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iwasbiggs

Difference between the partial() and d() in calculus?

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Well I''ve been going through multivariable calculus and I can''t help but wonder why there is a distinction between d and the partial (d on it''s side looking thing). Can anybody give a resonable explanation? I think that the d was adopted to give people an easier understanding or something at first... because I really see no difference between the two operators. ___________________________ Freeware development: ruinedsoft.com

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There is a big difference. The partial derivative assumes that everything except the differentiation variable is constant, so there is no dy, dz, dwhatever when differentiating over x. They are all 0, and @x/@x = dx/dx = 1, so that's the only thing remaining.

Cédric

[edited by - cedricl on October 20, 2002 6:07:29 PM]

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when you''re differentiating for more than one variable the gradient (d) is a sum of the partial differentials, so in that respect they''re very different. but if you were do find the partial differential and the ''full'' differential with respect to a single variable the process is exactly the same.

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quote:
Original post by joanusdmentia
when you''re differentiating for more than one variable the gradient (d) is a sum of the partial differentials, so in that respect they''re very different. but if you were do find the partial differential and the ''full'' differential with respect to a single variable the process is exactly the same.


Its not exactly true that the gradient is the sum of the partial derivatives. Its actually a vector, even if the original function is a scalar. The gradient is given by the del operator, which is:

del = d/dx*i + d/dy*j + d/dz*k

where i, j, and k are the basis vectors of an orthogonal coordinate system. For example, for a normal cartesian 3D space if you have a point (3,2,7), it could be represented as 3i+2j+7k, with i = (1,0,0), j = (0,1,0), and k = (0,0,1). The gradient turns scalar functions into a vector function.

So the gradient of f is:

grad(f) = del * f = df/dx*i + df/dy*j + df/dz*k



Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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To confuse things even more, there is also a thing called a "total" or "substantial" derivative, which uses a capital D instead of a small d. And it is defined in terms of partial derivatives. The total derivative shows up in computational fluid dynamics, and rarely in game development. It is defined exclusively a D?/Dt, never D?/dx, and applies only to vector functions.

For the curious, it is defined as:

D?/Dt = d?/dt + DotProduct(del, ?)

See my post above for the definition of del. The result of this, for f = a vector function of t, x, y, and z, is:

D?/Dt = d?/dt + projection of df/dx into direction i + projection of df/dy into direction j + projection of df/dz into direction k

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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quote:
Original post by cedricl
There is a big difference. The partial derivative assumes that everything except the differentiation variable is constant...
Cédric

[edited by - cedricl on October 20, 2002 6:07:29 PM]


Uh... When I did single variable calculus in High School and differentiated over one variable, eg. d(y * x)/dx = 1 * y, we treated the y as constant because y wasn't a function of x. The same rule applies to the partial of f as I have defined above.

So basically in my single variable calculus class they should have used a partial instead of a d? just because it had another variable? or do both operators work properly in this sense?

The operation of the two operators seems ambiguous still.

___________________________
Freeware development:
ruinedsoft.com

[edited by - iwasbiggs on October 21, 2002 2:30:42 AM]

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quote:
Original post by iwasbiggs
Uh... When I did single variable calculus in High School and differentiated over one variable, eg. d(y * x)/dx = 1 * y, we treated the y as constant because y wasn''t a function of x. The same rule applies to the partial of f as I have defined above.

If y is not a function of x already, then sure, never treat it as a variable. Well, you can treat it as a variable, but

d(y * x)/dx = dy/dx * x + y * dx / dx = 0

if dy / dx = 0

Graham, I''m working on the "total derivative" right now. We call it the divergence. Your statement:
quote:
grad(f) = del * f

that looks a lot like the divergence.

div(f) = del . f

When we write about the gradient vector, the del operator is directly in front of f

grad(f) = del f (without the space)

Anyway, that''s just notation. You forgot to mention the curl and the Laplacian

Cédric

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Guest Anonymous Poster
I guess it''s like with normal multiplication, the dot is usually omitted, but when it''s not it looks a lot like the dot product. There is no ambiguity though, since grad only operates on scalar functions.

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partial() reads "partial derivative", while d() reads "differential". The partial derivative of a function is another function. The differential of a function is not a function, but something called a 1-form. In one dimension, a 1-form just looks like a function times dx (like what you usually see under the integral sign). This makes partial() and d() virtually (but not exactly) the same for one variable calculus.

Most of the time, we should use partial() instead of d(), but there is a tradition in Math of using d() in one variable calculus. I would recommend using partial() even for the single variable case.


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quote:
Original post by grhodes_at_work
Its not exactly true that the gradient is the sum of the partial derivatives. Its actually a vector, even if the original function is a scalar.



Actually, it''s a one-form, not a vector. We just teach undergrads (particularly engineers) that it''s a vector, because it turns out that in 3 dimensions, one-forms and vectors have the same properties and there''s no point in confusing them!

quote:
Original post by grhodes_at_work
The gradient is given by the del operator, which is:

del = d/dx*i + d/dy*j + d/dz*k

where i, j, and k are the basis vectors of an orthogonal coordinate system.



This is where the ''lie'' comes in... i ,j and k are actually basis one-forms.

Why do I bother pointing this out? Because it makes a difference if you want to start applying these concepts in areas such as general relativity, or for taking derivatives on higher dimensional manifolds... not that this happens in game development very often!

Cheers,

Timkin

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quote:
Original post by Timkin
Actually, it''s a one-form, not a vector. We just teach undergrads (particularly engineers) that it''s a vector, because it turns out that in 3 dimensions, one-forms and vectors have the same properties and there''s no point in confusing them!



Good point! The higher-dimension manifold thing comes up very frequently in optimization problems, which we''ve dealt with.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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quote:
Original post by cedricl
Graham, I'm working on the "total derivative" right now. We call it the divergence. Your statement:
"grad(f) = del * f" - from grhodes_at_work

that looks a lot like the divergence.

div(f) = del . f

When we write about the gradient vector, the del operator is directly in front of f

grad(f) = del f (without the space)

Anyway, that's just notation. You forgot to mention the curl and the Laplacian



Ha ha! Glad to have a bunch of regulars to fill in the blanks, correct my mistakes, etc.!

The "del f" notation is probably more consistent. I actually wasn't thinking about divergence when I wrote that last night. Which is why I wrote the total derivative using a sort of DotProduct formula instead of "del . f". Its hard to know how people interpret attempts at ASCII-formatted math.

The total derivative has the divergence term in there (representing change due to advection/convection of a property with mass movement), but it also includes the time derivative (representing a local change in a property). Its called total since it includes changes that are occurring locally and changes that are happening since the new properties are being carried in with fluid mass that comes from another location.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

[edited by - grhodes_at_work on October 21, 2002 10:42:06 PM]

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