How can you explain $cot(alpha)= dfrac{d}{dtheta}cdot ln(r) = dfrac{1}{r} dfrac{dr}{dt}$ in a polar...












0














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







share|cite|improve this question















migrated from physics.stackexchange.com Dec 26 at 12:07


This question came from our site for active researchers, academics and students of physics.











  • 1




    I'm voting to move this to Mathematics SE
    – Aaron Stevens
    Dec 26 at 1:08
















0














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







share|cite|improve this question















migrated from physics.stackexchange.com Dec 26 at 12:07


This question came from our site for active researchers, academics and students of physics.











  • 1




    I'm voting to move this to Mathematics SE
    – Aaron Stevens
    Dec 26 at 1:08














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 at 12:25









Bernard

118k639112




118k639112










asked Dec 25 at 20:24









Anonymous I

8302826




8302826




migrated from physics.stackexchange.com Dec 26 at 12:07


This question came from our site for active researchers, academics and students of physics.






migrated from physics.stackexchange.com Dec 26 at 12:07


This question came from our site for active researchers, academics and students of physics.










  • 1




    I'm voting to move this to Mathematics SE
    – Aaron Stevens
    Dec 26 at 1:08














  • 1




    I'm voting to move this to Mathematics SE
    – Aaron Stevens
    Dec 26 at 1:08








1




1




I'm voting to move this to Mathematics SE
– Aaron Stevens
Dec 26 at 1:08




I'm voting to move this to Mathematics SE
– Aaron Stevens
Dec 26 at 1:08










2 Answers
2






active

oldest

votes


















2














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



$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$






share|cite|improve this answer





















  • Thx. That refers to a textbook I've found.
    – Anonymous I
    Dec 25 at 23:05



















1














I like this solution 



 How can you derivate it from this picture


enter image description here






share|cite|improve this answer





















  • 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.
    – FGSUZ
    Dec 26 at 10:11











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052883%2fhow-can-you-explain-cot-alpha-dfracdd-theta-cdot-lnr-dfrac1r%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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



$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$






share|cite|improve this answer





















  • Thx. That refers to a textbook I've found.
    – Anonymous I
    Dec 25 at 23:05
















2














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



$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$






share|cite|improve this answer





















  • Thx. That refers to a textbook I've found.
    – Anonymous I
    Dec 25 at 23:05














2












2








2






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



$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$






share|cite|improve this answer












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



$$cot{alpha}=frac{dot{r}}{rdot{theta}}=frac{1}{r}frac{dr}{dtheta}=frac{d}{dtheta}ln{r}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 at 22:14









G. Smith

2213




2213












  • Thx. That refers to a textbook I've found.
    – Anonymous I
    Dec 25 at 23:05


















  • Thx. That refers to a textbook I've found.
    – Anonymous I
    Dec 25 at 23:05
















Thx. That refers to a textbook I've found.
– Anonymous I
Dec 25 at 23:05




Thx. That refers to a textbook I've found.
– Anonymous I
Dec 25 at 23:05











1














I like this solution 



 How can you derivate it from this picture


enter image description here






share|cite|improve this answer





















  • 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.
    – FGSUZ
    Dec 26 at 10:11
















1














I like this solution 



 How can you derivate it from this picture


enter image description here






share|cite|improve this answer





















  • 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.
    – FGSUZ
    Dec 26 at 10:11














1












1








1






I like this solution 



 How can you derivate it from this picture


enter image description here






share|cite|improve this answer












I like this solution 



 How can you derivate it from this picture


enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 at 8:35









Eli

1212




1212












  • 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.
    – FGSUZ
    Dec 26 at 10:11


















  • 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.
    – FGSUZ
    Dec 26 at 10:11
















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.
– FGSUZ
Dec 26 at 10:11




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.
– FGSUZ
Dec 26 at 10:11


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052883%2fhow-can-you-explain-cot-alpha-dfracdd-theta-cdot-lnr-dfrac1r%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Questions related to Moebius Transform of Characteristic Function of the Primes

Image
2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
Read more

List of scandals in India

Image
List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more