Question

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` ?

Answer 1

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! /