# 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 ?

Sep 13, 2014

To find the equation of the tangent line, we need the slope
$m = \frac{\mathrm{dy}}{\mathrm{dx}}$ and the point of tangency $\left({x}_{o} , {y}_{o}\right)$.

Then the equation is the usual $y - {y}_{o} = m \left(x - {x}_{o}\right) .$

We have the parametric curve $x = {t}^{4} + 1 , y = {t}^{3} + t$,

so we compute $\frac{\mathrm{dx}}{\mathrm{dt}} = 4 {t}^{3} \mathmr{and} \frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2} + 1.$

The chain rule $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}$ says that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} .$

So we use the derivatives of the parametric equations:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {t}^{2} + 1}{4 {t}^{3}} .$ Now put in $t = - 1$ and find:

$m = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \cdot {\left(- 1\right)}^{2} + 1}{4 \cdot {\left(- 1\right)}^{3}} = \frac{3 + 1}{- 4} = - 1.$

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:
$y - {y}_{o} = m \left(x - {x}_{o}\right)$
$y - \left(- 2\right) = \left(- 1\right) \left(x - 2\right)$ or
$y + 2 = - x + 2$ or just plain old $y = - x$.

\ Another great answer from the modest dansmath! /