Binormal Vector: Direction at Point x

  • Thread starter Thread starter _N3WTON_
  • Start date Start date
  • Tags Tags
    Vectors
_N3WTON_
Messages
350
Reaction score
3

Homework Statement


Suppose that you draw a point "x" in the middle of a page, and a curve going through "x" in such a way that the curve is going straight upward at the point "x" and is curving to the left. In what direction does the binormal vector "B" (at "x") point?

Homework Equations



None

The Attempt at a Solution


Hello, I am having some trouble really visualizing this problem. I know about the right-hand rule, but I am unsure how to apply it here (or whether or not it does even apply here). My thought on the problem was that since if the curve is going straight upward at "x" and curving left, then the binormal vector would point out of the page, no?
 
Last edited:
Physics news on Phys.org
This sounds correct. Since the point is going straight up, the tangent vector is straight up, which means the normal vector points to the right and the binormal vector points out of the page.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top