Normal vector to a curve

In summary, the conversation discusses finding the unit vector normal to a given curve at a specific point, with the additional restriction that the vector cannot have components along a certain line. The conversation also includes some calculations and suggestions for how to approach the problem.
  • #1
steve23063
23
0
Hi I'm having trouble finishing this problem and I'd appreciate it if someone could help me out.

I need to find the unit vector normal to x=y^2=z^3 at the point (1,1,1) and it can't have components along the line x=y=z

Heres what I did so far:
1. dx=2ydy=3z^2dz
2. At (1,1,1) dx=2dy=3dz
3. differential length vector dl=dxax+(1/2)dxay+(1/3)dxaz
(ax is the unit vector in the x-direction etc)
4. Rewrote dl as dl=[ax+(1/2)ay+(1/3)az)]dx

If I use the dot product and I let an=normal vector then the dot product of dl and an should equal 0. OK that's as far as I can get :(
 
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  • #2
Let's see. If you if parametrize given curve you would get [tex]\gamma(t)=(t, t^\frac{1}{2}, t^\frac{1}{3})[/tex]. Tangent vector to a curve is given by [tex]\vec{\tau}=\frac{d\gamma(t)}{dt}=[1, \frac{1}{2}t^\frac{1}{2},\frac{1}{3}t^{-\frac{2}{3}}][/tex]. At point M=(1,1,1) it becomes [tex]\vec{\tau_M}=[1,\frac{1}{2},\frac{1}{3}][/tex]. Vector that is along x=y=z direction is given by [tex]\vec{a}=[1,1,1][/tex]. So, your vector should be normal to the curve at point M (which means it should be normal to [tex]\vec{\tau_M}[/tex] and should not be in direction of [tex]\vec{a}[/tex]. To be sure it is so, set your vector to be [tex]\vec{b}=\vec{\tau_m}\times \vec{a}[/tex]. Then just divide resulting vector by its magnitude and this is it.
 
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  • #3
Oh yes, excuse me. My fingers were faster then brain. Ill fix it immidiatelly
 
Last edited:
  • #4
To remain annoying, please add a minus sign to the differentiated y-component's exponent..:smile:
 
  • #5
Hmm, I don't see edit button under my post.
 

What is a normal vector to a curve?

A normal vector to a curve is a vector that is perpendicular to the tangent line at a given point on the curve. It represents the direction in which the curve is changing at that point.

How is a normal vector calculated?

A normal vector can be calculated by taking the derivative of the curve's equation and evaluating it at a specific point. This derivative represents the slope of the tangent line, and the normal vector is the negative reciprocal of this slope.

What is the significance of a normal vector to a curve?

A normal vector is important because it can be used to find the direction of acceleration for objects moving along the curve. It is also useful in finding the curvature of the curve at a specific point.

Can a curve have more than one normal vector?

Yes, a curve can have infinitely many normal vectors at different points along the curve. This is because the direction of the normal vector is constantly changing as the curve changes.

How is a normal vector used in real-world applications?

A normal vector is commonly used in physics and engineering to calculate forces and accelerations in curved paths. It is also used in computer graphics to create realistic lighting and shading effects on curved surfaces.

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