{"id":15743,"date":"2025-06-10T13:12:05","date_gmt":"2025-06-10T13:12:05","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=15743"},"modified":"2025-06-10T13:12:09","modified_gmt":"2025-06-10T13:12:09","slug":"the-standard-helix-has-parameterization","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-standard-helix-has-parameterization\/","title":{"rendered":"The standard helix has parameterization"},"content":{"rendered":"\n

The standard helix has parameterization . Find the curvature of the helix. Does the result agree with your intuition?<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

Okay, let’s find the curvature of the standard helix.<\/p>\n\n\n\n

The standard helix has the parameterization:<\/p>\n\n\n\n

        r(t)=(acos\u2061t,asin\u2061t,bt)\\mathbf{r}(t) = (a \\cos t, a \\sin t, bt)r(t)=(acost,asint,bt)<\/code>\n      <\/pre>\n\n\n\n
To find the curvature<\/p>\n\n\n\n
        \u03ba\\kappa\u03ba<\/code>\n      <\/pre>\n\n\n\n
, we can use the formula:<\/p>\n\n\n\n
        \u03ba=\u2223\u2223r\u2032(t)\u00d7r\u2032\u2032(t)\u2223\u2223\u2223\u2223r\u2032(t)\u2223\u22233\\kappa = \\frac{||\\mathbf{r}'(t) \\times \\mathbf{r}''(t)||}{||\\mathbf{r}'(t)||^3}\u03ba=\u2223\u2223r\u2032(t)\u2223\u22233\u2223\u2223r\u2032(t)\u00d7r\u2032\u2032(t)\u2223\u2223\u200b<\/code>\n      <\/pre>\n\n\n\n
First, we find the first and second derivatives:<\/p>\n\n\n\n
        r\u2032(t)=ddt(acos\u2061t,asin\u2061t,bt)=(\u2212asin\u2061t,acos\u2061t,b)\\mathbf{r}'(t) = \\frac{d}{dt}(a \\cos t, a \\sin t, bt) = (-a \\sin t, a \\cos t, b)r\u2032(t)=dtd\u200b(acost,asint,bt)=(\u2212asint,acost,b)<\/code>\n      <\/pre>\n\n\n\n
        r\u2032\u2032(t)=ddt(\u2212asin\u2061t,acos\u2061t,b)=(\u2212acos\u2061t,\u2212asin\u2061t,0)\\mathbf{r}''(t) = \\frac{d}{dt}(-a \\sin t, a \\cos t, b) = (-a \\cos t, -a \\sin t, 0)r\u2032\u2032(t)=dtd\u200b(\u2212asint,acost,b)=(\u2212acost,\u2212asint,0)<\/code>\n      <\/pre>\n\n\n\n
Next, we calculate the magnitude of the first derivative (the speed):<\/p>\n\n\n\n
        \u2223\u2223r\u2032(t)\u2223\u2223=(\u2212asin\u2061t)2+(acos\u2061t)2+b2=a2sin\u20612t+a2cos\u20612t+b2||\\mathbf{r}'(t)|| = \\sqrt{(-a \\sin t)^2 + (a \\cos t)^2 + b^2} = \\sqrt{a^2 \\sin^2 t + a^2 \\cos^2 t + b^2}\u2223\u2223r\u2032(t)\u2223\u2223=(\u2212asint)2+(acost)2+b2\u200b=a2sin2t+a2cos2t+b2\u200b<\/code>\n      <\/pre>\n\n\n\n
        \u2223\u2223r\u2032(t)\u2223\u2223=a2(sin\u20612t+cos\u20612t)+b2=a2(1)+b2=a2+b2||\\mathbf{r}'(t)|| = \\sqrt{a^2(\\sin^2 t + \\cos^2 t) + b^2} = \\sqrt{a^2(1) + b^2} = \\sqrt{a^2 + b^2}\u2223\u2223r\u2032(t)\u2223\u2223=a2(sin2t+cos2t)+b2\u200b=a2(1)+b2\u200b=a2+b2\u200b<\/code>\n      <\/pre>\n\n\n\n
Now, we calculate the cross product of the first and second derivatives:<\/p>\n\n\n\n
        r\u2032(t)\u00d7r\u2032\u2032(t)=\u2223ijk\u2212asin\u2061tacos\u2061tb\u2212acos\u2061t\u2212asin\u2061t0\u2223\\mathbf{r}'(t) \\times \\mathbf{r}''(t) = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ -a \\sin t & a \\cos t & b \\\\ -a \\cos t & -a \\sin t & 0 \\end{vmatrix}r\u2032(t)\u00d7r\u2032\u2032(t)=\u200bi\u2212asint\u2212acost\u200bjacost\u2212asint\u200bkb0\u200b\u200b<\/code>\n      <\/pre>\n\n\n\n
        =i((acos\u2061t)(0)\u2212b(\u2212asin\u2061t))\u2212j((\u2212asin\u2061t)(0)\u2212b(\u2212acos\u2061t))+k((\u2212asin\u2061t)(\u2212asin\u2061t)\u2212(acos\u2061t)(\u2212acos\u2061t))= \\mathbf{i}((a \\cos t)(0) - b(-a \\sin t)) - \\mathbf{j}((-a \\sin t)(0) - b(-a \\cos t)) + \\mathbf{k}((-a \\sin t)(-a \\sin t) - (a \\cos t)(-a \\cos t))=i((acost)(0)\u2212b(\u2212asint))\u2212j((\u2212asint)(0)\u2212b(\u2212acost))+k((\u2212asint)(\u2212asint)\u2212(acost)(\u2212acost))<\/code>\n      <\/pre>\n\n\n\n
