Results 1 to 5 of 5

Math Help - Line element definition by differentiating plane polar coordinates

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    4

    Line element definition by differentiating plane polar coordinates

    Hi folks, hopefully this is in the right forum. My textbook tells me I can use the product rule to go from the functions x = r cos phi and y = r sin phi to find expressions for dx and dy. I'm not clear whether this is partial differentiation (I thought I understood that!), or what I am differentiating with respect to! I have attached a jpg of the textbook section which should be clearer that typing it all out. I think I can get to the final expression, it's finding the middle two I don't understand. Can anyone help?



    My jpg refers to equation 3.4, this is simply dl^2 = dx^2 + dy^2.
    Attached Thumbnails Attached Thumbnails Line element definition by differentiating plane polar coordinates-plane-polar.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: Line element definition by differentiating plane polar coordinates

    taking the derivattive w/r to \phi ...

    \frac{d}{d\phi}(x = r\cos{\phi})

    \frac{dx}{d\phi} = -r\sin{\phi} + \cos{\phi} \cdot \frac{dr}{d\phi}

    dx = \cos{\phi} \cdot dr - r\sin{\phi} \cdot d\phi

    same idea for dy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,971
    Thanks
    1121

    Re: Line element definition by differentiating plane polar coordinates

    It's the "total differential". If f(x,y) is a function of two variables, and each of those variables is a function of the single variable t, x(t), y(t), (think of an object moving along some trajectory in the plane with t as time), then, by the chain rule
    \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}

    Since that is now a function of a single variable, we can write its "differential", df= \frac{df}{dt} dt:
    df= \frac{\partial f}{\partial x}\frac{dx}{dt} dt+ \frac{\partial f}{\partial y}\frac{dy}{dt} dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy
    (Which no longer has any dependence on t!)

    In particular, if x= r cos(\phi) and y= r sin(\phi), then dx= \frac{\partial x}{\partial r}dr+ \frac{\partial x}{\partial \phi}d\phi= cos(\phi)dr- r sin(\phi)\d\phi. And if y= r sin(\phi), then dy= \frac{\partial y}{\partial x}dr+ \frac{\partial y}{\partial \phi}d\phi= sin(\phi)dr+ r cos(\phi)d\phi.

    Squaring those,
    dx^2= cos^2(\phi)dr^2- 2r^2sin(\phi)cos(\phi)drd\phi+ r^2sin^2(\phi)d\phi^2 and
    dy^2= sin^2(\phi)dr^2+ 2r^2sin(\phi)cos(\phi)drd\phi+ r^2 cos^2(\phi)d\phi^2

    Adding, the " drd\phi" terms cancel while sin^2(\phi)+ cos^2(\phi)= 1 so that
    ds^2= dx^2+ dy^2= dr^2+ r^2 d\phi^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2011
    Posts
    4

    Re: Line element definition by differentiating plane polar coordinates

    Thanks very much for taking the time to answer chaps, having quoted one of your responses I can see the effort that goes into a properly formatted answer!

    Skeeter: Your response was the same as I had seen in another textbook; whilst undoubtedly correct, unfortunately I am not able to make sense of it. I don;'t quite see why dx/dphi is not as simple as -r sin phi. Entirely my failing, not yours!

    HallsofIvy: I have been able to follow quite a lot of your answer, but I still have a couple of questions about it. If you have time to explain further it would be much appreciated, if not it has at least got me to a working solution!

    Quote Originally Posted by HallsofIvy View Post
    It's the "total differential". If f(x,y) is a function of two variables,
    In my case I have x(r,phi) as one function of two variables, and y(r,phi) as a second function of two variables...

    and each of those variables is a function of the single variable t, x(t), y(t), (think of an object moving along some trajectory in the plane with t as time),
    I can't quite see how this works in my case. Aren't r and phi two independant variables? What is the analogy to your t variable in my example?


    then, by the chain rule
    \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}
    I get this part! This is partial differentiation as I understand it

    Since that is now a function of a single variable, we can write its "differential", df= \frac{df}{dt} dt:
    df= \frac{\partial f}{\partial x}\frac{dx}{dt} dt+ \frac{\partial f}{\partial y}\frac{dy}{dt} dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy
    (Which no longer has any dependence on t!)
    Is this 'thing' df the "total differential"? And does the operation where you "multiply though" by dt have a name (I realise it is not as simple as multiplying through!)

    In particular, if x= r cos(\phi) and y= r sin(\phi), then dx= \frac{\partial x}{\partial r}dr+ \frac{\partial x}{\partial \phi}d\phi= cos(\phi)dr- r sin(\phi)\d\phi. And if y= r sin(\phi), then dy= \frac{\partial y}{\partial x}dr+ \frac{\partial y}{\partial \phi}d\phi= sin(\phi)dr+ r cos(\phi)d\phi.

    Squaring those,
    dx^2= cos^2(\phi)dr^2- 2r^2sin(\phi)cos(\phi)drd\phi+ r^2sin^2(\phi)d\phi^2 and
    dy^2= sin^2(\phi)dr^2+ 2r^2sin(\phi)cos(\phi)drd\phi+ r^2 cos^2(\phi)d\phi^2

    Adding, the " drd\phi" terms cancel while sin^2(\phi)+ cos^2(\phi)= 1 so that
    ds^2= dx^2+ dy^2= dr^2+ r^2 d\phi^2
    That last bit all makes sense to me, I have never seen this type of differential manipulation before though which is why I am struggling!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: Line element definition by differentiating plane polar coordinates

    Skeeter: Your response was the same as I had seen in another textbook; whilst undoubtedly correct, unfortunately I am not able to make sense of it. I don;'t quite see why dx/dphi is not as simple as -r sin phi.
    if r were a constant, then \frac{d}{d\phi}(r \cos{\phi}) = -r\sin{\phi} ... however, r is not a constant, it is an implicit function of \phi .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 24th 2010, 04:33 AM
  2. Plane region in polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 27th 2010, 02:35 PM
  3. Replies: 3
    Last Post: February 16th 2010, 05:38 AM
  4. area element in polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 24th 2009, 05:53 AM
  5. Replies: 6
    Last Post: February 4th 2009, 12:12 AM

Search Tags


/mathhelpforum @mathhelpforum