# Normal vector to a curve

1. Feb 12, 2006

### steve23063

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 :(

2. Feb 12, 2006

### tomkeus

Let's see. If you if parametrize given curve you would get $$\gamma(t)=(t, t^\frac{1}{2}, t^\frac{1}{3})$$. Tangent vector to a curve is given by $$\vec{\tau}=\frac{d\gamma(t)}{dt}=[1, \frac{1}{2}t^\frac{1}{2},\frac{1}{3}t^{-\frac{2}{3}}]$$. At point M=(1,1,1) it becomes $$\vec{\tau_M}=[1,\frac{1}{2},\frac{1}{3}]$$. Vector that is along x=y=z direction is given by $$\vec{a}=[1,1,1]$$. So, your vector should be normal to the curve at point M (which means it should be normal to $$\vec{\tau_M}$$ and should not be in direction of $$\vec{a}$$. To be sure it is so, set your vector to be $$\vec{b}=\vec{\tau_m}\times \vec{a}$$. Then just divide resulting vector by its magnitude and this is it.

Last edited: Feb 12, 2006
3. Feb 12, 2006

### tomkeus

Oh yes, excuse me. My fingers were faster then brain. Ill fix it immidiatelly

Last edited: Feb 12, 2006
4. Feb 12, 2006

### arildno

To remain annoying, please add a minus sign to the differentiated y-component's exponent..

5. Feb 13, 2006

### tomkeus

Hmm, I don't see edit button under my post.