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

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this,

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!

 

[topsquark]

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

 

[haruspex]

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.

 

[jbriggs444]

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.

 

[ChiralSuperfields]

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!

 

[haruspex]

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:

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 $x=f^{-1}(y)$.