Parametric equation confusion of d2y/dx2

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Discussion Overview

The discussion revolves around the computation of the second derivative \( \frac{d^2y}{dx^2} \) using parametric equations, specifically with the equations \( x = t^2 \) and \( y = t^3 - 3t \). Participants explore the application of the chain rule and the reasoning behind the formula used for finding the second derivative in parametric form.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about deriving \( \frac{d^2y}{dx^2} \) and questions the necessity of using \( \frac{d}{dt} \) instead of simply deriving twice.
  • Another participant explains that the process involves multiple applications of the chain rule, reiterating the relationship \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \) and suggests that the second derivative can be derived similarly.
  • A later reply attempts to clarify by suggesting a change in notation to \( \omega \) for \( \frac{dy}{dx} \) and demonstrates the application of the chain rule to find \( \frac{d\omega}{dx} \).
  • One participant emphasizes the importance of ensuring that \( \frac{dt}{dx} \) is defined, indicating that \( \frac{dx}{dt} \) must be non-zero for the calculations to hold.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there remains confusion about the application of the chain rule and the interpretation of the derivatives involved. Some participants provide explanations while others continue to express uncertainty.

Contextual Notes

There are indications of missing assumptions regarding the definitions of the derivatives and the conditions under which the calculations are valid. The discussion also highlights the potential for misunderstanding in the application of the chain rule in parametric equations.

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x=t2 and y=t3-3t, find dy/dx and d2y/dx2

I understand how to get to dy/dx but an confused on how to get to d2y/dx2 can someone please explain in depth, I know the formula is (d/dt dy/dx)/(dx/dt) I don't understand where d/dt comes from and what it is. why do we not just derive it twice? I really do not understand the 2nd part at all.
 
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It's just several applications of the chain rule

\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \left(\frac{dy}{dt} / \frac{dx}{dt} \right)
from what your OP says you're familiar with this calculation. We're just going to do it again
\frac{d}{dx} \left( \frac{dy}{dx} \right)= \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx}= \frac{d}{dt} \left( \frac{dy}{dt}/\frac{dx}{dt} right) / \frac{dx}{dt}
 
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Office_Shredder said:
It's just several applications of the chain rule

\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \left(\frac{dy}{dt} / \frac{dx}{dt} \right)
from what your OP says you're familiar with this calculation. We're just going to do it again
\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx} = \frac{d}{dt} \left( \frac{dy}{dt}/\frac{dx}{dt} right) / \frac{dx}{dt}

Im sorry but i still have no idea : \.
 
It will perhaps become more clear if you switch your notation slightly. How about we denote dy/dx instead by ω. Okay, so ω is a function, and what we are looking for is dω/dx. We will apply the chain rule:

dω/dx=dω/dt * dt/dx=d/dt(ω)*dt/dx

Now we simply put back dy/dx for ω to get

d2y/dx2=d/dt(dy/dx)*dt/dx as claimed. Hope that clears it up :smile:
 
Office_Shredder said:
It's just several applications of the chain rule

\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \left(\frac{dy}{dt} / \frac{dx}{dt} \right)
from what your OP says you're familiar with this calculation. We're just going to do it again
\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx} = \frac{d}{dt} \left( \frac{dy}{dt}/\frac{dx}{dt} right) / \frac{dx}{dt}

Only one note of caution with this approach: make sure dt/dx is defined which means dx/dt is non-zero (i.e. t changes with respect to x at the point you are considering and hence x also changes with respect to y)
 

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