Hi, I'm trying to get better at problem-solving in math. However, I'm having trouble with these problems. Can someone please help? I attached a picture of the problems below.

 Jul 14, 2021

I'm not entirely sure about my answers, as I'm not the best with cylindrical coordinates, but I'll try my best

First one:

The distance can be represented, by the Pythagorean theorem, as:

\(\sqrt{(p\sin(\phi)\cos(\theta))^2+(p\sin(\phi)\sin(\theta))^2+(p\cos(\phi))^2}\\ \sqrt{p^2(\sin(\phi)\cos(\theta))^2+p^2(\sin(\phi)\sin(\theta))^2+p^2(\cos(\phi))^2}\\ =p\sqrt{(\sin(\phi)\cos(\theta))^2+(\sin(\phi)\sin(\theta))^2+(\cos(\phi))^2}\)

We just need to prove that the expression under the radical is always one (also, note that \(\sin^2(x)+\cos^2(x)=1\) for any x by the Pythagorean identity):

\((\sin(\phi)\cos(\theta))^2+(\sin(\phi)\sin(\theta))^2+(\cos(\phi))^2\\ =\sin^2(\phi)\cos^2(\theta)+\sin(\phi)^2\sin^2(\theta)+\cos^2(\phi)\\ =\sin^2(\phi)(\cos^2(\theta)+\sin^2(\theta))+\cos^2(\phi)\\ =\sin^2(\phi)+\cos^2(\theta)\\ =1\)

Because of that, the expression reduces to \(p\), which is our desired answer. (obviously if p<0, the distance would be negative, which is impossible)

Second one:

(a)  \((r, \theta, -z)\), because the z-axis determines the height, and reflecting through the xy-plane is basically inverting the height.

(b) \((r, \theta\pm \pi, z)\), because adding/subtracting pi radians makes no difference as they both reflect \((r, \theta)\) across the origin.

(c) I think it would be a combination of the previous two, namely \((r, \theta \pm \pi, -z)\)

also, I'm kinda confused on why my post is shown as being the second post, not the first.

 Jul 15, 2021
edited by textot  Jul 15, 2021

maybe because someone posted something before you but they deleted it?

Guest Jul 15, 2021

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