# Homework Help: Finding unit tangent and unit normal vectors

1. Jul 17, 2011

### wyosteve

1. The problem statement, all variables and given/known data

The graph shows the vector-valued function r(t). It is given that r(1) = <-1,1> and r'(1) = <1,2>. Find the unit tangent vector T(1) and unit normal vector N(1).
(the graph given is that of y = x^2 shifted 2 units left.)

2. Relevant equations

(1) T(t) = r'(t)/||r'(t)||
(2) N(t) = T'(t)/||T'(t)||

3. The attempt at a solution
After playing around for a bit I found r(t) = <t-2,t^2> so r'(t) = <1,2t>

Using this in eq. (1) I get T(t) = <1,2t>/sqrt(1+4t^2).

T(t) = <1/sqrt(1+4t^2),2t/sqrt(1+4t^2)>

T(1) = <1/sqrt5,2sqrt5> (this vector im fairly confident is correct)

T'(t) = <-4t/(4t^2+1)^(3/2),2/(4t^2+1)^(3/2)>

||T'(t)|| = sqrt((-4t/(4t^2+1)^(3/2))^2+(2/(4t^2+1)^(3/2))^2)

which reduces to sqrt(16t^2/(4t^2+1)^3+(4/(4t^2+1)^3))

Then using T'(t) and ||T'(t)|| in eq. (2) I get

N(t) = <-4t/(4t^2+1)^(3/2),2/(4t^2+1)^(3/2)>/sqrt(16t^2/(4t^2+1)^3+(4/(4t^2+1)^3)

N(1) = <-4/5^(3/2),2/5^(3/2)>/(4/25) or approx. <-2.236,1.118>

Drawing this vector on the graph it seems too big to be a unit vector. Can anyone spot any errors in my calculations? Am I even close? I feel like I may be making this harder then neccesary. Thanks in advance for any help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 18, 2011

### ehild

It is not needed to solve for arbitrary point of the curve. The point is given, and the tangent vector is r'(1) = <1,2> The normal vector is perpendicular to the tangent vector. Just make a vector <a,b> that is normal to <1,2>. What should be their scalar product?

ehild

3. Jul 18, 2011

### wyosteve

As I thought I was making this more diffucult then needed. Ok the scalar product needs to equal zero for the two vectors to be perpendicular. So <a,b> . <1,2> = 0
a+2b = 0. Or since im trying to find a unit vector should I be finding T(t) . N(t) = 0?
Which would be <a/sqrt5,2b/sqrt5> N(t) = <-2,1> N(1) = <-2/sqrt5,1/sqrt5>
Plotting this last vector it looks right.

4. Jul 18, 2011

### ehild

You are welcome.

ehild

5. Jul 18, 2011

### LCKurtz

Solving for a normal by using the dot product like that gives two possibilities for the direction of the Normal, in your case <-2,1> or <2,-1> normalized both work. Usually, if you are looking for THE normal it would be understood to be the normal pointing towards the direction the curve is bending (the concavity). So you need to make sure you pick the right one.

6. Jul 18, 2011

### ehild

LCKurtz is right, the unit normal vector to a curve in the plane x,y is defined so that the unit tangent vector T, the unit normal vector N and the unit vector along the z axis, ez define a right-handed coordinate system. Therefore N is obtained as the vector product N=ezxT. The vector product is perpendicular both to the z axis and to T, so it is also in the x,y plane, and normal to T, but it is unique unlike the one obtained from zero scalar product.
If T=(u,v,0) then N=(-v,u,0)

ehild

7. Jul 18, 2011

### wyosteve

good points, thanks everyone for the help.