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?

1 Answer
Cesareo R.
Sep 20, 2016

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

Explanation:

If
{(x=t^2+3t),(y=t^2-2t):}

them

{(dx/(dt)=2t+3),(dy/(dt)=2t-2):}

and

dy/dx=dy/(dt)dt/dx = (2t-2)/(2t+3)

Stationary points are those points that observe

dy/(dx)=0 So, for t=1 we have one such a point. x=4, y=-1

Qualification is done considering the signal of (d^2y)/(dx^2). In this case we have

d/(dx)(dy/(dx)) = (d^2y)/(dx^2) =1/(dx/(dt))^2d/(dt)(dy/(dt))+dy/(dx) d/(dt)(1/(dx/dt))=10/(3+2t)^3

For t = 1 we have (d^2y)/(dx^2) = 10/5^3 > 0 so the stationary point is a relative minimum point. The quadratic after parameter ellimination is a parabola

p(x,y) =x^2-10x - 15 y - 2 x y + y^2=0

Attached the conic plot.

enter image source here

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