# Derivative of Parametric Equation $x= t^3 -3t^2$ $y=2t^3 - 3t^2 -12t$

My parametric equation is:

$x= t^3 -3t^2$

$y=2t^3 - 3t^2 -12t$

Thus, $\frac {dy}{dx}= \frac{\frac {dy}{dt}}{\frac {dx}{dt}}$, which in the case of the above parametric equation, is:

$\frac {dy}{dx}= \frac{6t^2-6t-12}{3t^2 -6t} = \frac{6[t-2)(t+1)}{3t (t -2)} = \frac{6(t+1)}{3t}$.According to his, at t =2, the slope should be $\frac{18}{6}=3$.However, when I graph the parametric on my TI84, and check $\frac {dy}{dx}$ at t=2. the calculator says $\frac {dy}{dx} =2$.

Did I do something wrong mathematically or technologically?

• @DougM isn't the slope technically undefined at t=2 Sep 28, 2016 at 21:03
• I believe that the aspect ratio on TI 84 calculators is 3:2. So it could be that the slope on the calculator's graph is 2 but appears to be 3 because of the unequal scales on the $x$ and $y$ axes. Sep 28, 2016 at 21:25
• @qbert On second thought, I think you are right. $x(2) = -4$ and that is a local min. $x$ does not exist on the left side of $-4$. $\frac {dy}{dx}$ is a limit defined (or not defined) as x approaches $-4$ The limit as $x$ approaches $-4$ is not a 2-sided limit, $\frac {dy}{dx}$ does not exist. Sep 28, 2016 at 21:26
• @DougM I'm not sure, I was hoping you would know :). I think generally if you have a removable discontinuity you must define the derivative appropriately to be 3 at $t=2$ Sep 28, 2016 at 21:26 That local minimum at $y=-20$ is the point for $t=2$. The derivative looks at first glance as though it would be undefined, but you can see that the gradient actually tends to $3$ from either side, and $t=2$ would just be a hole. If you wanted to fill it in (I'm not sure what the convention is), $3$ is the correct answer—the TI must be messing up somehow.