Question

\[d(y(u,v))=\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv\]Why?

Answer 1

Or: why aren't there half signs in front of each term?

Answer 2

Right now I see the equation as\[dy=dy+dy=2dy\]

Answer 3

those dy's are not variables, its not like xy = xy/2+xy/2
there is product rule involved (fg)' = fg'+f'g

Answer 4

I don't understand how to expand the product rule into a function of 2 variables.

Answer 5

that is just like linear approximation on a function with 2 independent variable.

Answer 6

\[\frac{dz(a)y(b)}{d?}=\frac{dz}{da}y+\frac{dy}{db}z\]

Answer 7

What is the question mark, or is my premise incorrect?

Answer 8

Is the concept in the question a little like the gradient of a 2-variable function?

Answer 9

With du and dv infinitesimal unit vectors?

Answer 10

you know Taylor expansion of two independent variables?

Answer 11

No

Answer 12

oh ... this relation works our if \( \phi \) is function of x, y, z ... but what you wrote is more fundamental than this one.
\[ d\phi = \nabla \phi \cdot d\vec r \]

Answer 13

you know linear approximation right?

Answer 14

No, but google says the first 2 terms of a function's Taylor E = its linear approximation. Correct?

Answer 15

Single variable, that is.

Answer 16

\[ f(x) = f(a) + f'(a)(x-a) + ... \implies f(x) - f(a) = \Delta f = {df \over dx}\Delta x \]

Answer 17

\[f'(a)x-f'(a)a=f(x)-2f(a)\] I'm probably being an idiot, but I don't see it.

Answer 18

f(x) - f(a) = this is delta f ...
x - a = this is delta x

Answer 19

I understand that, but not the equalsarrow part

Answer 20

\[ f(x) = f(a) + f'(a)(x-a) \\
f(x) - f(a) = f'(x) (x-a) \\
\Delta f = f'(x) \Delta x\]

Answer 21

OK- I've got that (sorry, thought that equals arrow was 'equal to')

Answer 22

all right ... when x->a
delta f = df
delta x = dx
this is linear approximation for a single variable function.

do you know how to make linear approx for multivariate function?

Answer 23

No

Answer 24

use the same concept as single variable function ... use steps and ignore the dy dx (quadratic term) ... i'll give you in short.

Answer 25

yeah that's the one.

Answer 26

put f(x,y) - f(a,b) = df
x-a = dx
y-b = dy

Answer 27

Oh, I see.

Answer 28

I understand your answer algebraically, but is gradient a good conceptual model to hold about it?

Answer 29

consider single it variable x
\[ f(a+dx, b+dy) = f(a, b+dy) + {\partial f (a, b+dy)\over \partial x}_{x=a}\; dx --- (1)\]
consider it single variable in y
\[ f(a, b+dy) = f(a, b) + { \partial f \over \partial y} _{y=b}dy ---- (2)\\
{\partial f (a, b+dy)\over \partial x} dx = {\partial f (a, b)\over \partial x}_{x=a}dx + {\partial \; f\over \partial x \partial y}_{a,b} dx dy ---- (3)\]

put everything at (1) ... and drop the quadratic terms. you will get desired result.

Answer 30

probably this is more fundamental concept than gradient.

Answer 31

Is it a branch of calculus that gradient is a twig of, or are the concepts completely unrelated?

Answer 32

nothing is unrelated.

Answer 33

Call the function of intensity of relatedness between two mathematical concepts a and b h(a,b). Is\[\left \langle h(a,b) \right \rangle<h(gradient , \text{ 2 variable linear approximation})\]?

Answer 34

Taking the modulus of the values first, of course.

Answer 35

both are very useful technique ... you use gradient more on vector calculus .. you use this approximation everywhere. especially on physics.

Answer 36

So it's not used in vector calculus? What are some examples? Thanks for your time, but I've got to go.

Answer 37

it's used everywhere.