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 $y=t^2+t+1$ ?

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Answer 1 · 2015-02-25T20:12:48

The answer is:

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

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

$x=(-1)^3+2(-1)=-3$

$y=(-1)^2+(-1)+1=1$ .

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

$x'=3t^2+2$

$y'=2t+1$

so:

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

$y'(-1)=-2+1=-1$ .

And finally, remembering that the equation of a line given a point $P(x_P,y_P)$ and a direction $vecv(a,b)$ is:

$x=x_P+at$
$y=y_P+bt$

The solution is:

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

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