Finding Unit Normal to Curve Defined by Parametric Equation

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SUMMARY

The discussion focuses on computing the unit normal vector to a curve defined by the parametric equation \(\gamma(\theta) = 1 + 0.5 \times \cos(N \theta) (\cos(\theta), \sin(\theta))\) for \(0 \leq \theta \leq 2\pi\). The user seeks to determine the x and y components of the unit normal vector, \(n_x\) and \(n_y\), using the chain rule for differentiation. The equations provided detail the relationships between the derivatives of the curve and the normal vector components. The solution involves finding the tangent vector through differentiation and ensuring the normal vector is orthogonal to the tangent vector.

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nawidgc
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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 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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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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