How do you find the equation of the tangent to the curve #x=t^4+1#, #y=t^3+t# at the point where #t=-1# ?

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Answer 1 · 2014-09-13T21:32:42

To find the equation of the tangent line, we need the slope
#m = dy/dx# and the point of tangency #(x_o,y_o)#.

Then the equation is the usual # y-y_o = m(x - x_o).#

We have the parametric curve #x = t^4+1, y = t^3+t#,

so we compute #dx/dt = 4t^3 and dy/dt = 3t^2+1.#

The chain rule #dy/dt = dy/dx*dx/dt# says that #dy/dx = (dy/dt)/(dx/dt).#

So we use the derivatives of the parametric equations:

#dy/dx = (3t^2+1)/(4t^3).# Now put in #t = -1# and find:

#m = dy/dx = (3*(-1)^2+1)/(4*(-1)^3) = (3+1)/(-4) = -1.#

Also at #t=-1# the original equations give
#(x_o,y_o) = ((-1)^4+1,(-1)^3+(-1)) =(1+1,(-1)+(-1))=(2,-2)#

Now we put in the info for the tangent line:
#y-y_o=m(x-x_o)#
#y-(-2)=(-1)(x-2)# or
#y+2=-x+2# or just plain old #y=-x#.

\ Another great answer from the modest dansmath! /