1-Forms .... Bachman, Exercise 3.2, Part 2 ....

[Math Amateur]

I am reading David Bachman's book: "A Geometric Approach to Differential Forms" (Second Edition) ...

I need some help with Exercise 3.2, Part 2 ... Exercise 3.2, Part 2 ... reads as follows:View attachment 8607Bachman gives he answer to Exercise 3.2, Part 2 as \(\displaystyle dy = -4dx\) ... ... but ... I cannot understand how he gets this answer ..Can someone please help ...

Peter=========================================================================================It may help MHB
readers of the above post to have access to Bachman's Section 3.2 ... ... so I am providing the same .. ... as follows:View attachment 8608
View attachment 8609
View attachment 8610It may also help MHB
readers of the above post to have access to Bachman's Section 3.1 ... ... so I am providing the same .. ... as follows:View attachment 8611
View attachment 8612
Hope that helps ...

Peter

 

[jvicens]

Hello Peter,

Thank you for reaching out for help with Exercise 3.2, Part 2 from Bachman's book. I have taken a look at the section and I believe I can provide some assistance.

In this exercise, we are given the differential form ω = 2xydx + (x^2 - 2y)dy and we are asked to find the exact differential of ω. To do this, we need to use the definition of an exact differential, which states that a differential form ω is exact if and only if there exists a function f such that ω = df.

In order to find f, we need to integrate the given differential form ω. To do this, we need to use the properties of exact differentials, which state that the integral of an exact differential along a path from point a to point b is independent of the path chosen and depends only on the endpoints a and b.

Using this property, we can integrate ω along two different paths, one from (0,0) to (x,y) and another from (0,0) to (x,0). This gives us:

∫ω = ∫(2xydx + (x^2 - 2y)dy) = f(x,y) - f(x,0) = f(x,y) - f(0,0)

But since ω is an exact differential, this integral should be independent of the path chosen. Therefore, we can equate the two integrals and solve for f(x,y):

f(x,y) - f(0,0) = f(x,y) - f(x,0)

f(x,0) = f(0,0)

This means that f(x,y) is a constant with respect to y. Therefore, we can write f(x,y) = g(x) where g(x) is some function of x.

Now, we can use the definition of an exact differential again to find g(x). We have:

df = d(g(x)) = g'(x)dx

Comparing this with ω = 2xydx + (x^2 - 2y)dy, we can see that g'(x) = 2xy and g(x) = x^2y + h(y) where h(y) is some function of y.

Substituting this into f(x,y) = g(x), we have:

f(x,y)