Linearising an Equation to the Form y=mx+c
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I have been working on some data analysis stuff and I have to linearise this equation so I can plot it as a straight line with form y=mx+c
$$T=2pisqrt{frac{(k^2 + h^2)}{gh}}$$
Where, T will be the y and h the x, k is a constant but g is a variable.
But no matter how I've manipulated it I can't get just one h. any help or pointers would be appreciated.
graphing-functions data-analysis linearization
asked Nov 21 at 12:43
Ethilios
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I have been working on some data analysis stuff and I have to linearise this equation so I can plot it as a straight line with form y=mx+c
$$T=2pisqrt{frac{(k^2 + h^2)}{gh}}$$
Where, T will be the y and h the x, k is a constant but g is a variable.
But no matter how I've manipulated it I can't get just one h. any help or pointers would be appreciated.
graphing-functions data-analysis linearization
asked Nov 21 at 12:43
Ethilios
1
How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have been working on some data analysis stuff and I have to linearise this equation so I can plot it as a straight line with form y=mx+c
$$T=2pisqrt{frac{(k^2 + h^2)}{gh}}$$
Where, T will be the y and h the x, k is a constant but g is a variable.
But no matter how I've manipulated it I can't get just one h. any help or pointers would be appreciated.
graphing-functions data-analysis linearization
asked Nov 21 at 12:43
Ethilios
1
I have been working on some data analysis stuff and I have to linearise this equation so I can plot it as a straight line with form y=mx+c
$$T=2pisqrt{frac{(k^2 + h^2)}{gh}}$$
Where, T will be the y and h the x, k is a constant but g is a variable.
But no matter how I've manipulated it I can't get just one h. any help or pointers would be appreciated.
graphing-functions data-analysis linearization
graphing-functions data-analysis linearization
asked Nov 21 at 12:43
Ethilios
1
asked Nov 21 at 12:43
Ethilios
1
asked Nov 21 at 12:43
Ethilios
1
asked Nov 21 at 12:43
Ethilios
1
asked Nov 21 at 12:43
Ethilios
1
1
How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45
add a comment |
How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45
How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45
add a comment |
1 Answer
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0
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I don't believe this can be done with the information you gave us. These are two options:
The most obvious thing you can do is call $x = sqrt{(k^2+h^2)/gh}$ so that you have a model of the form $T = x/2pi$, but I'm not sure this is what you're looking for
If you plot this in log scale you'll see that if you around the model behave fairly linear for $h / k ll 1$ and $h /k gg 1$. You can Taylor expand the model in these two regimes up to linear order in log scale
answered Nov 21 at 15:13
caverac
12.5k21027
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I don't believe this can be done with the information you gave us. These are two options:
The most obvious thing you can do is call $x = sqrt{(k^2+h^2)/gh}$ so that you have a model of the form $T = x/2pi$, but I'm not sure this is what you're looking for
If you plot this in log scale you'll see that if you around the model behave fairly linear for $h / k ll 1$ and $h /k gg 1$. You can Taylor expand the model in these two regimes up to linear order in log scale
answered Nov 21 at 15:13
caverac
12.5k21027
add a comment |
up vote
0
down vote
I don't believe this can be done with the information you gave us. These are two options:
The most obvious thing you can do is call $x = sqrt{(k^2+h^2)/gh}$ so that you have a model of the form $T = x/2pi$, but I'm not sure this is what you're looking for
If you plot this in log scale you'll see that if you around the model behave fairly linear for $h / k ll 1$ and $h /k gg 1$. You can Taylor expand the model in these two regimes up to linear order in log scale
answered Nov 21 at 15:13
caverac
12.5k21027
add a comment |
up vote
0
down vote
up vote
0
down vote
I don't believe this can be done with the information you gave us. These are two options:
The most obvious thing you can do is call $x = sqrt{(k^2+h^2)/gh}$ so that you have a model of the form $T = x/2pi$, but I'm not sure this is what you're looking for
If you plot this in log scale you'll see that if you around the model behave fairly linear for $h / k ll 1$ and $h /k gg 1$. You can Taylor expand the model in these two regimes up to linear order in log scale
answered Nov 21 at 15:13
caverac
12.5k21027
I don't believe this can be done with the information you gave us. These are two options:
The most obvious thing you can do is call $x = sqrt{(k^2+h^2)/gh}$ so that you have a model of the form $T = x/2pi$, but I'm not sure this is what you're looking for
If you plot this in log scale you'll see that if you around the model behave fairly linear for $h / k ll 1$ and $h /k gg 1$. You can Taylor expand the model in these two regimes up to linear order in log scale
answered Nov 21 at 15:13
caverac
12.5k21027
answered Nov 21 at 15:13
caverac
12.5k21027
answered Nov 21 at 15:13
caverac
12.5k21027
answered Nov 21 at 15:13
caverac
12.5k21027
12.5k21027
add a comment |
add a comment |
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How does $g$ change with the other variables in the problem?
– caverac
Nov 21 at 13:23
you may consider a three dimension coordinate and assume z as g.
– sirous
Nov 21 at 14:07
@caverac I'm not sure however, I think (based on hints further on) that it is somehow incorporated into the gradient of the line. Later it suggests you can calculate and find the error for g from the gradient of the line of least squares which I plot based on the linearised data points. However I obviously can't get the linearised data points without the equation being linearised first.
– Ethilios
Nov 21 at 14:45