# Thread: Unit Tangent & Unit Normal Vectors

1. ## Unit Tangent & Unit Normal Vectors

Find the unit tangent vector.

r (t) = 4ti

r'(t) = 4i

||r'(t)|| = root {4^2} = root {16} = 4

So, T (t) = 4i/4 = i.

Yes? No?

I now need the unit normal vector.

I am stuck here.

I know that T (t) = i (see above).

I need to find T'(t) and ||T'(t)||.

What is the derivative of i?

2. ## Re: Unit Tangent & Unit Normal Vectors

Hi there,
So you are close with your result for a unit vector tangent. It is actually:

T(t) = (4t/(4|t|))i = (t/|t|)i

The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

N(t) = (t/|t|)j

I think this will work....good luck,

Craig

3. ## Re: Unit Tangent & Unit Normal Vectors

Originally Posted by clombard1973
Hi there,
So you are close with your result for a unit vector tangent. It is actually:

T(t) = (4t/(4|t|))i = (t/|t|)i

The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

N(t) = (t/|t|)j

I think this will work....good luck,

Craig
this completely misses the point that there are infinite normal vectors to a line.

4. ## Re: Unit Tangent & Unit Normal Vectors

Yes, indeed there are....so......

5. ## Re: Unit Tangent & Unit Normal Vectors

I was able to solve the problem.

T (t) = i

N (t) = undefined

6. ## Re: Unit Tangent & Unit Normal Vectors

Originally Posted by clombard1973
Hi there,
So you are close with your result for a unit vector tangent. It is actually:

T(t) = (4t/(4|t|))i = (t/|t|)i

The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

N(t) = (t/|t|)j

I think this will work....good luck,

Craig
Sorry but this is incorrect.