MHB Proof that the integral is independent from the path from A to B

[mathmari]

Hey!

I am facing some troubles understanding the proof of the following theorem.

Let $\overrightarrow{F}$ be a vector field, $\overrightarrow{F}=M\hat{\imath}+N\hat{\jmath}+P\hat{k}$, where $M,N,P$:continuous at a region $D$ ($M,N,P \in C^0(D)$) . Then a suficient and necesary condition so that the integral $\int_A^B{\overrightarrow{F}d \overrightarrow{R}}$ is independent from the path from $A$ to $B$ in $D$ is that there is a differentiable function $f$ such that $\overrightarrow{F}=\nabla f= \hat{\imath} \frac{\partial{f}}{\partial{x}}+\hat{\jmath} \frac{\partial{f}}{\partial{y}}+\hat{k} \frac{\partial{f}}{\partial{z}}$ in $D$.
In this case the value of the integral is given by:
$$\int_A^B \overrightarrow{F}d \overrightarrow{R}=f(B)-f(A)$$

The proof that I am given is the following:
View attachment 2552
$$\overrightarrow{R}(t)=(x(t), y(t), z(t)), t_1 \leq t \leq t_2$$

$\displaystyle{f(x(t), y(t), z(t))}$

$\displaystyle{\frac{df}{dt}=\frac{\partial{f}}{\partial{x}} \frac{dx}{dt}+\frac{\partial{f}}{\partial{y}} \frac{dy}{dt}+ \frac{\partial{f}}{\partial{z}} \frac{dz}{dt}}$

$\displaystyle{\Rightarrow \frac{df}{dt}= \nabla f \cdot \frac{d \overrightarrow{R}}{dt}}$

$\displaystyle{\overrightarrow{F} \cdot d \overrightarrow{R}=\nabla f \cdot \frac{d \overrightarrow{R}}{dt} dt=\frac{df}{dt}dt=\overrightarrow{F} \cdot d \overrightarrow{R}}$

$\displaystyle{\int_A^B \overrightarrow{F} d \overrightarrow{R}=\int_{t_1}^{t_2} \frac{df}{dt} dt=f(B)-f(A)}$ $\ \ \ \ (*)$
$\overrightarrow{F}(x,y,z)=\hat{\imath} M(x,y,z)+\hat{\jmath} N(x,y,z)+\hat{k} P(x,y,z)$

We will show that: $$\frac{\partial{f}}{\partial{x}}=M, \frac{\partial{f}}{\partial{y}}=N, \frac{\partial{f}}{\partial{z}}=P$$

View attachment 2553

$$x=x'+t \cdot h, y=y', z=z', 0 \leq t \leq 1$$
$$\frac{f(x'+h, y', z')-f(x', y', z')}{h}=\frac{1}{h} \int_B^{B'} \overrightarrow{F} d \overrightarrow{R}=\frac{1}{h} \int_0^1 M(x'+t \cdot h, y', z') \cdot h dt= \int_0^1 M(x'+t \cdot h, y', z') dt$$
$$h \rightarrow 0$$
$$\frac{\partial{f}}{\partial{x}}=M(x,y,z)$$

In a similar way can show that
$$\frac{\partial{f}}{\partial{y}}=N(x,y,z), \frac{\partial{f}}{\partial{z}}=P(x,y,z)$$

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At the beginning (till the point $(*)$ ) we suppose that there is a differentiable function $f$ such that $\overrightarrow{F}=\nabla f$ and we show that the integral $\int_A^B \overrightarrow{F} d \overrightarrow{R}$ is equal to $f(B)-f(A)$. Or have I understood it wrong? (Wondering)

Then I haven't understood the proof from the point
"$\overrightarrow{F}(x,y,z)=\hat{\imath} M(x,y,z)+\hat{\jmath} N(x,y,z)+\hat{k} P(x,y,z)$
We will show that: $\frac{\partial{f}}{\partial{x}}=M, \frac{\partial{f}}{\partial{y}}=N, \frac{\partial{f}}{\partial{z}}=P$ $ \dots \dots \dots "$ (Worried)

Could you explain it to me? (Wondering)

 

[I like Serena]

Hey! (Blush)

At the beginning (till the point $(*)$ ) we suppose that there is a differentiable function $f$ such that $\overrightarrow{F}=\nabla f$ and we show that the integral $\int_A^B \overrightarrow{F} d \overrightarrow{R}$ is equal to $f(B)-f(A)$. Or have I understood it wrong? (Wondering)

You've understood perfectly! (Happy)