        =i(absin\u2061t)\u2212j(abcos\u2061t)+k(a2sin\u20612t+a2cos\u20612t)= \\mathbf{i}(ab \\sin t) - \\mathbf{j}(ab \\cos t) + \\mathbf{k}(a^2 \\sin^2 t + a^2 \\cos^2 t)=i(absint)\u2212j(abcost)+k(a2sin2t+a2cos2t)<\/code>\n      <\/pre>\n\n\n\n
        =(absin\u2061t,\u2212abcos\u2061t,a2(sin\u20612t+cos\u20612t))=(absin\u2061t,\u2212abcos\u2061t,a2)= (ab \\sin t, -ab \\cos t, a^2(\\sin^2 t + \\cos^2 t)) = (ab \\sin t, -ab \\cos t, a^2)=(absint,\u2212abcost,a2(sin2t+cos2t))=(absint,\u2212abcost,a2)<\/code>\n      <\/pre>\n\n\n\n
Finally, we calculate the magnitude of the cross product:<\/p>\n\n\n\n
        \u2223\u2223r\u2032(t)\u00d7r\u2032\u2032(t)\u2223\u2223=(absin\u2061t)2+(\u2212abcos\u2061t)2+(a2)2||\\mathbf{r}'(t) \\times \\mathbf{r}''(t)|| = \\sqrt{(ab \\sin t)^2 + (-ab \\cos t)^2 + (a^2)^2}\u2223\u2223r\u2032(t)\u00d7r\u2032\u2032(t)\u2223\u2223=(absint)2+(\u2212abcost)2+(a2)2\u200b<\/code>\n      <\/pre>\n\n\n\n
        =a2b2sin\u20612t+a2b2cos\u20612t+a4=a2b2(sin\u20612t+cos\u20612t)+a4= \\sqrt{a^2 b^2 \\sin^2 t + a^2 b^2 \\cos^2 t + a^4} = \\sqrt{a^2 b^2(\\sin^2 t + \\cos^2 t) + a^4}=a2b2sin2t+a2b2cos2t+a4\u200b=a2b2(sin2t+cos2t)+a4\u200b<\/code>\n      <\/pre>\n\n\n\n
        =a2b2(1)+a4=a2b2+a4=a2(b2+a2)= \\sqrt{a^2 b^2(1) + a^4} = \\sqrt{a^2 b^2 + a^4} = \\sqrt{a^2(b^2 + a^2)}=a2b2(1)+a4\u200b=a2b2+a4\u200b=a2(b2+a2)\u200b<\/code>\n      <\/pre>\n\n\n\n
Assuming<\/p>\n\n\n\n
        a>0a > 0a>0<\/code>\n      <\/pre>\n\n\n\n
, this is<\/p>\n\n\n\n
        aa2+b2a \\sqrt{a^2 + b^2}aa2+b2\u200b<\/code>\n      <\/pre>\n\n\n\n
.<\/p>\n\n\n\n
Now, substitute these magnitudes into the curvature formula:<\/p>\n\n\n\n
        \u03ba=aa2+b2(a2+b2)3=a(a2+b2)1\/2(a2+b2)3\/2=a(a2+b2)1\/2\u22123\/2=a(a2+b2)\u22121\\kappa = \\frac{a \\sqrt{a^2 + b^2}}{(\\sqrt{a^2 + b^2})^3} = \\frac{a (a^2 + b^2)^{1\/2}}{(a^2 + b^2)^{3\/2}} = a (a^2 + b^2)^{1\/2 - 3\/2} = a (a^2 + b^2)^{-1}\u03ba=(a2+b2\u200b)3aa2+b2\u200b\u200b=(a2+b2)3\/2a(a2+b2)1\/2\u200b=a(a2+b2)1\/2\u22123\/2=a(a2+b2)\u22121<\/code>\n      <\/pre>\n\n\n\n
        \u03ba=aa2+b2\\kappa = \\frac{a}{a^2 + b^2}\u03ba=a2+b2a\u200b<\/code>\n      <\/pre>\n\n\n\n
The curvature of the standard helix<\/strong><\/p>\n\n\n\n
        r(t)=(acos\u2061t,asin\u2061t,bt)\\mathbf{r}(t) = (a \\cos t, a \\sin t, bt)r(t)=(acost,asint,bt)<\/code>\n      <\/strong><\/pre>\n\n\n\n
is \u03ba=aa2+b2\\kappa = \\frac{a}{a^2 + b^2}\u03ba=a2+b2a\u200b<\/code> .<\/strong><\/p>\n\n\n\n
Explanation and Intuition:<\/strong><\/p>\n\n\n\n
The result<\/p>\n\n\n\n
        \u03ba=aa2+b2\\kappa = \\frac{a}{a^2 + b^2}\u03ba=a2+b2a\u200b<\/code>\n      <\/pre>\n\n\n\n
indicates that the curvature of the standard helix is constant, meaning it bends the same amount at every point. This makes sense intuitively because the helix is symmetric and uniform; its shape doesn’t change as you move along it.<\/p>\n\n\n\n
Let’s consider how the parameters<\/p>\n\n\n\n
        aaa<\/code>\n      <\/pre>\n\n\n\n
and<\/p>\n\n\n\n
        bbb<\/code>\n      <\/pre>\n\n\n\n
affect the curvature:<\/p>\n\n\n\n
    \n
