{"id":210,"date":"2018-10-16T01:27:37","date_gmt":"2018-10-16T01:27:37","guid":{"rendered":"http:\/\/kb.shoelace.biz\/?p=210"},"modified":"2025-11-08T15:37:12","modified_gmt":"2025-11-08T15:37:12","slug":"vector-analysis-parabolic-coordinates","status":"publish","type":"post","link":"https:\/\/kb.shoelace.biz\/?p=210","title":{"rendered":"Vector Analysis – Parabolic Coordinates"},"content":{"rendered":"

There are many orthogonal\u00a0curvilinear coordinate systems<\/a>. One of those is parabolic coordinates. In this post I will look at how parabolic coordinates are related to Cartesian coordinates in 3 dimensional space. I will determine unit vectors in parabolic coordinates in terms of Cartesian coordinate unit vectors. I will determine the arc length of a moving particle following a trajectory as a function of time. Why? That’s one thing I have not determined. Here we go.<\/p>\n

Cartesian & Parabolic Coordinate Relationship<\/h4>\n

Three dimensional parabolic coordinates are made up of (u,v,\u03a6) components. Cartesian components are (x,y,z). They are related by the following equations:<\/p>\n

$$ x = uvcos(\\phi),\u00a0 y = uvsin(\\phi),\u00a0 z=\\frac{1}{2}(u^{2}-v^{2}) $$<\/p>\n

how did ‘they’ come up with these? Well, in two dimensions (u and v) coordinates are confocal parabolas. In three, u and v are confocal paraboloids rotated about an axis of symmetry (z axis) creating an azimuthal angle\u00a0\u03a6 where:<\/p>\n

$$ tan(\\phi)= \\frac{y}{x} $$<\/p>\n

In other words, try reading this<\/a> and see if you can make sense of it on your own. Often, even when surrounded by people, I feel alone.<\/p>\n

Determining Parabolic Unit Vectors in Terms of Cartesian Coordinates<\/h4>\n

To determine the unit vectors in parabolic coordinates I will need to:<\/p>\n

    \n
  1. Create a position vector from the origin in Cartesian coordinates using the (x,y,z) parabolic equalities above.<\/li>\n
  2. \u00a0Differentiate that position vector with respect to each parabolic coordinate (u,v,\u03a6). these functions will serve as the parabolic basis vectors.<\/li>\n
  3. Normalize the basis vectors to get unit vectors with magnitude equal to 1. This step is equivalent to finding the normalizing scale factors.<\/li>\n
  4. Try to mask my emotional void with an unconvincing smile.<\/li>\n<\/ol>\n

    Position in Cartesian coordinates:<\/p>\n

    $$\\bar{r} = x\\bar{i}+y\\bar{j}+z\\bar{k} $$<\/p>\n

    Using the equations above to get the parabolic equivalent:<\/p>\n

    $$ \\bar{r} = uvcos(\\phi)\\bar{i}+uvsin(\\phi)\\bar{j}+\\frac{1}{2}(u^{2}-v^{2})\\bar{k} $$<\/p>\n

    Differentiating the vector with respect to u,v, and \u03a6:<\/p>\n

    $$ \\frac{\\partial \\bar{r}}{\\partial u} = vcos(\\phi)\\bar{i}+vsin(\\phi)\\bar{j}+u\\bar{k} $$<\/p>\n

    $$ \\frac{\\partial \\bar{r}}{\\partial v} = ucos(\\phi)\\bar{i}+usin(\\phi)\\bar{j}-v\\bar{k} $$<\/p>\n

    $$ \\frac{\\partial \\bar{r}}{\\partial \\phi} = -uvsin(\\phi)\\bar{i}+uvcos(\\phi)\\bar{j} $$<\/p>\n

    The above partial derivatives can be normalized by dividing by their respective magnitudes. Calculating the magnitudes:<\/p>\n

    $$\\left \\| \\frac{\\partial \\bar{r}}{\\partial v} \\right \\| = \\sqrt{u^2+v^2}$$<\/p>\n

    $$\\left \\| \\frac{\\partial \\bar{r}}{\\partial u} \\right \\| = \\sqrt{u^2+v^2}$$<\/p>\n

    $$\\left \\| \\frac{\\partial \\bar{r}}{\\partial \\phi} \\right \\| = uv$$<\/p>\n

    So with these magnitudes, I can divide each partial by its magnitude to get the unit vectors:<\/p>\n

    $$ \\hat{e}_{u} = \\frac{\\frac{\\partial \\bar{r}}{\\partial u}}{\\left \\| \\frac{\\partial \\bar{r}}{\\partial u} \\right \\|} = \\frac{vcos(\\phi)\\bar{i}+vsin(\\phi)\\bar{j}+u\\bar{k}}{\\sqrt{u^2+v^2}}$$<\/p>\n

    $$\\hat{e}_{v} = \\frac{\\frac{\\partial \\bar{r}}{\\partial v}}{\\left \\| \\frac{\\partial \\bar{r}}{\\partial v} \\right \\|} = \\frac{ucos(\\phi)\\bar{i}+usin(\\phi)\\bar{j}-v\\bar{k}}{\\sqrt{u^2+v^2}}$$<\/p>\n

    $$\\hat{e}_{\\phi} = \\frac{\\frac{\\partial \\bar{r}}{\\partial \\phi}}{\\left \\| \\frac{\\partial \\bar{r}}{\\partial \\phi} \\right \\|} = \\frac{-uvsin(\\phi)\\bar{i}+uvcos(\\phi)\\bar{j}}{uv}$$<\/p>\n

