How can you explain $cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}$ in a polar...
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This alinea is about the $$cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}.$$ Where does the $ln (r)$ come from? How can you derive it from that picture? I want to use that for the $$nabla f(r,theta)= dfrac {partial f}{partial r}vec e_r + dfrac {1}{r}dfrac{partial f}{partial theta}vec e_theta.$$ I know what the definition is of the gradient (i.e. it's a partial derivative of the function with respect to every component in a particular dimension for my case $x,y,z$). The only thing is that I don't know how to convert that to polar coordinates. Can someone please clarify this?
EDIT: I know that the derivative of $ln(x)= 1/x$. So bringing in $ln r$ would be convenient.
coordinate-systems derivatives
edited Dec 26 '18 at 12:25
Bernard
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
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migrated from physics.stackexchange.com Dec 26 '18 at 12:07
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
This alinea is about the $$cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}.$$ Where does the $ln (r)$ come from? How can you derive it from that picture? I want to use that for the $$nabla f(r,theta)= dfrac {partial f}{partial r}vec e_r + dfrac {1}{r}dfrac{partial f}{partial theta}vec e_theta.$$ I know what the definition is of the gradient (i.e. it's a partial derivative of the function with respect to every component in a particular dimension for my case $x,y,z$). The only thing is that I don't know how to convert that to polar coordinates. Can someone please clarify this?
EDIT: I know that the derivative of $ln(x)= 1/x$. So bringing in $ln r$ would be convenient.
coordinate-systems derivatives
edited Dec 26 '18 at 12:25
Bernard
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
$endgroup$
migrated from physics.stackexchange.com Dec 26 '18 at 12:07
This question came from our site for active researchers, academics and students of physics.
1
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I'm voting to move this to Mathematics SE
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– Aaron Stevens
Dec 26 '18 at 1:08
add a comment |
$begingroup$
This alinea is about the $$cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}.$$ Where does the $ln (r)$ come from? How can you derive it from that picture? I want to use that for the $$nabla f(r,theta)= dfrac {partial f}{partial r}vec e_r + dfrac {1}{r}dfrac{partial f}{partial theta}vec e_theta.$$ I know what the definition is of the gradient (i.e. it's a partial derivative of the function with respect to every component in a particular dimension for my case $x,y,z$). The only thing is that I don't know how to convert that to polar coordinates. Can someone please clarify this?
EDIT: I know that the derivative of $ln(x)= 1/x$. So bringing in $ln r$ would be convenient.
coordinate-systems derivatives
edited Dec 26 '18 at 12:25
Bernard
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
$endgroup$
This alinea is about the $$cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}.$$ Where does the $ln (r)$ come from? How can you derive it from that picture? I want to use that for the $$nabla f(r,theta)= dfrac {partial f}{partial r}vec e_r + dfrac {1}{r}dfrac{partial f}{partial theta}vec e_theta.$$ I know what the definition is of the gradient (i.e. it's a partial derivative of the function with respect to every component in a particular dimension for my case $x,y,z$). The only thing is that I don't know how to convert that to polar coordinates. Can someone please clarify this?
EDIT: I know that the derivative of $ln(x)= 1/x$. So bringing in $ln r$ would be convenient.
coordinate-systems derivatives
coordinate-systems derivatives
edited Dec 26 '18 at 12:25
Bernard
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
edited Dec 26 '18 at 12:25
Bernard
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
edited Dec 26 '18 at 12:25
Bernard
121k740116
edited Dec 26 '18 at 12:25
Bernard
121k740116
edited Dec 26 '18 at 12:25
Bernard
121k740116
121k740116
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
asked Dec 25 '18 at 20:24
Anonymous1967Anonymous1967
8302826
8302826
migrated from physics.stackexchange.com Dec 26 '18 at 12:07
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Dec 26 '18 at 12:07
This question came from our site for active researchers, academics and students of physics.
