General form of Newton II -- Not understanding this step in the derivation

In summary: If ##\Delta y\rightarrow 0## then ##\Delta x## can go to 0 or it can go to some other value. (In the example above ##\Delta x## actually goes to 2.)-DanIn summary, we cannot take the limit as Δy approaches zero in order to derive equation 9.9 from 9.8 because, while Δy going to zero may result in Δx going to zero, it could also result in Δx approaching some other value. Therefore, the limit must be taken as Δx approaches zero in order to accurately determine the derivative.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1684126577290.png

Does someone please know how do we derive equation 9.9 from 9.8? Do we take the limits as t approach's zero for both sides? Why not take limit as momentum goes to zero?

Many thanks!
 
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  • #2
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 326570
Does someone please know how do we derive equation 9.9 from 9.8? Do we take the limits as t approach's zero for both sides? Why not take limit as momentum goes to zero?

Many thanks!
Are you going from 9.9 to 9.8 or from 9.8 to 9.9?

9.8 to 9.9:
When you define
##y^{\prime}(x) \approx \dfrac{ \Delta y}{ \Delta x}##

which variable do we take the limit of as we pass to the exact definition?

9.9 to 9.8:
This is a definition of one way to approximate the derivative.

-Dan
 
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  • #3
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

Does someone please know how do we derive equation 9.9 from 9.8? Do we take the limits as t approach's zero for both sides? Why not take limit as momentum goes to zero?
It is standard differential calculus. If the momentum function is differentiable then, necessarily, as ##dt\rightarrow 0## ##d\vec p\rightarrow 0##. The converse is not necessarily true.
Since ##dt\geq 0##, the limit is taken from above.
 
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  • #4
haruspex said:
Since ##dt\geq 0##, the limit is taken from above.
For a derivative to exist, the two sided limit must exist. The limit is taken from both sides. ##dt## may be negative.

There is often a prejudice toward the future. Predictions for past behavior are less useful. We want to know what will happen next. But it is just a prejudice, not something inherent in the definitions.
 
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  • #5
topsquark said:
Are you going from 9.9 to 9.8 or from 9.8 to 9.9?

9.8 to 9.9:
When you define
##y^{\prime}(x) \approx \dfrac{ \Delta y}{ \Delta x}##

which variable do we take the limit of as we pass to the exact definition?

9.9 to 9.8:
This is a definition of one way to approximate the derivative.

-Dan
Thank you for your replies @topsquark, @haruspex and @jbriggs444 !

@topsquark You are correct, I was curious in going from equation 9.8 to 9.9.

According to calculus textbooks we take the limit as ##\Delta x## approach's zero. However, why can't we take the limit as ##\Delta y## approach's zero because as ##\Delta y## goes to zero so dose ##\Delta x## correct?

Many thanks!
 
  • #6
ChiralSuperfields said:
why can't we take the limit as ##\Delta y## approach's zero because as ##\Delta y## goes to zero so dose ##\Delta x## correct?
No. As I wrote:
haruspex said:
The converse is not necessarily true.
Consider a quartic with local minima at ##x=\pm 1, y=1##. If we start with ##(x,y)=(-1,1), \Delta x=2.1,\Delta y=0.1##, say, then as ##\Delta y\rightarrow 0##, ##\Delta x\rightarrow 2##.
The key point is that ##y=f(x)## does not necessarily have a unique inverse function ##x=f^{-1}(y)##.
 
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