+0  
 
+1
57
3
avatar+21 

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
 #2
avatar+420 
+2

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
 #3
avatar
0

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

Guest Jul 15, 2021

17 Online Users

avatar