The reciprocal of a derivative

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  • #1
B4ssHunter
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if dX/dY is the rate of change of X with respect to Y
say that dX/dY = 3
now would it be correct if i say that the rate of change of Y with respect to X = 1/3 = dY/dX ?
 

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  • #2
arildno
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Yes, it is. (but there are many pitfalls in more general situations!)

In YOUR case, suppose you are at point (x_0,y_0), where y_0=Y(x_0), and where dY/dX=3.

That means that in a neigbourhood of (x_0,y_0), Y(x) can be approximated by:
[tex]Y\approx{y}_{0}+\frac{dY}{dx}|_{x=x_{0}}(x-x_{0})[/tex]
That is, you function looks like the straight line:
[tex]y=y_{0}+3(x-x_{0})[/tex]
Locally, you can invert this relation, solving x in terms of "y", and we may write:
[tex]x=x_{0}+\frac{1}{3}(y-y_{0})[/tex]

But, this is "of course", the same as saying, roughly, that dX/dY=1/(dY/dX)=1/3
 
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  • #3
B4ssHunter
178
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Yes, it is. (but there are many pitfalls in more general situations!)

In YOUR case, suppose you are at point (x_0,y_0), where y_0=Y(x_0), and where dY/dX=3.

That means that in a neigbourhood of (x_0,y_0), Y(x) can be approximated by:
[tex]Y\approx{y}_{0}+\frac{dY}{dx}|_{x=x_{0}}(x-x_{0})[/tex]
That is, you function looks like the straight line:
[tex]y=y_{0}+3(x-x_{0})[/tex]
Locally, you can invert this relation, solving x in terms of "y", and we may write:
[tex]x=x_{0}+\frac{1}{3}(y-y_{0})[/tex]

But, this is "of course", the same as saying, roughly, that dX/dY=1/(dY/dX)=1/3

okay i understand but i have two questions
what does the underscore you used at the beginning mean ?
in (x_0 ) for instance
and second , in a linear equation , we describe the slope as the coefficient of (y-y0) Or (x-x0) ?
or is it such that the rate of change of y with respect to X is the co-efficient of ( x - xnaught ) and the rate of change of X with respect to y in another equation is the co-efficient of (y-y0) ?
also does this only apply to linear equations ?
 
  • #4
arildno
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I use (x_0, y_0) to denote a specific point VALUE, to distinguish from the VARIABLES (x,y)
 

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