Vector field and differential form confusion

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Homework Statement
We have xy-plane, which is has a mapping (x,y). Another map is (u,v) and the transformation of coordinates are u=4x+3y and v=x+y.

1. Write vector field \frac{\partial}{\partial u} in the map (x,y)

2. Write the 1-form du in the map (x,y)
Relevant Equations
u=4x+3y and v=x+y.
Here is a picture of the solution I made :

1697563788231.png
So my question is: Are these right and how do they differ from each other?
 
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In a formal Mathematical sense, they're duals to each other. Differential forms are evaluated at Vector Fields to produce numbers.
 
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## \text { The first thing: } ##
## \frac { \partial } { \partial u } \text { is not a vector field. } ##

## \text { The second thing: } ##
## \text { The expression } \frac { \partial } { \partial u } = \frac { \partial u } { \partial x } \frac { \partial } { \partial x } + \frac { \partial u } { \partial y } \frac { \partial } { \partial y } \text { is not a correct expression. } ##
## \text { It should be } \frac { \partial } { \partial u } = \frac { \partial x } { \partial u } \frac { \partial } { \partial x } + \frac { \partial y } { \partial u } \frac { \partial } { \partial y } \text { where } x \text { and } y \text { are functions of } u \text { and } v \text { . } ##
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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