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I want to find the relative extrema for the following function.

$f(\theta,\phi)=AR\cos\theta\sin\phi + BR\sin\theta\sin\phi + CR\cos\phi $

$A,B,C,R$ are constants

In a function $g(x,y)$ using cartesian coordinates, you can find critical points by setting the gradient equal to zero and solving for $x$ and $y$. Component wise, it would be solving the partial derivatives $g_x=0$ and $g_y=0$.

What happens with spherical coordinates? Can I do the same thing component wise ($f_\theta=0$ and $f_\phi=0$)? There's a specific form of the gradient in spherical coordinates so I'm not sure if I need to do something differently.

Pris
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  • Do we assume $R>0$? – Thomas Andrews Aug 20 '14 at 21:15
  • Yes, R is greater than zero – Pris Aug 20 '14 at 21:17
  • I think you have to use that special form of the gradient. There tends to be a misconcetion about spherical, cylindrical, polar... when you plot in one of these you tend to actually never leave the xyz coordinate system. No one really plots in the real polar system with r on one axis and $\theta $ on the other. It's because of the special transformation between these and the xyz that allows us to graph spherical the way we do. So $ f_\theta $ is the partial derivative in the REAL spherical coordinate system which is not where you plotted the function. – dylan7 Aug 20 '14 at 21:26
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    http://en.wikipedia.org/wiki/Gradient#Cylindrical_and_spherical_coordinates – rlartiga Aug 20 '14 at 21:33
  • Are you sure it is $BR\sin\theta\cos\phi$ and not $BR\sin\theta\sin\phi$? – Jack D'Aurizio Aug 20 '14 at 23:28
  • you're right, it was a typo – Pris Aug 21 '14 at 09:27

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As you said in the preamble, $f(\theta,\phi)$ is just a (periodic) function of two variables. It comes from the parametrization of a sphere, but this knowledge is just ours. So we find the stationary points of $f(\theta,\phi)$ as if we were looking for the stationary points of a smooth function $g(x,y)$. If $$f(\theta,\phi) = A\cos\theta\sin\phi + B\sin\theta\sin\phi + C\cos\phi,$$ then: $$\nabla f = (-A\sin\theta\sin\phi+B\cos\theta\sin\phi,A\cos\theta\cos\phi+B\sin\theta\cos\phi-C\sin\phi)$$ and the first component of the gradient vanishes if $\sin\phi=0$ or if $\tan\theta=\frac{B}{A}$. By imposing that the second component vanishes too, we can find the stationary points. Notice that, due to symmetry, it is way easier to find the stationary points by regarding our problem as a constrained optimization problem: $$\min_{(x,y,z)\in S^2} Ax+By+Cz,\qquad \max_{(x,y,z)\in S^2} Ax+By+Cz$$ Lagrange multipliers (or the Cauchy Schwarz inequality) give that stationary points occur when $(x,y,z)=\lambda(A,B,C)$, and since $x^2+y^2+z^2=1$ we must have $\lambda=\frac{1}{\sqrt{A^2+B^2+C^2}}$, so: $$\min = -\sqrt{A^2+B^2+C^2},\qquad \max=\sqrt{A^2+B^2+C^2}.$$

Jack D'Aurizio
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    In calculating $\nabla f$, you didn't use the spherical coordinate specific form... Is this because we don't care about the direction vectors in finding the stationary points, but just the magnitude? – Pris Aug 21 '14 at 09:32
  • Yes, we just need to find the points in which $f_\theta$ and $f_\phi$ vanish. – Jack D'Aurizio Aug 21 '14 at 20:18