Can't wrap my head around Cartesian to Polar?

Question

Hello! I'd preface this by saying that I understand the concepts of cartesian and polar coordinates, but I am experiencing a major brain fart with regards to converting cartesian velocities to polar in relation to velocity fields.

I have attached an image from a video I was watching. What my brain is refusing to understand is how this can be the case.
Surely the i component of r is r*cos(theta). But here the video says that the i component multiplied by cos(theta) is the i component of x.

Something is just not clicking in my brain, and I'm sure it's completely obvious but nevertheless I am very confused.

For context, I am trying to derrive the polar forms of potential functions for a fluid velocity field. But this relies on me understanding that Vr = u*cos(theta), and not the other way around.

Can anyone maths my brain into order? Thanks.

? · Answer

in vector form cartesian  velocities as a function of time are  (f'(t),g'(t)) by standard formula

x=f(t) = r*cos α
y = g(t) = r*sin α

with r and α as functions of t

x = r(t)*cos α(t)
y = r(t)*sin α(t)

dx/dt = r'(t)*cos α(t) - r(t)*α'(t)*sin α(t)
dy/dt = r'(t) sin α(t) + r(t)*a'(t)*cos α(t)

As a check consider a particle descending at from (0,1) at a constant velocity of 1 unit down and 1 unit right per second

(t, 1-t)

velocity vector
(1,-1)

r = sqrt(t^2+ (1- t)^2)
α(t) = arctan((1-t)/t)

r'(t)= (2t-1)/sqrt(t^2 + (1-t)^2)
a'(t)= -1/t^2 /(1+ ((1-t)/t)^2)

for t=1/2

sin is  1/sqrt(2) = sqrt(2)/2
cos is 1/sqrt(2) = sqrt(2)/2
r(t) is  sqrt(1/2)
r'(t) is 0
α'(t) is -2

dx/dt = sqrt(1/2) *-( -2*sqrt(2)/2) =sqrt(1) =1
dy/dt = sqrt(1/2) * -2*sqrt(2)/2 = -sqrt(1) = -1

or (cos theta(t),sin theta(t)) for constant angular motion

velocity (-sin theta(t) ,cos theta(t))
 
r = 1
theta = t
theta (t) = t
theta'(t) =1

for t = pi/4

dx/dt = -1*1*sin theta = -1/sqrt(2)
dy/dt = 1*1*cos theta = 1/sqrt(2)

Trending News

Can't wrap my head around Cartesian to Polar?

1 Answer