# Find the Unit Vector

1. Sep 4, 2008

### JeffNYC

A) Find the unit vector parallel to tan(x) at (Pi/4, 1)

B) Find the unit vector normal to tan(x) at (Pi/4, 1)

dy/dx tanx = sec2x and sec2x evaluated at x = Pi/4 is 2. So the slope of the parallel line is 2, but how do I then derive the unit vector?

2. Sep 5, 2008

### Defennder

One way is to parametrise the path r(t) corresponding to the graph of tan(x). Then find r'(t). Another way is to note that dy/dx gives you the relationship between the y and x components of the unit vector. Draw out the dy/dx triangle to see why. Now you should be able to find it.

B)Now you've got the vector you should be able to find the unit vector perpendicular to it.

3. Sep 5, 2008

### JeffNYC

Defennder,

Can you elaborate on what y' tells me about the y and x components of the unit vector, or point me to the triangle you mentioned?

4. Sep 5, 2008

### HallsofIvy

dy/dx is the ratio of the y-component of the tangent vector to the x-component. Coupling that with the requirement that the length be 1 lets you determine both. For example, if dy/dt= vy/vx= 2 then vy= 2vx. If then $\sqrt{v_x^2+ v_y^2}= 1$, we have vx2+ vy2= vx2+ 4vx2= 5vx[/sub]2= 5 so $v_x= 1/\sqrt{5}$ and $v_y= 2/\sqrt{5}$.