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! /

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