Finding Tangent Line and Normal of a parametric equation

**maxgunn555** · January 13th 2012, 05:25 AM

For $\delta \mathbb{R}\rightarrow \mathbb{R}^2,\ t\ =\ \delta (t)\ =\ (cosh(2t-2),sin(2\pi t^2))$ find the tangent line and the normal line passing through $\delta(t_0),t_0\ =\ 1$

So i think the tangent line is $T\delta (t_0)\ =\ \delta' (t_0)s\ +\ \delta (t_0)$

However Im clueless as to how to find the normal line.

**maxgunn555** · January 13th 2012, 06:12 AM

it's ok apparently, after pluggin in the to into the tangent equation to get Tdelta(to) = (x,y)s + (a,b) the normal line is simply Ndelta(to) = (-y,x)s + (a,b).
ps it was meant to say deltaR rightarrow R to the power of 2 not that weird symbol.

**HallsofIvy** · January 14th 2012, 07:02 AM

$\delta'(t_0)$ will be a two dimensional vector- the normal will be perpendicular to that so you just need a vector perpendicular to $\delta'(t_0)$ . In two dimensions, that's easy. A normal to the vector <a, b> is <b, -a>.

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