Help with understanding inexact differential

[granzer]

In

How is equation 145 giving a direction(ie gradient) and not a slope?.

Also here

how is equation 147 arrived at?

Any help would be much appreciated.

 

[granzer]

Can someone please explain this to me. Haven't been able to find an answer .

 

[Mark44]

In

View attachment 214932

How is equation 145 giving a direction(ie gradient) and not a slope?.Also here

View attachment 214933

how is equation 147 arrived at?

Any help would be much appreciated.

I can just barely read the images you posted.
For your first question, in the US, we tend to call $\frac {dy}{dx}$ a slope; in Europe, people tend to call this a gradient. I prefer to reserve the term gradient to functions such as this: If z = f(x, y), then $\nabla z = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$

If y = 3x, then y' or ($\frac{dy}{dx}$) = 3. This gives a direction in the sense that from any point on the graph of this line, you can get to another point by going right 1 unit and then up 3 units.

Same idea for your second question.

 

[granzer]

I can just barely read the images you posted.
For your first question, in the US, we tend to call $\frac {dy}{dx}$ a slope; in Europe, people tend to call this a gradient. I prefer to reserve the term gradient to functions such as this: If z = f(x, y), then $\nabla z = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$

If y = 3x, then y' or ($\frac{dy}{dx}$) = 3. This gives a direction in the sense that from any point on the graph of this line, you can get to another point by going right 1 unit and then up 3 units.

Same idea for your second question.

@Mark44 Hello Sir,

Thank you so much. The answers u gave have cleared my doubt about equation 147 and I completely agree with you. I was wondering if it should be given as a slope. And yes going right 1 step and going up 3 step gives a direction from the point. But it is also true going 3 step down and one step back we can get another point. So the direction is both front and back and not a particular direction that would be given by a gradient. But if we take gradient to mean slope here then my doubt is cleared.

But with regard to eq 147, it is given:

$∂σ/∂x=X'∂σ/∂y=X'Y'/τ$

I am not understanding how #X′Y′/τ# is got here.
Any clarification on this would be really helpful.
So sorry about the image quality. The image I uploaded was very good. Don't know how to upload the image with that resolution here. Here is the image I uploaded https://i.stack.imgur.com/APHR9.png

Also, the first image is https://i.stack.imgur.com/cj7YG.png