# Given that a curve C is represented parametrically by x=t²/2 + 3t y=t²-2t how do you find the value of dy/dx and show that d²2y/dx²=8/(t+3)^3 and Show that C has only one stationary point and that this is a minimum?

Sep 20, 2016

Has a local minimum at $x = 4 , y = - 1$

#### Explanation:

If
$\left\{\begin{matrix}x = {t}^{2} + 3 t \\ y = {t}^{2} - 2 t\end{matrix}\right.$

them

$\left\{\begin{matrix}\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t + 3 \\ \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - 2\end{matrix}\right.$

and

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{2 t - 2}{2 t + 3}$

Stationary points are those points that observe

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ So, for $t = 1$ we have one such a point. $x = 4 , y = - 1$

Qualification is done considering the signal of $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$. In this case we have

$\frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dt}}} ^ 2 \frac{d}{\mathrm{dt}} \left(\frac{\mathrm{dy}}{\mathrm{dt}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dt}} \left(\frac{1}{\frac{\mathrm{dx}}{\mathrm{dt}}}\right) = \frac{10}{3 + 2 t} ^ 3$

For $t = 1$ we have $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{10}{5} ^ 3 > 0$ so the stationary point is a relative minimum point. The quadratic after parameter ellimination is a parabola

$p \left(x , y\right) = {x}^{2} - 10 x - 15 y - 2 x y + {y}^{2} = 0$

Attached the conic plot.