1
$begingroup$
I'm voting to move this to Mathematics SE
$endgroup$
– Aaron Stevens
Dec 26 '18 at 1:08
add a comment |
1
$begingroup$
I'm voting to move this to Mathematics SE
$endgroup$
– Aaron Stevens
Dec 26 '18 at 1:08
1
1
$begingroup$
I'm voting to move this to Mathematics SE
$endgroup$
– Aaron Stevens
Dec 26 '18 at 1:08
$begingroup$
I'm voting to move this to Mathematics SE
$endgroup$
– Aaron Stevens
Dec 26 '18 at 1:08
add a comment |
2 Answers
2
active
oldest
votes
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The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $mathbf{r}$ is its position vector from $O$ to $P$, then $alpha$ is the angle between its velocity vector $dot{mathbf{r}}$ (along the tangent line $t$) and the position vector $mathbf{r}$. So
$$cos{alpha}=frac{dot{mathbf{r}}}{|dot{mathbf{r}}|}cdotfrac{mathbf{r}}{|mathbf{r}|}$$
In polar coordinates, the position vector is
$$mathbf{r}=rhat{mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$dot{mathbf{r}}=dot{r}hat{mathbf{r}}+rdot{hat{mathbf{r}}}=dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}.$$
Thus we have
$$cos{alpha}=frac{dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}}{|dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}|}cdothat{mathbf{r}}=frac{dot{r}}{sqrt{dot{r}^2+r^2dot{theta}^2}}$$
which implies
$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
$endgroup$
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
add a comment |
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I like this solution
How can you derivate it from this picture
answered Dec 26 '18 at 8:35
EliEli
1234
$endgroup$
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $mathbf{r}$ is its position vector from $O$ to $P$, then $alpha$ is the angle between its velocity vector $dot{mathbf{r}}$ (along the tangent line $t$) and the position vector $mathbf{r}$. So
$$cos{alpha}=frac{dot{mathbf{r}}}{|dot{mathbf{r}}|}cdotfrac{mathbf{r}}{|mathbf{r}|}$$
In polar coordinates, the position vector is
$$mathbf{r}=rhat{mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$dot{mathbf{r}}=dot{r}hat{mathbf{r}}+rdot{hat{mathbf{r}}}=dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}.$$
Thus we have
$$cos{alpha}=frac{dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}}{|dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}|}cdothat{mathbf{r}}=frac{dot{r}}{sqrt{dot{r}^2+r^2dot{theta}^2}}$$
which implies
$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
$endgroup$
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
add a comment |
$begingroup$
The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $mathbf{r}$ is its position vector from $O$ to $P$, then $alpha$ is the angle between its velocity vector $dot{mathbf{r}}$ (along the tangent line $t$) and the position vector $mathbf{r}$. So
$$cos{alpha}=frac{dot{mathbf{r}}}{|dot{mathbf{r}}|}cdotfrac{mathbf{r}}{|mathbf{r}|}$$
In polar coordinates, the position vector is
$$mathbf{r}=rhat{mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$dot{mathbf{r}}=dot{r}hat{mathbf{r}}+rdot{hat{mathbf{r}}}=dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}.$$
Thus we have
$$cos{alpha}=frac{dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}}{|dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}|}cdothat{mathbf{r}}=frac{dot{r}}{sqrt{dot{r}^2+r^2dot{theta}^2}}$$
which implies
$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
$endgroup$
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
add a comment |
$begingroup$
The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $mathbf{r}$ is its position vector from $O$ to $P$, then $alpha$ is the angle between its velocity vector $dot{mathbf{r}}$ (along the tangent line $t$) and the position vector $mathbf{r}$. So
$$cos{alpha}=frac{dot{mathbf{r}}}{|dot{mathbf{r}}|}cdotfrac{mathbf{r}}{|mathbf{r}|}$$
In polar coordinates, the position vector is
$$mathbf{r}=rhat{mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$dot{mathbf{r}}=dot{r}hat{mathbf{r}}+rdot{hat{mathbf{r}}}=dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}.$$
Thus we have
$$cos{alpha}=frac{dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}}{|dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}|}cdothat{mathbf{r}}=frac{dot{r}}{sqrt{dot{r}^2+r^2dot{theta}^2}}$$
which implies
$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
$endgroup$
The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $mathbf{r}$ is its position vector from $O$ to $P$, then $alpha$ is the angle between its velocity vector $dot{mathbf{r}}$ (along the tangent line $t$) and the position vector $mathbf{r}$. So
$$cos{alpha}=frac{dot{mathbf{r}}}{|dot{mathbf{r}}|}cdotfrac{mathbf{r}}{|mathbf{r}|}$$
In polar coordinates, the position vector is
$$mathbf{r}=rhat{mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$dot{mathbf{r}}=dot{r}hat{mathbf{r}}+rdot{hat{mathbf{r}}}=dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}.$$
Thus we have
$$cos{alpha}=frac{dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}}{|dot{r}hat{mathbf{r}}+rdot{theta}hat{mathbf{theta}}|}cdothat{mathbf{r}}=frac{dot{r}}{sqrt{dot{r}^2+r^2dot{theta}^2}}$$
which implies
$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
answered Dec 25 '18 at 22:14
G. SmithG. Smith
2664
2664
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
add a comment |
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
$begingroup$
Thx. That refers to a textbook I've found.
$endgroup$
– Anonymous1967
Dec 25 '18 at 23:05
add a comment |
$begingroup$
I like this solution
How can you derivate it from this picture
answered Dec 26 '18 at 8:35
EliEli
1234
$endgroup$
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
add a comment |
$begingroup$
I like this solution
How can you derivate it from this picture
answered Dec 26 '18 at 8:35
EliEli
1234
$endgroup$
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
add a comment |
$begingroup$
I like this solution
How can you derivate it from this picture
answered Dec 26 '18 at 8:35
EliEli
1234
$endgroup$
I like this solution
How can you derivate it from this picture
answered Dec 26 '18 at 8:35
EliEli
1234
answered Dec 26 '18 at 8:35
EliEli
1234
answered Dec 26 '18 at 8:35
EliEli
1234
answered Dec 26 '18 at 8:35
EliEli
1234
1234
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
add a comment |
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
$begingroup$
Nice answer, it's simple and useful. But please, you should not post formulae in pictures. Type them instead using Mathjax. There are many reasons, including optimisation of algorithms and helping users whose device doesn't display them well.
$endgroup$
– FGSUZ
Dec 26 '18 at 10:11
add a comment |
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1
$begingroup$
I'm voting to move this to Mathematics SE
$endgroup$
– Aaron Stevens
Dec 26 '18 at 1:08