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#.