How to linearise translated and dilated surd functions?

[Stanley_physics]

So, in Physics, we were to learning about linearising data of a practical. We know that if the graph of the data of a practical or experiments represents a surd function (y=x^0.5), then it can be linearised by graphing y^2 against x. Therefore, y^2 would be directly proportional to x. However, if you have a translated and dilated surd, would this still be possible. For example, in the equation y = 4 + 9(23+7x)^0.5, is y^2 still directly proportional to x? If not, what would be proportional in this case, and how would i graph it to get a perfectly linear line? This is a difficult concept for me to grasp. Help would be veryy appreciated.

Thanks.

 

[haruspex]

in the equation y = 4 + 9(23+7x)^0.5, is y^2 still directly proportional to x?

That is easy to check. You have an expression for y. What does that give for y²?
How can you rearrange the equation so that when you square both sides the square root terms disappear?

 

[Stanley_physics]

Yeah but when i try to expand after squaring both sides, i get a two x values. For example, continuing from the example i gave above, if i square both sides, i get y^2 =567x + 1879 +72(√7x+23).
I don't know how to work with that.
Also, when i square after rearranging, i get:
162x = y^2 -8y -1867.
Don't know how i can use that to linearise my graph either.

 

[haruspex]

if i square both sides, i get y^2 =567x + 1879 +72(√7x+23).

Right, so y² doesn't do it.

when i square after rearranging, i get:
162x = y^2 -8y -167.

You expanded too much. Look at the steps you took to get that. If you can't spot the line that does it, please post all your steps.

 

[Stanley_physics]

Couldn't find it.

y = 4 + 9(23+2x)^0.5
y-4 = 9(23+2x)^0.5
(y-4)^2 = (9(23+2x)^0.5)^2
y^2 - 8y +16 = 1863 +162x
y^2 -8y - 1867 = 162x

 

[haruspex]

Couldn't find it.

y = 4 + 9(23+2x)^0.5
y-4 = 9(23+2x)^0.5
(y-4)^2 = (9(23+2x)^0.5)^2
y^2 - 8y +16 = 1863 +162x
y^2 -8y - 1867 = 162x

Take a closer look at your third line. It's nearly there. The fourth has overshot.

 

[Stanley_physics]

(y-4)^2 / 81 = 23 + 7x.
Got it.
But the problem is, how can i graph this to create a linear relationship. Thats what i am stuck on.

 

[haruspex]

(y-4)^2 / 81 = 23 + 7x.
Got it.
But the problem is, how can i graph this to create a linear relationship. Thats what i am stuck on.

To linearise a relationship for graphing, you just need to find two functions, f(x), g(y), such that g(y) and f(x) have a linear relationship. Then you plot ... what do you think?

 

[Stanley_physics]

Sorry, i did not understand that. How does that work?

 

[haruspex]

Sorry, i did not understand that. How does that work?

If f(x) and g(y) have a linear relationship, what do you expect to see if you plot a graph of g(y) against f(x)?

 

[Stanley_physics]

A straight line.
But i am rather looking to do something like this;
Kepler's third law for example, it can be graphed as shown in the picture.
I want to do something like that to form a linear relationship rather than graphing the functions themselves if you get what i mean.

 

[haruspex]

A straight line.

Right.

But i am rather looking to do something like this

Yes, I believe I understand what you are trying to do.
Look at your post #9. You have a function of y on the left and a function of x on the right, and a linear relationship between them.
But in that equation, there are no unknown constants. More generally, you do have unknown constants that you are trying to determine by graphing. Clearly it is no use choosing the f and g functions in such a way that they contain unknown constants, because in that case you would not be able to plot the values of f(x) and g(y).
In your post #9 equation, this presents no difficulty if any of the 81, the 23 or the 7 is really an unknown, but it would be a problem if the 4 is unknown.