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Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by

$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$

$t \in (0, \frac{\pi}{2})$

a) Compute the arc length of $ \gamma (t). $

b) Show that the curve cuts each "parallel" in a constant angle, by computing $(\gamma (t))'$ and $\psi_{\phi}$

c) Make a diagram of the curves

So the formula for arc length would be $$ \begin{equation} \ell_\gamma = \int_0^1 \sqrt{ \left ( \frac{dr}{dt} \right )^2 + r^2\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 + r^2 \left ( \frac{d \theta}{dt} \right )^2} \, dt. \end{equation}$$

In my case, as $$\phi'(t) = \frac 1 {2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right) \sin\left(\frac{\pi}{4} - \frac{t}{2}\right)}, $$ and $$ \theta ' (t) = -1 ,$$ then

\begin{align*} \ell_\gamma &= \int_0^1 \sqrt{ \sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 + \left ( \frac{d \theta}{dt} \right )^2} \, dt \\ &= \int_0^1 \sqrt{ \sin^2 \theta \left (\frac 1 {2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 + 1} \, dt\\ &= \int_0^1 \sqrt{ \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 + 1} \, dt. \end{align*}

I am stuck here. Before I go further, would this integral be correct? That $\sin^2(\theta)$ confuses me maybe, I don't know if I should change the argument to something else.

For exercise b) Could someone verify this reasoning?

I want the angle between two curves on the sphere, which is defined as the angle between the tangent vectors of this curves at the point of intersection. So I want to find $\alpha$ such that $\alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) $, because that is what the angle between two vectors in a vector space is. So if $\beta (t) = (t, \theta_0)$ is a curve that describes the parallel curve in a sphere, with $\theta_0$ constant, Its tangents are given by $(1, 0),$. I already have the derivatives of $\gamma$ so $$ \alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) = cos^{-1}(\frac{<(1, 0) \dot (\frac{1}{sin{\frac{\pi}{2}- t}}, -1)>}{(1) \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$ $$ = cos^{-1}(\frac{1}{sin{\frac{\pi}{2}- t}}{ \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$.

Thanks in advance.

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  • $\begingroup$ $\theta'(t)=-1$ And $\sin\theta=\sin(\pi/2-t)$ $\endgroup$ – Rafa Budría May 25 '17 at 19:55
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    $\begingroup$ For b), Choose a new path $\beta(t)$: $\theta=\theta_0$ constant and $\phi=t$ and compute the dot product of its tangent with the tangent of $\gamma$ $\endgroup$ – Rafa Budría May 25 '17 at 20:02
  • $\begingroup$ The double-angle formula for the sine says $2\sin\alpha\cos\alpha = \sin(2\alpha).$ An instance of that says $ 2\cos\left( \frac\pi4 - \frac t 2\right) \sin\left(\frac\pi4 - \frac t2\right) = \sin\left( \frac\pi2 - t\right). $ And then recall that $\sin\left( \frac\pi2 - t\right) = \cos t. \qquad$ $\endgroup$ – Michael Hardy May 25 '17 at 20:58
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    $\begingroup$ These are rhumb lines or Loxodromes of constant $\alpha$. Cylindrical coordinates derivation is much simpler. $ r = a \,sech( \theta \cot \alpha )\,, z = a \,tanh( \theta \cot \alpha )\,$ $\endgroup$ – Narasimham May 25 '17 at 21:03
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    $\begingroup$ @MichaelHardy sorry, I just rechecked this, yes, by usual spherical coordinates I meant $0$ minor or equal than $\theta$ & minor or equal than $2\pi$ , $\phi$ starts in $- \pi/2$ from north pole, ends in the south pole at $\pi/2$, which is what you suggested $\endgroup$ – Trux May 26 '17 at 1:02
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\begin{align} & \int_0^1 \sqrt{ \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt] = {} & \int_0^1 \sqrt{\left(\frac{\sin\theta}{\cos t} \right)^2+1} \, dt = \int_0^1 \sqrt{ \left( \frac{\cos t}{\cos t} \right)^2 + 1} \, dt = \int_0^1 \sqrt 2 \, dt = \sqrt 2. \end{align} The first two equalities are trigonometric identities: \begin{align} & 2\cos\left( \frac \pi 4 - \frac t 2 \right) \sin \left( \frac \pi 4 - \frac t 2 \right) = \sin\left( \frac \pi 2 - t \right) = \cos t \\[10pt] & \end{align}

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  • $\begingroup$ thanks! I got the same result but did something a little different, which I will answer in the post just for fun $\endgroup$ – Trux May 26 '17 at 1:06
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\begin{align} & \int_0^1 \sqrt{ \left(\frac{\sin (\frac{\pi}{2} - t) }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt] \end{align}

As $$\sin^2(\frac \pi 4 - \frac t 2)cos^2(\frac \pi 4 - \frac t 2) = \frac{1 - \cos(\pi - 2t)}{8},$$ then

\begin{align} & \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -\cos(2(\frac{\pi}{2} - t ))}\right) + 1} \, dt & = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -[\cos^2(\frac{\pi}{2} - t )- \sin^2(\frac{\pi}{2} - t ))]}\right) + 1} \, dt \end{align}

\begin{align} = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{2\sin^2(\frac{\pi}{2} - t)}\right) + 1} \, dt = \sqrt2 \\[10pt] \end{align}

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