The gravitational field at a mass point produced by an infinite plane

[3102]

I'm reading:
http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S4


1. In the link it says:
$2\pi\rho d\rho$ is the area of the ring of the radius $\rho$ and width $d\rho$, if $d\rho \ll \rho$.

Why is this true??

2. A bit further down in the text it says:
Since $r^2 = \rho^2 + a^2$, $\rho\,d\rho = r\,dr$.

How do you get this result??

3. Even further down, after "But we see that", it says:
$r^2 = ... = a^2 + R^2 - 2Rx$ Then: $2r\,dr=-2R\,dx$

How do you get that?

It would be great if you answered my questions and if you recommended me some book or link that shows how to get these mathematical results.

 

[Nugatory]

I'm reading:
http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S41. In the link it says:
$2\pi\rho d\rho$ is the area of the ring of the radius $\rho$ and width $d\rho$, if $d\rho \ll \rho$.

Why is this true??

The area of the ring is the difference between the area of a circle with radius $\rho$ and the area of a circle with radius $\rho+d\rho$; the ring is what's left if I cut the smaller circle out of the bigger circle. The condition that $d\rho \ll \rho$ allows us to ignore the $(d\rho)^2$ terms.

2. A bit further down in the text it says:
Since $r^2 = \rho^2 + a^2$, $\rho\,d\rho = r\,dr$.

How do you get this result??

Differentiate both sides with respect to $\rho$, and a bit of algebra will take you the rest of the way.

3. Even further down, after "But we see that", it says:
$r^2 = ... = a^2 + R^2 - 2Rx$ Then: $2r\,dr=-2R\,dx$

How do you get that?

(Note that Feynman has moved on to a different problem here)
The same trick as in #2 works here.

It would be great if you answered my questions and if you recommended me some book or link that shows how to get these mathematical results.

These results are all based on introductory-level calculus, and any decent intro textbook will give you the necessary background.

However, it takes a fair amount of practice (not hard, just practice) to get to where you can reach into the mathematical toolbox and pull out the right technique without thinking about it. The exercises in a calculus-based but pragmatic mechanics textbook such as Kleppner and Kolenkow would be a good start.

 

[3102]

Thanks for your fast and great answer, but unfortunately I don't understand everything.


ad question1) Thanks, I completely understand your advice.

ad question2) I don't understand what's going on here. I've used your advise and differentiated both sides with respect to $\rho$.
On the right side I've got $f(\rho)=\rho^2+a^2$ this is easy to differentiate: $f'(\rho)=2\rho$, since $a^2$ is a constant.

The left side isn't that easy: $f(\rho)=r(\rho)^2$.

$$f'(\rho)=\dfrac{r}{dr}\cdot\dfrac{dr}{d\rho}$$

$$f'(\rho)=1\cdot\dfrac{dr}{d\rho}$$



If I know set both sides together I get: $1\cdot\dfrac{dr}{d\rho}=2\rho$. Then I multiply both sides with $dp$ and divide them by $2$: $\frac{1}{2} \cdot dr = 2\rho \, d\rho$. This is not the same as $r\,dr=\rho\,d\rho$. What did I do wrong?

ad question3) I think: If I'm able to understand question2, I will also understand question3.

 

[Nugatory]

What did I do wrong?

$\frac{d}{d\rho}(r^2) = 2r\frac{dr}{d\rho}$

$\frac{d}{d\rho}(\rho^2+a^2) = 2\rho$

 

[PJC01]

1. The statement is true because as the width of the ring, $d\rho$, approaches zero, it can be considered as a point on the ring. In this case, the area of the ring becomes the circumference of the circle, which is given by $2\pi\rho$. Multiplying this by the infinitesimal width, $d\rho$, gives the area of the ring.

2. This result can be obtained by using basic trigonometry. In the right triangle formed by the hypotenuse, $r$, and the sides, $\rho$ and $a$, we can use the Pythagorean theorem to get $r^2 = \rho^2 + a^2$. Then, by differentiating both sides with respect to $r$, we get $2r\,dr = 2\rho\,d\rho$. Rearranging this gives us $\rho\,d\rho = r\,dr$.

3. This result follows from the previous result, where we have shown that $\rho\,d\rho = r\,dr$. Rearranging this gives us $2r\,dr = 2\rho\,d\rho$. Substituting this into the equation $r^2 = a^2 + R^2 - 2Rx$, we get $r^2 = a^2 + R^2 - 2Rx = a^2 + R^2 - 2R\rho\,d\rho$. Simplifying this equation gives us $2r\,dr=-2R\,dx$.

As for recommended resources, I would suggest looking into calculus textbooks or online resources that cover basic trigonometry and differentiation. These concepts are fundamental in understanding the mathematical results mentioned in the link provided. Some good resources could include Khan Academy, MIT OpenCourseWare, or textbooks such as "Calculus" by James Stewart or "A First Course in Calculus" by Serge Lang.