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  1. MHB Apprentice

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    #1
    Hello. I study differential geomety but I think this is difficult. Please help me.

    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..

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    #2
    Quote Originally Posted by bw0young0math View Post
    Hello. I study differential geomety but I think this is difficult. Please help me.

    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 $1\cdot \mathbf X_u(t,1) + 0 \cdot \mathbf X_v(t,1)$.

  3. MHB Apprentice

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    #3 Thread Author
    Thank you so much. So I understand this! Thank you!

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