Find the equation of the tangent to a curve at a point

Question

Consider the parametric curve: $x(t)=1+4t-t^2$, $y(t)=2-t^3$ for $t\in\mathbb{R}$.

What is the equation of the tangent line at the point $(x(1),y(1))$?

I got the slope by using the equation $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ and I found that $y = -\frac{3}{2}x + 7$. Is it correct?

Answer 1

There is room for improvement of the typographical layout of your equations. But it looks as if at $t=1$ the curve goes through the point $(4,1)$ with 'velocity' $(2, -3)$. Your line seems to have the right slope and go through the point in question.

Answer 2

Yes, it is correct.

Let $y=mx+q$ be the equation of the tangent at $(x(t), y(t))$. Then
$$m=\frac{y'(t)}{x'(t)}=\frac{-3t^2}{4-2t}\quad\mbox{and}\quad q=y(t)-mx(t).$$ Therefore, for $t=1$, we obtain $m=-3/2$, $q=1-(-3/2)\cdot 4=1+6=7$ and the tangent line is $$y=-\frac{3}{2} x+7.$$