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ODE now made me think about derivatives and partial derivatives

by flyingpig
Tags: derivatives, partial
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flyingpig
#1
Jun14-11, 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 z-axis.

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 (single-variable 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
[tex] 2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0[/tex]
as opposed to
2x
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HallsofIvy
#2
Jun14-11, 05:18 AM
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[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]
flyingpig
#3
Jun14-11, 05:29 AM
P: 2,568
Quote Quote by HallsofIvy View Post
[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.
I thought that was what the partial derivative with respect to x is

HallsofIvy
#4
Jun14-11, 05:31 AM
Math
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PF Gold
P: 39,293
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.
lavinia
#5
Jun14-11, 05:35 AM
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Quote Quote by flyingpig View Post
I should mention that many of these problems assume that F(x,y(x)).
Right.

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.
flyingpig
#6
Jun14-11, 05:43 AM
P: 2,568
Quote Quote by lavinia View Post
Right.

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.
If I were to graph all three of those "derivatives" what would they look like? How do yuo even graph Fx alone?
flyingpig
#7
Jul5-11, 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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