Skip to main content

[Community Question] Linear-algebra: Vector Calculus proof part derivation.

One of our user asked:

I am following a proof in my vector calculus book but I am getting stuck.

Let T(s) be the the unit tangent vector at s. and let k(s) = norm( T’(s)) here T’(s) is orthogonal to T(s) and let N(s) be the Unit vector such that T’(s)=k(s)*N(s)

See photoenter image description here

Now, in the proof it says:

Differentiate N(s) • T(s) =0 Gives N’(s)•T(s)+N(s)•T’(s) =0

Hence N’(s)•T(s)=-k(s) But I don’t see how this Is derived?

Also , written on mobile as this is the only access I have at the moment, please go easy on the syntax.


Comments

Popular posts from this blog

[Community Question] Linear-algebra: non-negative matrix satisfying two conditions

One of our user asked: A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $B\ge 0$ . I want to find a non-negative matrix $B$ satisfying the following two conditions: (1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix. (2) There is a non-zero and non-negative vector $\vec{d}$ such that $(I-B)^{-1}\vec{d}\ge 0$ . I tried all the $2\times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).