How to Find the Unit Vector for a Parallel Line at a Given Point on tan(x)?

[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?

 

[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.

 

[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?

 

[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 let's you determine both. For example, if dy/dt= v_y/v_x= 2 then v_y= 2v_x. If then $\sqrt{v_x^2+ v_y^2}= 1$, we have v_x²+ v_y²= v_x²+ 4v_x²= 5v^{x[/sub]2= 5 so $v_x= 1/\sqrt{5}$ and $v_y= 2/\sqrt{5}$.}