Answer
$$y=-\frac{3}{2}x+7$$
Work Step by Step
Recall, to find the equation of a line, we need to find the slope of the line and a point that lies on the line.
Now, we want to find the equation of the line tangent to the curve at the point where $t=1$.
So, we know the slope of this line is the derivative of the curve at the point where $t=1$, and because this line is tangent to the curve at the point where $t=1$, we know the point where $t=1$ lies on this line.
So, let us first find the derivative at $t=1$. The curve is given by the parametric equations $$x=1+4t-t^2$$ and $$y=2-t^3.$$ So the derivative at $t=1$ is $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}},$$ evaluated at $t=1$
So, we find the derivatives of each of the parametric equations with respect to $t$. $$\frac{dx}{dt}=\frac{d}{dt}(1)+\frac{d}{dt}(4t)-\frac{d}{dt}(t^2)=0+4-2t=4-2t$$ and $$\frac{dy}{dt}=\frac{d}{dt}(2)-\frac{d}{dt}(t^3)=0-3t^2 =-3t^2.$$ Then $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3t^2}{4-2t}.$$ Next we plug in $t=1$ to get $$\frac{-3(1)^2}{4-2(1)}=\frac{-3}{4-2}=\frac{-3}{2}.$$ Thus, the slope of our line is $-\frac{3}{2}$.
Now, to find the point that lies on the line, we plug $t=1$ into our parametric equations. We get $$x=1+4t-t^2=1+4(1)-1^2=1+4-1=4$$ and $$y=2-t^3=2-1^3=2-1=1.$$ Thus, our point is $(4,1)$.
Finally, we are ready to find the equation of our line. We will use the formula $$y-y_{1}=m(x-x_{1}),$$ where $m$ is the slope and $(x_{1},y_{1})$ is the point that lies on the line.
For us, $m=-\frac{3}{2}$ and $(x_{1},y_{1})=(4,1)$ We plug these into the equation to get $$y-1=-\frac{3}{2}(x-4).$$ Now we solve for $y$. We distribute the $-\frac{3}{2}$ on the right hand side of the equation to get $$y-1=-\frac{3}{2}x+\frac{3}{2}(4)=-\frac{3}{2}x+6,$$ Then, add one to both sides to get $$y=-\frac{3}{2}x+7.$$
