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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
1) What's a good 3d graphing program? Preferably GUI and open source.

2) I'll have to show you:
Code:
f(x, y) = 5x^2 - 3xy + y^2
d/dx f = 10x - 3y + dy/dx(-3x + 2y) Right?
    

From here, it's confusing. I know dy/dx can be rewritten as d/dx y, so does that mean that:
Code:
dy/dx(-3x + 2y) == d/dx y(-3x + 2y)
d/dx y(-3x + 2y) == d/dx(-3xy + 2y^2)
d/dx(-3xy + 2y^2) == d/dx(-3xy) + d/dx(2y^2)
d/dx(-3xy) + d/dx(2y^2) == -3y + d/dx(2y^2)

... and finally:
d/dx f = 10x - 6y + d/dx(2y^2) ???
    


3) Assuming I'm wrong, can you show me the step by step solution to the above problem, please?
Post 09 Nov 2010, 00:16
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Explanations are presented in the picture attached below (PDF version of it you may see here):
Image
Smile
Post 09 Nov 2010, 11:35
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
Thanks man. I appreciate the time you must have taken to give such a detailed explanation.

What would make x and y dependant? By dependant, do you mean one is used in the others definition. Like y is defined in terms of x for example?

Should the planes z=10x-3y and z=-3x+2y be tangent to z=5x^2-3xy+y^2? The graph I made with Maxima seems to contradict my assumption that they should be. You can see how z=10x-3y crosses into the bottom of the "dip" in z=5x^2-3xy+y^2 to create a flat place and a slight change in color.


Description:
Filesize: 100.12 KB
Viewed: 3134 Time(s)

image.png


Post 09 Nov 2010, 22:49
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ctl3d32



Joined: 30 Dec 2009
Posts: 204
Location: Brazil
ctl3d32
It is tangent. You see this glitch bacause your curve has too few polygons. Try increasing the number of polygons and you will see it is tangent.
Post 09 Nov 2010, 23:14
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
clt3d32, it is a graph of a function and it's indefinite derivative. I was graphing f(x,y), d/dx f(x,y), and d/dy f(x,y) which aren't tangent. I meant to graph f(x,y) and a definite derivative for a point. I've got a function now to graph d/dx tangents but I'm having trouble with the concept of tangents in y... I post a complete graph when I've got both working(If I do...).
Post 10 Nov 2010, 01:19
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
How 'bout now?

Is there a general form for this? Given df/dx and df/dy.


Description:
Filesize: 99.88 KB
Viewed: 3103 Time(s)

image.png


Post 10 Nov 2010, 05:26
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Some explanations are presented in the following pictures (printer-friendly PDF document you may see here Wink ):
ImageImage
Post 13 Nov 2010, 00:15
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ctl3d32



Joined: 30 Dec 2009
Posts: 204
Location: Brazil
ctl3d32
Nice! You've made me remember some calculus.
Post 13 Nov 2010, 01:59
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DarkAlchemist



Joined: 08 Oct 2010
Posts: 108
DarkAlchemist
"I know dy/dx can be rewritten as d/dx y"

Wouldn't that simplify down to x/y since d resolves to 1 so we do not need d at all in that case?
Post 13 Nov 2010, 09:06
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
DarkAlchemist wrote:
"I know dy/dx can be rewritten as d/dx y"

Wouldn't that simplify down to x/y since d resolves to 1 so we do not need d at all in that case?
d is the operator of differentiation, and can't cancel out. It comes from Gottfried Leibniz's notation for derivatives. It is the equivalent to Newton's prime notation, ei f(x) = x^2, f'(x) = 2x.
Post 13 Nov 2010, 14:55
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