danny_boy

Hey,

Y = [ax + b/x]^1/2

Now, I made this:

Y = ax^1/2 + (b^1/2 / x^1/2)

Now, I would plot Y vs. X^1/2 for a linear plot, with a slope of a. But whats my intercept ?

Or can someone else suggest another way to linearize, or what the intercept is for the plot I just stated.

Thanks!

rlbond86

no! bad!

you can't distribute an exponent!
square both sides to remove the square root, then work from there.

AngelicVampire

Y^2 = Ax + b/x

So we have a line which does strange things at the origin (when x = 0 b/x = ??) and from then on we have a square relationship that will tend towards Y^2 = Ax (as x-> inf b/x -> 0).

danny_boy

If I square both sides, I get:

Y^2 = aX + b/X

So, I can plot Y^2 vs. X, where the slope is a, and the intercept is b/x..is that correct ?

AngelicVampire

No, both terms include X so there can be no intercept other than the origin for a truly linear equation, however this isn't linear.

normally you have y = mx + c, which gives the intercept as c, in this case we have two X terms, when we evaluate this as X -> 0 then we get 0a + b/+-0 (a very small number), as such at the origin the actual value is undefined, similar to a tan curve at various points.

For large values of X aX dominates the equation and we have a function which says Y^2 = aX or Y = (aX)^1/2, and again for negative X, at the origin the b/X dominates and we have the graph rapidly shooting off to +ve/-ve infinity for +0/-0.

filtherton

When i first read this i thought the op was referring to the linearization where you find the equation for the tangent line at any given point on the graph:

L(x) = f(x1)+f'(x)*(x-x1)

But now i don't really have any clue what the op needs.

AngelicVampire

I thought he was just trying to graph the equation...

danny_boy

Hey,

What I need to do with the original equation, is somehow manipulate it so that it is linear, and I need to specify the slope and y-intercept.

So, So, I can plot Y^2 vs. X, where the slope is a, and the intercept is b/x..

That plot does in fact yield a linear line....and the intercept does work out to be b/x.

Unless I've messed something up...?

AngelicVampire

Well b/x is +/- infinity as x tends to 0... that equation is in no way straight.

Input (8x+8/x)^.5 in this url: http://www.coolmath.com/graphit/

That will show you what the function should look like

danny_boy

I see I see,

So, I am still at a loss how to linearize my original equation...and what the intercept and slope should be..

The whole aim of the question was to linearize a given equation, and specify its slope and int...

filtherton

What class is this for?
Can you use logarithms?

AngelicVampire

http://www.cbu.edu/~rprice/lectures/lineariz.html

Talks about Taylor expansions and suchlike to approximate a non-linear equation with a linear one.

danny_boy

This is an engineering course...

spindles

The Y intercept is simple, isn't it?

solve for X when Y is 0

Y^2=Ax + B/x

0 = Ax + B/x
-Ax = B/x
-Ax^2 = B
x^2 = B/-A
x = (B/-A)^1/2

__________________
who hid my keyboard's PANIC button?

Amano

Are you trying to fit data to a model to solve for a and b?

If so, you could do this:

y^2=ax+b/x
xy^2=ax^2+b

Plot xy^2 vs. x^2. You'll get an intercept of b, and a slope of a.

OR....

y^2/x=a+b/x^2

Plot y^2/x vs 1/x^2. You'll get a intercept of a and a slope of b.