Unit normal to curve

  • Thread starter nawidgc
  • Start date
  • #1
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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 any one 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.
 

Answers and Replies

  • #2
lavinia
Science Advisor
Gold Member
3,236
623
find the tangent to the curve by differentiation then solve for a unit vector whose inner product with the tangent is zero.
 
Last edited:

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