Find an equation tangent to the curve at the point corresponding to the given value of the parameter

Question

$x = 1 +4t -t^2$, $y = 2 - t^3$, at $t=1$

$\frac{dy}{dx}$ $= \frac{-3}{2}$ at t = 1.

Where do I go from here?

Answer 1

The equation of the tangent is just:

$$y- y(1) = y'(1) (x - 1)$$

where $y'(1)$ is $dy/dx$ at $1$. We get:

$$y = -\frac32 x + 7$$

Another way to solve the problem: let $f(t)= (1+4t - t^2, 2 - t^3)$. We have $f'(1) = (2,-3)$ is a direction vector of the tangent line $T$, and $T$ passes through $f(1) = (4,1)$, so a parameterization of $T$ is given by:

$$\begin{cases} x = 2t + 4 \\ y = -3t + 1 \end{cases}$$

Answer 2

So you have the slope of your line, and a point it goes through. Using $$ y-y_0=m(x-x_0) $$ And inputting $$ (x_0,y_0)=(1+4-1,2-1)=(4,1), m=-3/2 $$ You find the equation $$ y-1=-\frac{3}{2}(x-4)=3*2-\frac{3}{2}x=6-\frac{3}{2}x\Rightarrow y=7-\frac{3}{2}x $$