    Now we can use the above unit vectors to calculate arc lengths of functions defined in parabolic coordinates.<\/p>\n

    Determining Arc Length of a Particle Moving as a Function of Time<\/h4>\n

    The function defined\u00a0parametrically in parabolic coordinates:<\/p>\n

    $$\u00a0u(t) = t, v(t) = t, \\phi = 2t $$<\/p>\n

    We will need to define the arc length\u00a0s\u00a0<\/em>in parabolic coordinates. I understand that arc length can be found by integrating along a differential displacement over time. To do this, the magnitude of the arc length can be equated to the square root of the dot product of the velocity vector.<\/p>\n

    $$s = \\int_{a}^{b}ds = \\int_{a}^{b}\\frac{ds}{dt}dt = \\int_{a}^{b}\\sqrt{d\\bar{r}\\cdot d\\bar{r}}$$<\/p>\n

    The next step is then to find what that integral evaluates to. So what is the differential position vector dr? I can use the partial with respect to each parabolic coordinate multiplied by the scale factor to find it.<\/p>\n

    $$ d\\bar{r} = h_{u}\\frac{\\partial u}{\\partial t}\\hat{e}_{u} + h_{v}\\frac{\\partial v}{\\partial t}\\hat{e}_{v}+h_{\\phi}\\frac{\\partial \\phi}{\\partial t}\\hat{e}_{\\phi} $$<\/p>\n

    $$ ds = \\sqrt{h_{u}^2(\\frac{\\partial u}{\\partial t})^2\\hat{e}_{u} + h_{v}^2(\\frac{\\partial v}{\\partial t})^2\\hat{e}_{v}+h_{\\phi}^2(\\frac{\\partial \\phi}{\\partial t})^2\\hat{e}_{\\phi}} $$<\/p>\n

    My head is going to explode here if we don’t make some progress. A couple steps into this calculation you get something that looks like this:<\/p>\n

    $$s = \\int_{a}^{b}\\sqrt{1^2(v^2+u^2)+1^2(u^2+v^2)+2(u^2v^2)} $$<\/p>\n

    We can replace the v’s and u’s with t using the parametric equations above. Lets integrate with respect to t from 0 to 1 so I can actually get an arc length value out. Moving along another couple steps:<\/p>\n

    $$s = \\int_{0}^{1}\\sqrt{4t^2+4t^4}dt$$<\/p>\n

    Not a particularly pleasant integral, but also not unworkable. That is the same as:<\/p>\n

    $$\u00a0s = \\int_{0}^{1}2t\\sqrt{1+t^2}dt $$<\/p>\n

    This will probably be the thousandth time I have had to do a u substitution.<\/p>\n

    An aside: I have never actually done a u substitution in my career by hand. I have never earned a penny doing u substitutions. I have never connected with another individual regarding u substitutions. U substitutions have never come in handy in my day to day life.\u00a0 Yet here we are at 6:15PM on a Monday evening in October, writing up blog posts requiring u substitutions.<\/p>\n

    I think in this case we will use u equal to 1 plus t squared.<\/p>\n

    $$u = 1 + t^2, du = 2tdt$$<\/p>\n

    $$s = \\int \\sqrt{u}du = \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} = \\frac{2}{3}(1+t^2)^{\\frac{3}{2}})|_{0}^{1} = \\frac{2}{3}(2^\\frac{3}{2}-1)$$<\/p>\n

    There we have it. That answer represents the length along the arc of the parametrically defined function in parabolic coordinates from time t equal to 0 to time t equal to 1.<\/p>\n

     <\/p>\n

     <\/p>\n","protected":false},"excerpt":{"rendered":"

    There are many orthogonal\u00a0curvilinear coordinate systems. One of those is parabolic coordinates. In this post I will look at how parabolic coordinates are related to Cartesian coordinates in 3 dimensional space. I will determine unit vectors in parabolic coordinates in terms of Cartesian coordinate unit vectors. I will determine the arc length of a moving…<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-210","post","type-post","status-publish","format-standard","hentry","category-applied-mathematics"],"_links":{"self":[{"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/posts\/210","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=210"}],"version-history":[{"count":1,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/posts\/210\/revisions"}],"predecessor-version":[{"id":970,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=\/wp\/v2\/posts\/210\/revisions\/970"}],"wp:attachment":[{"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=210"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=210"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kb.shoelace.biz\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=210"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}