Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 3

$y = -x$

$x = t^3+1$ $y = t^4+t$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t^3+1}{3t^2}$ When $t=-1$: $x = t^3+1$ $x = (-1)^3+1 = 0$ $y = t^4+t$ $y = (-1)^4+(-1) = 0$ $\frac{dy}{dx} = \frac{4(-1)^3+1}{3(-1)^2} = -1$ We can find the equation of the tangent line: $(y-0) = -1(x-0)$ $y = -x$