Parametric Derivative Derivation (1 Viewer)

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How does one conclude that [tex] \frac{d^{2} y}{dx^{2}} = \frac{dy\'/dt}{dx/dt} [/tex]?

Thanks
 
[tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
[tex]= \frac{dy'}{dx} = \frac{dy}{dx^2}[/tex]

Which is clearly incorrect.
 
Last edited:

HallsofIvy

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cronxeh said:
Well if you googled "parametric derivative" you would've stumbled on your answer in about 3 seconds:
http://www.mathwords.com/p/parametric_derivative_formulas.htm
And, you will notice that what you wrote was incorrect. What is true is that [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex] not the second derivative.
That follows from the chain rule.
 
I didn't write
[tex]\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
. I wrote [tex]\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy'}{dt}}{\frac{dx}{dt}}[/tex]
 
Yes. Oops.
[tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
[tex]= \frac{dy'}{dx} = y''[/tex]
 

HallsofIvy

Science Advisor
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Then is y' the derivative with respect to x or t? If with respect to t, then your equation is wrong. If with respect to x, then you are back to the first derivative case.
 
[tex]y'' = \frac{d^2 y}{dx^2}[/tex]
 

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