How do you find the normal unit vector to a parametric curve?

[member 508213]

Homework Statement

I am supposed to find the tangent and normal unit vector to r(t)=<2sint,5t,cost>.

Homework Equations

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The Attempt at a Solution

r1(t)=<2cost,5,-sint> which is a tangent vector to the curve, and then to make it a unit vector I would multiply by 1/(sqrt(4cos^2t+25+sin^2t).

Finding the normal unit vector is a little bit more confusing for me. I understand that it is supposed to be the derivative of the tangent vector divided by the magnitude (to make it length 1) but I am surprised because taking the derivatives of <2cost/sqrt(4cos^2t+25+sin^2t), 5/sqrt(4cos^2t+25+sin^2t),-sint/sqrt(4cos^2t+25+sin^2t)> and then turning that into a unit vector is extremely messy. Would it be possible to instead simply take the derivative of the tangent vector (not the unit tangent vector) r1(t)=<2cost,5,-sint> and get r2(t)=<-2sint,0,-cost> and then make that into a unit vector by dividing by sqrt(4sin^2t+cos^2t) or would this be incorrect.

Thanks

 

[SteamKing]

Homework Statement

I am supposed to find the tangent and normal unit vector to r(t)=<2sint,5t,cost>.

Homework Equations

.

The Attempt at a Solution

r1(t)=<2cost,5,-sint> which is a tangent vector to the curve, and then to make it a unit vector I would multiply by 1/(sqrt(4cos^2t+25+sin^2t).

Finding the normal unit vector is a little bit more confusing for me. I understand that it is supposed to be the derivative of the tangent vector divided by the magnitude (to make it length 1) but I am surprised because taking the derivatives of <2cost/sqrt(4cos^2t+25+sin^2t), 5/sqrt(4cos^2t+25+sin^2t),-sint/sqrt(4cos^2t+25+sin^2t)> and then turning that into a unit vector is extremely messy. Would it be possible to instead simply take the derivative of the tangent vector (not the unit tangent vector) r1(t)=<2cost,5,-sint> and get r2(t)=<-2sint,0,-cost> and then make that into a unit vector by dividing by sqrt(4sin^2t+cos^2t) or would this be incorrect.

Thanks

No, the unit normal vector is defined as the derivative of the unit tangent vector, which derivative is then divided again by its magnitude:

http://mathwiki.ucdavis.edu/Calculu.../The_Unit_Tangent_and_the_Unit_Normal_Vectors

You are correct in the fact that taking the derivative of the unit tangent vector can get pretty messy, as the link above acknowledges.

 

[member 508213]

No, the unit normal vector is defined as the derivative of the unit tangent vector, which derivative is then divided again by its magnitude:

http://mathwiki.ucdavis.edu/Calculu.../The_Unit_Tangent_and_the_Unit_Normal_Vectors

You are correct in the fact that taking the derivative of the unit tangent vector can get pretty messy, as the link above acknowledges.

Thank you very much, I have an additional question as well. I am supposed to find the curvature of these curves as well and there are 2 formulas I know of:
k=lTprimel/lvl and k=lvxal/lvl^3 for messy examples such as this one am I correct in assuming that using the second formula would make finding the curvature much easier?

 

[HallsofIvy]

It helps a little bit to observe that $4 cos^2(t)+ 25+ sin^2(t)= 3cos^2(t)+ 25+ cos^2(t)+ sin^2(t)= 3 cos^2(t)+ 26$

 

[SteamKing]

Thank you very much, I have an additional question as well. I am supposed to find the curvature of these curves as well and there are 2 formulas I know of:
k=lTprimel/lvl and k=lvxal/lvl^3 for messy examples such as this one am I correct in assuming that using the second formula would make finding the curvature much easier?

This article gives several different formulas for the curvature:

http://mathworld.wolfram.com/Curvature.html

There are formulas derived especially for curves which are defined using parametric equations.