Finding Unit Normal to Curve Defined by Parametric Equation

In summary, the conversation discusses finding the unit normal and its x and y components for a curve defined by a parametric equation. One method suggested is to use the chain rule to find the x and y components, but the challenge lies in finding the dr / dn term. Another suggestion is to find the tangent to the curve through differentiation and solve for a unit vector that is perpendicular to the tangent.
  • #1
nawidgc
25
0
Hi,

I have a curve defined by following parametric equation

\begin{equation}
\gamma(\theta) = 1 + 0.5 \times \cos (N \theta) (\cos(\theta),\sin(\theta)), 0 \leq \theta \leq 2 \pi \
\end{equation}
where N is an integer. x and y coordinate of any point on the curve are simply

\begin{align}
x &= \gamma (1) \nonumber \\
y &= \gamma (2)
\end{align}

Question is how do I compute a unit normal (n) to above curve in eq. (1) and its x (n_x) and y (n_y) components?
One way to find x and y components of normal n would be to find dx / dn and dy / dn respectively. dx / dn and dy / dn can in turn be found using chain rule as follows -

\begin{align}
n_x &= \frac{dx}{dn} \nonumber \\
& = \frac{dx}{dr} \frac{dr}{dn} \nonumber \\
& = \frac{dx}{d\theta}\frac{d\theta}{dr}\frac{dr}{dn}
\end{align}

and similarly for n_y


\begin{align}
n_y &= \frac{dy}{dn} \nonumber \\
& = \frac{dy}{dr} \frac{dr}{dn} \nonumber \\
& = \frac{dy}{d\theta}\frac{d\theta}{dr}\frac{dr}{dn}
\end{align}


While first two terms on rhs of equations (3) and (4) can be found easily, how to find the dr / dn term? Or if anyone has an easy way to compute the normal components of the unit normal using the parametric equation of the curve, it'd be a great help.

Many thanks for help.
 
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  • #2
find the tangent to the curve by differentiation then solve for a unit vector whose inner product with the tangent is zero.
 
Last edited:

Related to Finding Unit Normal to Curve Defined by Parametric Equation

1. What is a unit normal to a curve defined by a parametric equation?

A unit normal to a curve is a vector that is perpendicular to the tangent vector of the curve at a given point. It represents the direction in which the curve is "curving" at that point.

2. How is the unit normal vector calculated for a parametric curve?

The unit normal vector can be calculated by taking the cross product of the tangent vector and the second derivative of the curve. This will result in a vector that is perpendicular to both and has a magnitude of 1, making it a unit vector.

3. Why is it important to find the unit normal to a curve?

The unit normal vector is important because it can be used to calculate the curvature of the curve at a given point. It can also be used in various applications, such as computer graphics and physics, to determine the direction of forces acting on an object moving along the curve.

4. Can the unit normal vector change along a parametric curve?

Yes, the unit normal vector can change along a parametric curve. This is because the direction of the tangent vector changes as the parameter changes, which in turn affects the direction of the unit normal vector.

5. Are there any limitations to finding the unit normal to a curve using a parametric equation?

One limitation is that the parametric equation must be defined in terms of a single parameter, such as time. If the curve is defined using multiple parameters, the unit normal vector cannot be accurately calculated. Additionally, if the curve is not smooth or has sharp corners, the unit normal vector may not exist at those points.

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