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#?

Question

Emilia Aguilar · Accepted Answer

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#.

Adam Adkins · Answer

undefined To find the equation of the tangent line to the curve described by the parametric equations $x = t^3 + 2t$ and $y = t^2 + t + 1$ at the point where $t = -1$, follow these steps:

1. Substitute $t = -1$ into the parametric equations to find the coordinates of the point on the curve where the tangent line touches.
2. Compute the derivatives $dx/dt$ and $dy/dt$ with respect to $t$.
3. Evaluate $dx/dt$ and $dy/dt$ at $t = -1$ to find the slope of the tangent line.
4. Use the point-slope form of the equation of a line ($y - y_1 = m(x - x_1)$), where $m$ is the slope and $(x_1, y_1)$ is the point of tangency, to find the equation of the tangent line.