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If $\dfrac{dy}{dt} = 2$ and $\dfrac{dt}{dx}=5$, then chain rule gives $\dfrac{dy}{dx}=2*5=10$.

However if we define $\dfrac{dy}{dt}=2$ to be the speed of a person in a moving train relative to train,
and $\dfrac{dt}{dx}=5$ to be the speed of train relative to ground, this gives speed of person relative to ground as $\dfrac{dy}{dx}=2+5=7$.

I feel I'm missing something.. Any help?

AgentS
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  • What are your variables $x$, $y$ and $t$? It seems $t$ and $x$ both represent time in your definition. Define them clearly to see whats happening. – UserA Nov 22 '19 at 21:39
  • $x$ refers to ground, $y$ to the person inside the train, and $t$ to the train. I see how this is ambiguous. I'll try to fix by definiing these variables as positions... one sec:) – AgentS Nov 22 '19 at 21:41
  • ground, person and train are not variables. Distance and time are variables. – Paul Nov 22 '19 at 21:44
  • Ahh right! I was just looking for a simple real example to better understand chain rule. Looks the relative speeds is just a simple addition as answered by @oskar .. doesn't require chain rule at all. Hmm.. – AgentS Nov 22 '19 at 21:47
  • If $dy/dt$ is measured in feet per second, say, the units of $dt/dx$ are seconds per foot. That’s not a speed. – amd Nov 23 '19 at 03:45

1 Answers1

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If you define $ \frac{dy}{dt} $ as the speed of a person relative to the train, you define that persons position $ y $ relative to the train as a function of $ t $.

The trains speed relative to the ground cannot be defined as $ dt / dy $, it would make no sense to define the trains time $ t $ as a function of the persons position $ y $.

Rather define the speed of the train as $ \frac{dy'}{dt} $, where $ y' $ is the trains position relative to the ground.

Then the persons speed is, (without taking relativity into account),

$$ \frac{dy}{dt} + \frac{dy'}{dt} $$

Oskar
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