Derivative to function ratio
Clash Royale CLAN TAG#URR8PPP
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Is there a physical meaning for derivative of the function to function ratio? That is, this quantity,
$$
Q(x) = frac1f(x)fracdf(x)dx
$$
Like for instance, if $f$ is the potential energy, this would be work to potential energy ratio.
Or even what can we say about $Q(x)$, say when $Q(x)<0$.
derivatives
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up vote
1
down vote
favorite
Is there a physical meaning for derivative of the function to function ratio? That is, this quantity,
$$
Q(x) = frac1f(x)fracdf(x)dx
$$
Like for instance, if $f$ is the potential energy, this would be work to potential energy ratio.
Or even what can we say about $Q(x)$, say when $Q(x)<0$.
derivatives
This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
1
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a physical meaning for derivative of the function to function ratio? That is, this quantity,
$$
Q(x) = frac1f(x)fracdf(x)dx
$$
Like for instance, if $f$ is the potential energy, this would be work to potential energy ratio.
Or even what can we say about $Q(x)$, say when $Q(x)<0$.
derivatives
Is there a physical meaning for derivative of the function to function ratio? That is, this quantity,
$$
Q(x) = frac1f(x)fracdf(x)dx
$$
Like for instance, if $f$ is the potential energy, this would be work to potential energy ratio.
Or even what can we say about $Q(x)$, say when $Q(x)<0$.
derivatives
asked 16 hours ago
user2167741
979
979
This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
1
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago
add a comment |Â
This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
1
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago
This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
1
1
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
It is also $frac d(ln(f(x))dx$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.
add a comment |Â
up vote
1
down vote
This would give
$$
intlimits_x_0^x !Q(xi) , dxi = ln fracf(x)f(x_0) iff \
f(x) = f(x_0) exp intlimits_x_0^x ! Q(xi) , dxi
$$
An example would be a radioactive decay process ($Q$: decay constant)
Another: light absorption ($Q$: attenuation coefficient).
add a comment |Â
up vote
1
down vote
I know in Physics they use that $fracdNdt=Nk$
for various uses so you could use it to show whether a relationship is exponential or not
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
add a comment |Â
up vote
1
down vote
If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.
For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) , e^rt$.
But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It is also $frac d(ln(f(x))dx$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.
add a comment |Â
up vote
1
down vote
It is also $frac d(ln(f(x))dx$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is also $frac d(ln(f(x))dx$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.
It is also $frac d(ln(f(x))dx$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.
answered 15 hours ago
Ross Millikan
275k21183348
275k21183348
add a comment |Â
add a comment |Â
up vote
1
down vote
This would give
$$
intlimits_x_0^x !Q(xi) , dxi = ln fracf(x)f(x_0) iff \
f(x) = f(x_0) exp intlimits_x_0^x ! Q(xi) , dxi
$$
An example would be a radioactive decay process ($Q$: decay constant)
Another: light absorption ($Q$: attenuation coefficient).
add a comment |Â
up vote
1
down vote
This would give
$$
intlimits_x_0^x !Q(xi) , dxi = ln fracf(x)f(x_0) iff \
f(x) = f(x_0) exp intlimits_x_0^x ! Q(xi) , dxi
$$
An example would be a radioactive decay process ($Q$: decay constant)
Another: light absorption ($Q$: attenuation coefficient).
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This would give
$$
intlimits_x_0^x !Q(xi) , dxi = ln fracf(x)f(x_0) iff \
f(x) = f(x_0) exp intlimits_x_0^x ! Q(xi) , dxi
$$
An example would be a radioactive decay process ($Q$: decay constant)
Another: light absorption ($Q$: attenuation coefficient).
This would give
$$
intlimits_x_0^x !Q(xi) , dxi = ln fracf(x)f(x_0) iff \
f(x) = f(x_0) exp intlimits_x_0^x ! Q(xi) , dxi
$$
An example would be a radioactive decay process ($Q$: decay constant)
Another: light absorption ($Q$: attenuation coefficient).
edited 15 hours ago
answered 15 hours ago
mvw
30.1k22150
30.1k22150
add a comment |Â
add a comment |Â
up vote
1
down vote
I know in Physics they use that $fracdNdt=Nk$
for various uses so you could use it to show whether a relationship is exponential or not
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
add a comment |Â
up vote
1
down vote
I know in Physics they use that $fracdNdt=Nk$
for various uses so you could use it to show whether a relationship is exponential or not
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I know in Physics they use that $fracdNdt=Nk$
for various uses so you could use it to show whether a relationship is exponential or not
I know in Physics they use that $fracdNdt=Nk$
for various uses so you could use it to show whether a relationship is exponential or not
answered 15 hours ago
Henry Lee
48210
48210
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
add a comment |Â
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it.
â user2167741
15 hours ago
1
1
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability.
â Henry Lee
15 hours ago
add a comment |Â
up vote
1
down vote
If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.
For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) , e^rt$.
But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
add a comment |Â
up vote
1
down vote
If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.
For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) , e^rt$.
But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.
For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) , e^rt$.
But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.
If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.
For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) , e^rt$.
But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.
answered 13 hours ago
Hans Lundmark
32.7k563109
32.7k563109
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
add a comment |Â
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it?
â user2167741
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $int df/f=int r(t) , dt$, etc.
â Hans Lundmark
12 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time.
â user2167741
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-)
â Hans Lundmark
11 hours ago
add a comment |Â
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This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient.
â Nick
15 hours ago
1
I also learned that this is called 'logarithmic derivative'. en.wikipedia.org/wiki/Logarithmic_derivative
â user2167741
15 hours ago