# How do you find the equation of a line tangent to the curve at point t=-1 given the parametric equations x=t^3+2t and y=t^2+t+1?

Feb 25, 2015

$x = - 3 + 5 t$
$y = 1 - t$.

First of all let's find the cartesian coordinates of the point with $t = - 1$,

$x = {\left(- 1\right)}^{3} + 2 \left(- 1\right) = - 3$

$y = {\left(- 1\right)}^{2} + \left(- 1\right) + 1 = 1$.

Than, let's find avector that is the direction of the tangent, putting $t = - 1$ in:

$x ' = 3 {t}^{2} + 2$

$y ' = 2 t + 1$

so:

$x ' \left(- 1\right) = 3 + 2 = 5$

$y ' \left(- 1\right) = - 2 + 1 = - 1$.

And finally, remembering that the equation of a line given a point $P \left({x}_{P} , {y}_{P}\right)$ and a direction $\vec{v} \left(a , b\right)$ is:

$x = {x}_{P} + a t$
$y = {y}_{P} + b t$

The solution is:

$x = - 3 + 5 t$
$y = 1 - t$.