Second derivative in parametric equations

  • Thread starter Karol
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  • #1
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Homework Statement


Snap2.jpg

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
 

Answers and Replies

  • #2
Orodruin
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You cannot do it like that. Use the chain and Leibniz rules.
 
  • #3
Ray Vickson
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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
 
  • #4
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$$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}\left( \frac{1-3t^2}{1-2t} \right)$$
$$=\frac{d \left( \frac{1-3t^2}{1-2t} \right)/dt}{dx/dt}=\frac{d \left( \frac{1-3t^2}{1-2t} \right)/dt}{1-2t}$$
It comes out right, -2
Thank you Orodruin and Ray
 
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