  • Parameter ‘a’:<\/strong> This represents the radius of the cylinder the helix wraps around. If aaa<\/code> is large, the helix is wide. If aaa<\/code> is small, it’s narrow. Our intuition suggests a wider helix (larger aaa<\/code> ) should be less curved. The formula \u03ba=aa2+b2\\kappa = \\frac{a}{a^2 + b^2}\u03ba=a2+b2a\u200b<\/code> confirms this: as aaa<\/code> increases (while bbb<\/code> is fixed), the denominator a2+b2a^2+b^2a2+b2<\/code> grows faster than the numerator aaa<\/code> , causing \u03ba\\kappa\u03ba<\/code> to decrease. For example, compare a=1,b=1a=1, b=1a=1,b=1<\/code> ( \u03ba=1\/2\\kappa = 1\/2\u03ba=1\/2<\/code> ) to a=10,b=1a=10, b=1a=10,b=1<\/code> ( \u03ba=10\/101\u22480.1\\kappa = 10\/101 \\approx 0.1\u03ba=10\/101\u22480.1<\/code> ).<\/li>\n\n\n\n
  • Parameter ‘b’:<\/strong> This parameter relates to the “pitch” or vertical steepness of the helix. If bbb<\/code> is large, the helix rises quickly and is more stretched out vertically. If bbb<\/code> is small, it rises slowly and is more compressed vertically, becoming closer to a circle. Our intuition suggests a steeper, more stretched-out helix (larger bbb<\/code> ) should be less curved. The formula confirms this: as bbb<\/code> increases (while aaa<\/code> is fixed), the denominator a2+b2a^2 + b^2a2+b2<\/code> increases, causing \u03ba\\kappa\u03ba<\/code> to decrease.<\/li>\n<\/ul>\n\n\n\n
    Consider the extreme cases:<\/p>\n\n\n\n
      \n
    • If b=0b=0b=0<\/code> , the parameterization becomes r(t)=(acos\u2061t,asin\u2061t,0)\\mathbf{r}(t) = (a \\cos t, a \\sin t, 0)r(t)=(acost,asint,0)<\/code> , which is a circle of radius aaa<\/code> in the xy-plane. The curvature formula gives \u03ba=aa2+02=aa2=1a\\kappa = \\frac{a}{a^2 + 0^2} = \\frac{a}{a^2} = \\frac{1}{a}\u03ba=a2+02a\u200b=a2a\u200b=a1\u200b<\/code> . This matches the well-known curvature of a circle of radius aaa<\/code> , agreeing perfectly with intuition.<\/li>\n\n\n\n
    • If a=0a=0a=0<\/code> , the parameterization becomes r(t)=(0,0,bt)\\mathbf{r}(t) = (0, 0, bt)r(t)=(0,0,bt)<\/code> , which is a straight line along the z-axis. The curvature formula gives \u03ba=002+b2=0\\kappa = \\frac{0}{0^2 + b^2} = 0\u03ba=02+b20\u200b=0<\/code> (as long as b\u22600b \\neq 0b\ue020=0<\/code> ). This also matches the intuition that a straight line has zero curvature.<\/li>\n<\/ul>\n\n\n\n
      The formula<\/p>\n\n\n\n
              \u03ba=aa2+b2\\kappa = \\frac{a}{a^2 + b^2}\u03ba=a2+b2a\u200b<\/code>\n      <\/pre>\n\n\n\n
      therefore aligns well with the geometric intuition of how the radius and the steepness of a helix affect its bending.<\/p>\n","protected":false},"excerpt":{"rendered":"
      The standard helix has parameterization . Find the curvature of the helix. Does the result agree with your intuition? The correct answer and explanation is: Okay, let’s find the curvature of the standard helix. The standard helix has the parameterization: r(t)=(acos\u2061t,asin\u2061t,bt)\\mathbf{r}(t) = (a \\cos t, a \\sin t, bt)r(t)=(acost,asint,bt) To find the curvature \u03ba\\kappa\u03ba , […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-15743","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15743","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=15743"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15743\/revisions"}],"predecessor-version":[{"id":15744,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15743\/revisions\/15744"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=15743"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=15743"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=15743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}