Proportional Relationship between y, x, and z

[santa]

if y\propto x at z constant

and y\propto z at x constant

then

y\propto xz



why not

y^2\propto xz


thank you

 

[mrandersdk]

if y^2 \propto xz

then you would get

y \propto \sqrt{xz}

so keeping fx. x constant you have

y \propto \sqrt{z}

which is wrong. Maybe a proof could go like this:

assume:

y \propto y for constant z

and

y \propto z for constant x

this must meen that we can write

y(x,z) = f(z) x for some function f and
y(x,z) = g(x) z for some function g

then

g(x) x = f(x) x so for x different from zero you have

g(x) = f(x)

that is

y(x,z) = f(z) x
y(x,z) = f(x) z

so

y(x,1) = f(1) x
y(x,1) = f(x) 1

from which you get

f(x)= f(1) x, inserting this you have

y(x,z) = f(1) z x

which is to say

y(x,z) \propto z x

maybe the proof is flawed did it pretty sloppy.

 

[quadraphonics]

why not

y^2\propto xz

Because then you'd have y\propto \sqrt{x} for a fixed z. However, you can have y \propto x f(x) where f(z) is just-about-any function of z.

 

[santa]

thanks

but a have another


Definition of directly proportional - can k be negative?

In almost all textbooks, "directly proportional" is defined by saying
that a is directly proportional to b if and only if a = kb for some
constant k. That's perfectly sensible, but taking the definition
literally, it would seem to imply that any k will do, even negatives.

However, in every example that I have seen to illustrate the concept,
the term "directly proportional" is always applied to the relationship
between two positive quantities or two negative quantities--never
between a positive quantity and a negative quantity.

 

[Hootenanny]

Yes the constant of proportionality can take any value, positive, negative, real, complex.

 

[santa]

ok the constant of Hooke's_law

F=-KX

k positive, negative, real, complex. or not

 

[Hootenanny]

ok the constant of Hooke's_law

F=-KX

k positive, negative, real, complex. or not

Real and positive.

 

[esbo]

if y\propto x at z constant

and y\propto z at x constant

then

y\propto xz



why not

y^2\propto xz


thank you

Would that not lead to:-

y\propto y^2

Irregardless of the first two statements.

Or have I grossly missed the point