The Ratio of Total Derivatives

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Discussion Overview

The discussion centers around the relationship between the total derivatives of two functions, C(y(t), r(t)) and I(y(t), r(t)), and whether the ratio of their total derivatives can be expressed as the derivative of one function with respect to the other. The scope includes mathematical reasoning and technical explanation.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant questions if the ratio of total derivatives $$\frac{\frac{dC}{dt}}{\frac{dI}{dt}}$$ can be equated to $$\frac{dC}{dI}$$.
  • Another participant suggests that while the relationship can hold, careful consideration is needed regarding the definitions and the functions involved.
  • The same participant elaborates on defining functions f(t) and u(t) for C and I, respectively, and discusses the implications of expressing t as a function of u.
  • They provide a mathematical formulation showing how the derivatives relate through the chain rule, indicating that $$\frac{dF}{du}$$ can be expressed in terms of $$\frac{df}{dt}$$ and $$\frac{du}{dt}$$.

Areas of Agreement / Disagreement

Participants appear to agree on the potential validity of the relationship between the derivatives, but there is no consensus on the conditions under which it holds or the implications of the definitions involved.

Contextual Notes

Participants note the need for careful consideration of the definitions and the relationships between the functions, but do not resolve the specific conditions or assumptions required for the claims made.

Ahmed Mehedi
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TL;DR
Total Derivative
If we have two functions C(y(t), r(t)) and I(y(t), r(t)) can we write $$\frac{\frac{dC}{dt}}{\frac{dI}{dt}}=\frac{dC}{dI}$$?
 
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Ahmed Mehedi said:
Summary:: Total Derivative

If we have two functions C(y(t), r(t)) and I(y(t), r(t)) can we write $$\frac{\frac{dC}{dt}}{\frac{dI}{dt}}=\frac{dC}{dI}$$?
Essentially yes, but you need to be careful that it all makes sense. In this case we can define:
$$f(t) = C(y(t), r(t)) \ \ \text{and} \ \ u(t) = I(y(t), r(t))$$
Then ##\frac{df}{dt}## and ##\frac{du}{dt}## are well defined. You also have to imagine that you express ##t## as a function of ##u##, so that we have a further function:
$$F(u) = f(t(u))$$
Then:
$$\frac{dF}{du} = \frac{df}{dt} \frac{dt}{du} = \frac{df/dt}{du/dt}$$
 
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PeroK said:
Essentially yes, but you need to be careful that it all makes sense. In this case we can define:
$$f(t) = C(y(t), r(t)) \ \ \text{and} \ \ u(t) = I(y(t), r(t))$$
Then ##\frac{df}{dt}## and ##\frac{du}{dt}## are well defined. You also have to imagine that you express ##t## as a function of ##u##, so that we have a further function:
$$F(u) = f(t(u))$$
Then:
$$\frac{dF}{du} = \frac{df}{dt} \frac{dt}{du} = \frac{df/dt}{du/dt}$$
Thanks a lot! You have been very helpful!
 

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