ODE now made me think about derivatives and partial derivativesby flyingpig Tags: derivatives, partial 

#1
Jun1411, 04:21 AM

P: 2,568

1. The problem statement, all variables and given/known data
Let's say I have a function for a circle [tex]x^2 + y^2 = C[/tex] where C is a constant. Then this is a cylinder with the zaxis. Now in my ODE book, we would normally define it as [tex]F(x,y) = C = x^2 + y^2[/tex] as a level surface. Now my question is about what the partial derivative with respect to x mean as opposed to (singlevariable calculus) derivative with respect to x mean. Am I losing anything if I take one derivative over the other? I should mention that many of these problems assume that F(x,y(x)). [tex]\frac{\partial F}{\partial x} = 2x[/tex] [tex]\frac{\partial F}{\partial y} = 2y[/tex] [tex]\frac{\mathrm{d} F}{\mathrm{d} x} = 2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0[/tex] So now my question is, what exactly is this 



#2
Jun1411, 05:18 AM

Math
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Thanks
PF Gold
P: 38,879

[tex]2x+ 2y\frac{dy}{dx}[/tex]
is the rate of change of the function [itex]f(x, y(x))= x^2+ y(x)^2[/itex] with respect to x it measures how fast f(x,y(x)) changes as x changes. Of course that will depend upon exactly how y(x) changes as x changes and that is what dy/dx tells you. Suppose y(x)= 3x. Then [itex]d(x^2+ y^2)/dx= 2x+ 2y dy/dx= 2x+ 2y(3)= 2x+ 2(3x)(3)= 20x[/itex].That is exactly the same as if you had replaced y with 3x from the start: [itex]x^2+ (3x)^2= x^2+ 9x^2= 10x^2[/itex] so [itex]df/dx= 20x[/tex] 



#3
Jun1411, 05:29 AM

P: 2,568





#4
Jun1411, 05:31 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

ODE now made me think about derivatives and partial derivatives
No, the partial derivative of f with respect to x is the rate of change as x change assuming y does not change.




#5
Jun1411, 05:35 AM

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P: 1,716

given a small change in x, dx, the change is x^2 is close to 2xdx and the change in y^2 is close to 2ydy. But dy = (dy/dx)dx. If y were independent of x then dy/dx would be zero. This would just mean that x can change without a change in y. But the constraint C = x^2 + y^2 tells you that y is a function of x, at least locally. 



#6
Jun1411, 05:43 AM

P: 2,568





#7
Jul511, 12:49 AM

P: 2,568

Actualyl could I get a unit representation?
Say F(x,y) was a position function of time (perhaps x(t) = x and y = y(t)) what would the different derivatives tell me? 


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