Physical meaning of equations of equilibrium

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The discussion focuses on understanding the physical meaning of the equation of equilibrium involving the term "σy + (∂σy/∂y)(δy)." The second part of the equation represents the partial derivative of σy with respect to y, multiplied by a small change in y, indicating that σy may depend on other variables. The context of the equation is crucial for interpreting its physical significance, particularly in applications like aerodynamics. Participants emphasize the importance of grasping the underlying concepts of calculus and differentiation rather than relying solely on memorization. A recommendation is made to revisit basic calculus to better understand these mathematical principles.
Aravind Raj
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From the equations of equilibrium
"σy+(∂σy/∂y)*(lower case delta y)" is the force acting along y direction. can anybody explain the physical meaning of the second part of the force where we multiply delta and del?
 
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Welcome to PF;
What is your education level?

Do you mean: $$\sigma_y + \left[\frac{\partial \sigma_y}{\partial y}\right]\delta y$$ ... in which case, "we" do not "multiply delta and del".

##\delta y## is one variable and reads: "a small change in y"
##\partial## implies a partial derivative. It means that ##\sigma_y## may depend on variables other than y
In the above expression, the partial derivative of ##\sigma_y## with respect to y is being multiplied by a small change in y. This product is added to ##\sigma_y##.
The physical meaning depends on context.

There are many equations of equilibrium: where did you find that particular expression?
 
Simon Bridge said:
Welcome to PF;
What is your education level?

Do you mean: $$\sigma_y + \left[\frac{\partial \sigma_y}{\partial y}\right]\delta y$$ ... in which case, "we" do not "multiply delta and del".

##\delta y## is one variable and reads: "a small change in y"
##\partial## implies a partial derivative. It means that ##\sigma_y## may depend on variables other than y
In the above expression, the partial derivative of ##\sigma_y## with respect to y is being multiplied by a small change in y. This product is added to ##\sigma_y##.
The physical meaning depends on context.

There are many equations of equilibrium: where did you find that particular expression?
I know we don't multiply del and delta, typo while expressing my doubt. Thanks for explaining it. Its a part of derivation for equation of equilibrium. Book: Aircraft structures by MegsonALSO CAN U SUGGEST ANY BOOKS TO UNDERSTAND BASICS OF CALCULUS, DIFFERENTIATION ?
 
OK - so it's probably something to do with airflows ... in the text of the book by the equation it should have told you what the different symbols mean.
A partial derivative is not the same as "a part of derivative" - the maths of aircraft is usually very complicated.

Note: you can take an online beginning calculus course - best value for money.
I cannot possibly advise you on which one or which text will help you because I don't know your education background so far.
Have a look at: http://nrich.maths.org/4722 see if it's too easy of too hard.
 
Simon Bridge said:
OK - so it's probably something to do with airflows ... in the text of the book by the equation it should have told you what the different symbols mean.
A partial derivative is not the same as "a part of derivative" - the maths of aircraft is usually very complicated.

Note: you can take an online beginning calculus course - best value for money.
I cannot possibly advise you on which one or which text will help you because I don't know your education background so far.
Have a look at: http://nrich.maths.org/4722 see if it's too easy of too hard.
I'm doing my 3rd year Aeronautical engineering. The way I've been taught in my school and college, I can solve almost all the differential equations, I know the formulae ,partial differentiation etc. I don't know the physical meanings of those differentiations. so i want a book that can help me with understanding the applications and physical meaning.
 
The physical meaning depends on the context ... like all maths.

For instance - if the function represents a surface, then the slope of the surface in a particular direction is the derivative of the function along that direction.

Put most simply:
The derivative is how much something is changing with the variable the derivative is taken against.
What that means depends on the something and the variable.

Simplest example is that the time-derivative of the displacement is the velocity, and the time-derivative of the velocity is the acceleration.
More tricky: the space-derivative of work is the force - which is also the time derivative of momentum.See the problem with answering your question?
I don't think there are any books covering this topic - it's too variable.
 
Simon Bridge said:
The physical meaning depends on the context ... like all maths.

For instance - if the function represents a surface, then the slope of the surface in a particular direction is the derivative of the function along that direction.

Put most simply:
The derivative is how much something is changing with the variable the derivative is taken against.
What that means depends on the something and the variable.

Simplest example is that the time-derivative of the displacement is the velocity, and the time-derivative of the velocity is the acceleration.
More tricky: the space-derivative of work is the force - which is also the time derivative of momentum.See the problem with answering your question?
I don't think there are any books covering this topic - it's too variable.
okay. thanks. I will go thru the basics again and understand the concepts instead of just memorising. :D
 
That's the idea - you can go a long way in your education on memorization skills alone but usually this trips people up at senior undergrad level at the latest.
If you studied pure science the "memorization" approach would have tripped you up much sooner.

The trick is to see the maths as a language.
 

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