linearizing equations - Tilted Forum Project Discussion Community

danny_boy · 01-13-2006, 07:41 AM

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 · 01-13-2006, 10:05 AM

no! bad!

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

AngelicVampire · 01-13-2006, 10:12 AM

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 · 01-13-2006, 03:17 PM

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 · 01-13-2006, 09:30 PM

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 · 01-14-2006, 04:39 AM

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 · 01-14-2006, 12:02 PM

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

danny_boy · 01-15-2006, 10:52 AM

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 · 01-16-2006, 07:08 AM

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 · 01-16-2006, 11:03 AM

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 · 01-16-2006, 11:52 AM

What class is this for?
Can you use logarithms?

AngelicVampire · 01-17-2006, 08:34 AM

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 · 01-17-2006, 02:18 PM

This is an engineering course...

spindles · 01-17-2006, 09:26 PM

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

Amano · 01-18-2006, 08:18 AM

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.

01-13-2006, 07:41 AM	#1 (permalink)
danny_boy Crazy	linearizing equations 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!

01-17-2006, 09:26 PM	#14 (permalink)
spindles Mine is an evil laugh Location: Sydney, Australia	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?

01-13-2006, 10:05 AM	#2 (permalink)
rlbond86 Insane	no! bad! you can't distribute an exponent! square both sides to remove the square root, then work from there.

01-13-2006, 10:12 AM	#3 (permalink)
AngelicVampire Insane	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).

01-13-2006, 03:17 PM	#4 (permalink)
danny_boy Crazy	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 ?

01-13-2006, 09:30 PM	#5 (permalink)
AngelicVampire Insane	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.

01-14-2006, 04:39 AM	#6 (permalink)
filtherton Junkie Location: In the land of ice and snow.	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.

01-14-2006, 12:02 PM	#7 (permalink)
AngelicVampire Insane	I thought he was just trying to graph the equation...

01-15-2006, 10:52 AM	#8 (permalink)
danny_boy Crazy	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...?

01-16-2006, 07:08 AM	#9 (permalink)
AngelicVampire Insane	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

01-16-2006, 11:03 AM	#10 (permalink)
danny_boy Crazy	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...

01-16-2006, 11:52 AM	#11 (permalink)
filtherton Junkie Location: In the land of ice and snow.	What class is this for? Can you use logarithms?

01-17-2006, 08:34 AM	#12 (permalink)
AngelicVampire Insane	http://www.cbu.edu/~rprice/lectures/lineariz.html Talks about Taylor expansions and suchlike to approximate a non-linear equation with a linear one.

01-17-2006, 02:18 PM	#13 (permalink)
danny_boy Crazy	This is an engineering course...

01-18-2006, 08:18 AM	#15 (permalink)
Amano Insane	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.