Then I haven't understood the proof from the point
"$\overrightarrow{F}(x,y,z)=\hat{\imath} M(x,y,z)+\hat{\jmath} N(x,y,z)+\hat{k} P(x,y,z)$
We will show that: $\frac{\partial{f}}{\partial{x}}=M, \frac{\partial{f}}{\partial{y}}=N, \frac{\partial{f}}{\partial{z}}=P$ $ \dots \dots \dots "$ (Worried)

Could you explain it to me? (Wondering)

Apparently they want to show that the components of $\overrightarrow{F}$ match the partial derivatives of $f$.
But to be fair, I don't understand why they're being so difficult about it, because I believe this is obvious from the definition of $f$. (Doh)

Apparently they take an infinitesimal step in the x direction using $f$, convert it to the corresponding infinitesimal path integral with $\overrightarrow{F}$, and then convert it to the M component of $\overrightarrow{F}$, showing these are the same. (Nerd)

 

[mathmari]

Hey! (Blush)

You've understood perfectly! (Happy)

$\displaystyle{\int_A^B \overrightarrow{F} d \overrightarrow{R}=\int_{t_1}^{t_2} \frac{df}{dt} dt=f(B)-f(A)}$

Shouldn't it be $\displaystyle{\int_{t_1}^{t_2} \frac{df}{dt} dt=f(t_2)-f(t_1)}$

How can we get $f(B)-f(A)$ ? (Wondering)

Apparently they want to show that the components of $\overrightarrow{F}$ match the partial derivatives of $f$.
But to be fair, I don't understand why they're being so difficult about it, because I believe this is obvious from the definition of $f$. (Doh)

Apparently they take an infinitesimal step in the x direction using $f$, convert it to the corresponding infinitesimal path integral with $\overrightarrow{F}$, and then convert it to the M component of $\overrightarrow{F}$, showing these are the same. (Nerd)

Now we know that $\displaystyle{\int_B^{B'}\overrightarrow{F} d \overrightarrow{R}=f(B')-f(B)=f(x'+h, y', z')-f(x', y', z')}$$$\frac{f(x'+h, y', z')-f(x', y', z')}{h}=\frac{1}{h} \int_B^{B'} \overrightarrow{F} d \overrightarrow{R}$$
Why is this equal to $$\frac{1}{h} \int_0^1 M(x'+t \cdot h, y', z') \cdot h dt$$
where there is only $M$, and not $N$ or $P$ ? (Wondering)

 

[I like Serena]

$\displaystyle{\int_A^B \overrightarrow{F} d \overrightarrow{R}=\int_{t_1}^{t_2} \frac{df}{dt} dt=f(B)-f(A)}$

Shouldn't it be $\displaystyle{\int_{t_1}^{t_2} \frac{df}{dt} dt=f(t_2)-f(t_1)}$

How can we get $f(B)-f(A)$ ? (Wondering)

It's the same thing.
The notation is a bit sloppy.
When $f(t)$ is written, it should really be $f(\overrightarrow R(t))$.

So $f(t_1)$ is really $f(\overrightarrow R(t_1)) = f(A)$.

Now we know that $\displaystyle{\int_B^{B'}\overrightarrow{F} d \overrightarrow{R}=f(B')-f(B)=f(x'+h, y', z')-f(x', y', z')}$

$$\frac{f(x'+h, y', z')-f(x', y', z')}{h}=\frac{1}{h} \int_B^{B'} \overrightarrow{F} d \overrightarrow{R}$$
Why is this equal to $$\frac{1}{h} \int_0^1 M(x'+t \cdot h, y', z') \cdot h dt$$
where there is only $M$, and not $N$ or $P$ ? (Wondering)

The point $B'$ has been selected in such a way that it has the same y and z coordinates as $B$.
That is why only the x coordinate is changed by a small amount.
So $d\overrightarrow R$ is a vector aligned with $\hat \imath$ at every point of the path from $B$ to $B'$.
As a result the dot product with $\overrightarrow F$ shows only $M$. (Mmm)

 

[mathmari]

It's the same thing.
The notation is a bit sloppy.
When $f(t)$ is written, it should really be $f(\overrightarrow R(t))$.

So $f(t_1)$ is really $f(\overrightarrow R(t_1)) = f(A)$.

The point $B'$ has been selected in such a way that it has the same y and z coordinates as $B$.
That is why only the x coordinate is changed by a small amount.
So $d\overrightarrow R$ is a vector aligned with $\hat \imath$ at every point of the path from $B$ to $B'$.
As a result the dot product with $\overrightarrow F$ shows only $M$. (Mmm)

Ahaa! Ok! I got it! Thank you for explaining it to me! (Handshake) (Smile)