What is the geometric interpretation of the partial derivative?

gulsen
Messages
215
Reaction score
0
Say, E is dependent to x,y,z. I'm expecting it's derivative at x_0,y_0,z_0 to be

dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} E(x_0+\Delta x, y_0+\Delta y,z_0+\Delta z) - E(x_0,y_0,z_0)

But with following definition, it's not the thing above:

dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} \frac{E(x_0+\Delta x, y_0,z_0) - E(x_0,y_0,z_0)}{\Delta x} \Delta x + \frac{E(x_0, y_0+\Delta y,z_0) - E(x_0,y_0,z_0)}{\Delta y} \Delta y + \frac{E(x_0, y_0,z_0+\Delta z) - E(x_0,y_0,z_0)}{\Delta z} \Delta z = \frac{\partial E}{\partial x}dx + \frac{\partial E}{\partial y}dy + \frac{\partial E}{\partial z}dz.

Now, which is correct? (and why?!?)
 
Last edited:
Physics news on Phys.org
Neither one. The differential (not "derivative") is
dE= \left(\lim_{\Delta x\rightarrow0}\frac{E(x_0+\Delta x,y_0,z_0)-E(x_0,y_0,z_0)}{\Delta x}\right)dx+ \left(\lim_{\Delta y\rightarrow 0}\frac{E(x_0,y_0+\Delta y,z_0)-E(x_0,y_0,z_0)}{\Delta y}\right)dy+ \left(\lim_{\Delta z\rightarrow 0}\frac{E(x_0,y_0,z_0+\Delta z)-E(x_0,y_0,z_0)}{\Delta z}\right)dz
That's the same as your second formula except for the "d" rather than "\Delta" in the numerator.

Why would you "expect" the first? In one variable, would you "expect" the differential of f(x) to be
df= lim_{\substack{\Delta x\rightarrow 0}}f(x_0+\Delta x)- f(x_0)
If f is any continuous function, that's 0!
 
Last edited by a moderator:
Whoops, just a typo anyway.

Why would you "expect" the first?

That was just a guesswork based on intuition and my ignorance, I simply don't know. So tell me, why would you expect the latter?
 
Well, at least you could point some links.
 
Or you could look at the definition of the differential in any calculus book:

df(x,y,z)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz
which, if you go back to the definition of the partial derivatives, is the formula I gave.
 
Whoops, we've got a communication problem here, it seems.
I know that this is the definition, and I've written it before you as well. My question is: why? What geometry/mathematical quantity does this correntponds to?
 
From what I've found, it should be like this:
df(x,y,z)= \frac{\partial f}{\partial x}dx \vec{i} + \frac{\partial f}{\partial y}dy \vec{j} + \frac{\partial f}{\partial z}dz \vec{k}
But this way, for the magnitude of this df, we got to multiply (dot product) df with itself and square root...
 
gulsen said:
Whoops, we've got a communication problem here, it seems.
I know that this is the definition, and I've written it before you as well. My question is: why? What geometry/mathematical quantity does this correntponds to?
Just picture f as a function of two variables instead of three; z = f(x,y). Thus f can be represented by a surface S in 3 space. The total differential df in the equation
df(x,y)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy
would represent the change in the z component of the tangent plane T of f at a point P at coordinates (x,y,z) as we go to some other point P' at (x+delta x,y+delta y,f(x+delta x,y+delta y)). In this case, since delta x and delta y are independent variables, we can say dx=delta x and dy=delta y; i.e dx and dy can be actual distance values. Notice that df is related to the z component of points in T but delta f is related to the z component of points in the surface that f generates.

It's even easier to picture if you just consider f as function of one variable: y=f(x). Though you lose the "partial" nature of the derivitive, the geometric picture you were looking for is still there. For example, you can picture some curved line that f generates and some tangent line T at a point P at (x, f(x)). If you take another point on the curve P' at (x+delta x, f(x+delta x)), dy would represent the verticle distance between a horizontal line through P and the tangent line T. In other words, dy would represent the distance that T changes in the y direction for some change dx=delta x in the x direction. We could imagine an animal called delta y (as distinct from dy) that represents the distance from the horizontal line through point P to the CURVE (rather than the tangent line T).

Why would I even begin to set here and type all that?
 
Back
Top