Question

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?

Answer 1

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.