Finding Tangent Line and Normal of a parametric equation

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

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

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

Re: Finding Tangent Line and Normal of a parametric equation

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.

Re: Finding Tangent Line and Normal of a parametric equation

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