Second derivative in parametric equations

[Karol]

Homework Statement

Only the second part

Homework Equations

Second derivative:
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}$$

The Attempt at a Solution

$$dx=(1-2t)\,dt,~~dy=(1-3t^2)\,dt$$
Do i differentiate the differential dt?
$$d^2x=(-2)\,dt^2,~~d^2y=(-6)t\,dt^2$$
$$\frac{d^2y}{dx^2}=\frac{d^2y/dt^2}{d^2x/dt^2}=\frac{-6t}{-2}=...=3$$
The answer should be -2

 

[Orodruin]

You cannot do it like that. Use the chain and Leibniz rules.

 

[Ray Vickson]

Homework Statement

View attachment 210022
Only the second part

Homework Equations

Second derivative:
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}$$

The Attempt at a Solution

$$dx=(1-2t)\,dt,~~dy=(1-3t^2)\,dt$$
Do i differentiate the differential dt?
$$d^2x=(-2)\,dt^2,~~d^2y=(-6)t\,dt^2$$
$$\frac{d^2y}{dx^2}=\frac{d^2y/dt^2}{d^2x/dt^2}=\frac{-6t}{-2}=...=3$$
The answer should be -2

Set $y' = dy/dx = y'(t)$ and then comput $d^2 y/dx^2 = dy'/dx$ paremetrically

 

[Karol]

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}\left( \frac{1-3t^2}{1-2t} \right)$$
$$=\frac{d \left( \frac{1-3t^2}{1-2t} \right)/dt}{dx/dt}=\frac{d \left( \frac{1-3t^2}{1-2t} \right)/dt}{1-2t}$$
It comes out right, -2
Thank you Orodruin and Ray