45').
But I can't understand why the direction of tangent vector of a'(t) is (1,0).

If you can explain with some pictiure, I really appreciate..

2. Originally Posted by bw0young0math

this is the problem.

Let surface X(u,v)=(u,v,u^2+v^2). Then a point p=(1,1,2) in on X.
principal vector at (1,1,2) is (1,1)&(1,-1). And principal curvature at (1,1) is 2/27. And pricipal curvature at (1,-1) is 2/3.

Let a curve a(t)=(t,1,t^2+1) on X. Then find normal curvature kn at (1,1,2) on a(t)⊆X .

I have a question.
1.Why the principal vector is not 3-dimension but 2- dimension?
If the principal vector above are not correct, how can I find the pricipal vector?
In fact, the suface is very smooth like surface of revolution but not suface of revolution. So I think that kn can't be applied latitude&longitude.( If the surface is revolution, I wanted Euler's thm using latitude&longitude as principal vector for finding kn.)
How can I find the principal vector? If you can explain with some pictiure, I really appreciate..

2.And how can I find the kn?

3.In fact, I have a solution. But I think the solution is a litte strange.
Because, it says, the direction of tangent vector of a'(t) is (1,0)& the angle between a'(t)& principal vector is π/4 (45').
But I can't understand why the direction of tangent vector of a'(t) is (1,0).

If you can explain with some pictiure, I really appreciate..
Hey bw0young0math!

1. The principal curvature is a linear combination of $\mathbf X_u$ and $\mathbf X_v$.
As such it can be represented either as a 3D vector, or as the tuple of linear combination factors.
In other words, your principal vectors are $\mathbf X_u + \mathbf X_v$ respectively $\mathbf X_u - \mathbf X_v$.

Note that $(u,v) \mapsto u\mathbf X_u + v\mathbf X_v$ is the tangential plane.

2. Curvature can be calculated with either
$$\kappa = \left\| \d{\mathbf T}{s} \right\|$$
where $\mathbf T$ is the tangent vector, reparametrized to be of unit speed.

Or we can use:
$$\kappa = \frac{\|\mathbf a' \times \mathbf a''\|}{\|\mathbf a'\|^3}$$

3. The tangent vector of $\mathbf a$ is $\mathbf a' = (1, 0, 2t)$.
And we have $\mathbf X_u = (1,0,2u)$ and $\mathbf X_v = (0,1,2v)$.
In other words, we can express the tangent as $1\cdot \mathbf X_u(t,1) + 0 \cdot \mathbf X_v(t,1)$.