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

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One of these formulas is the one you mentioned: k=lvxal/lvl^3In this case, using that formula might make the calculation a little easier, but it ultimately depends on the specific curve and the values of t you are working with. It's always a good idea to check multiple formulas to see which one makes the most sense for your particular problem.
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## Homework Statement

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

.

## 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

Austin said:

## Homework Statement

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

.

## 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.

SteamKing said:
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?

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

Austin said:
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?

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

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

## 1. What is a normal unit vector to a curve?

A normal unit vector to a curve is a vector that is perpendicular to the tangent of the curve at a specific point. It has a magnitude of 1 and points in the direction of the curve's outward curvature.

## 2. How is a normal unit vector to a curve calculated?

To calculate a normal unit vector to a curve, you first need to find the tangent vector at the specific point on the curve. Then, using the tangent vector, you can use the cross product with the unit vector in the z-direction (0,0,1) to find the normal vector. Finally, divide the normal vector by its magnitude to get the unit vector.

## 3. Why is the normal unit vector important in studying curves?

The normal unit vector is important because it helps to determine the curvature of a curve at a specific point. It also helps in determining the direction in which the curve is bending, which can be useful in various applications such as computer graphics and navigation.

## 4. Can a normal unit vector be negative?

Yes, a normal unit vector can be negative. This means that it points in the opposite direction of the curve's outward curvature. It is important to note that the magnitude of a normal unit vector is always 1, regardless of its direction.

## 5. How does a normal unit vector relate to the normal vector of a curve?

The normal unit vector is a scaled version of the normal vector of a curve. The normal vector has a magnitude equal to the curvature of the curve at a specific point, while the normal unit vector has a magnitude of 1. The normal unit vector is often used in calculations and applications because of its standardized magnitude.

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