Horizontal and Vertical tangent lines for a parametric curve
Consider the parametric curve in the xy-plane given by:
x = t^2 -2t + 4 and y = 2t^3 + 3t^2 -36t -4
Find all equations of the lines which are either horizontal or vertical
tangents to the given curveOkay. The horizontal tangent is when $dx/dt = 0$ The vertical tangent is when $dy/dt = 0$ I think..
$dx/dt$ = 2t -2
0 = 2t - 2 => t = 1;
$dy/dt = 6t^2 + 6t - 36$
$6t^2 + 6t - 36 = 0$
$y = -48, -31$
Should I plug these values back into their corresponding equations? Like: t=1 for the x related eq'. So, when t=1, x is 3.
I don't really know what i'm doing
$\endgroup$ 11 Answer
$\begingroup$The horizontal tangent line is indeed at the point corresponding to $t=1$. You can easily find the point $(x,y)$ of tangency by plugging in $1$ for $t$.
We get that the point of tangency is $(3,-35)$. So the horizontal tangent line has equation $y=-35$.
For vertical tangency, solve $6t^2+6t-36=0$, or more simply $t^2+t-6=0$. The quadratic factors as $(t+3)(t-2)$. So the $t$ at which we have a vertical tangent line are $t=-3$ and $t=2$.
Separately, for each of these, find the point of tangency by substituting. The two vertical tangent lines will have equations of the shape $x=k$. One of them is the line $x=4$. You can find the other one.
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