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

AI Thread Summary
The discussion focuses on the derivation of equation 9.9 from 9.8 in the context of Newton's second law. Participants clarify that the limit should be taken as Δx approaches zero, not Δy, because the relationship between the two is not necessarily direct. It is emphasized that while a differentiable momentum function implies that dP approaches zero as dt approaches zero, the converse does not hold true. The conversation also touches on the importance of two-sided limits for derivatives and the implications of taking limits in different directions. The key takeaway is the necessity of understanding the relationship between variables when deriving equations in calculus.
member 731016
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!
 
Physics news on Phys.org
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
 
  • Like
Likes member 731016
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.
 
  • Like
Likes topsquark and member 731016
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.
 
  • Like
Likes member 731016, DaveE and nasu
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!
 
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)##.
 
  • Like
Likes member 731016, topsquark and hutchphd
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top