# Trying to use polar coordinates to find the distance between two points

• SamRoss
In your case, A = \frac{3}{\sqrt{2}}.The derivative of A\sec\theta is A\sec\theta \tan\theta. In your case, A = \frac{3}{\sqrt{2}}.
SamRoss
Gold Member
Homework Statement
Trying to get a better understanding of invariant distance. As an example, I want to find the distance between (3,0) and (0,3) (in Cartesian coordinates) but using polar coordinates.
Relevant Equations
##{ds}^2={dx}^2+{dy}^2={dr}^2+r^2{d\theta}^2##
##{dx}^2+{dy}^2=3^2+3^2=18##

##{dr}^2+r^2{d\theta}^2=0^2+3^2*(\theta/2)^2\neq18##

I have a feeling that what I'm doing wrong is just plugging numbers into the polar coordinate formula instead of treating it as a curve. For example, I naively plugged in 3 for r even though I know the radius would change along the line from (3,0) to (0,3) (Cartesian coordinates). How can this be done correctly? Also, I am not looking to use the ##{r_1}^2+{r_2}^2-2{r_1}{r_2}cos(\theta_1-\theta_2)## formula which I know would solve the problem easily.

The drdr and dθdθ are differentials, not coordinate differences. The expression dsds is the distance between two infinitesimally separated points. You can integrate it along any curve to get its length - not just the straight line. In your case you (inadvertently) computed the length of a quarter circle

SamRoss said:
Problem Statement: Trying to get a better understanding of invariant distance.
What is "invariant distance"? Is it a standard name for a mathematical topic?

How can this be done correctly?
My guess is that your are trying to find the arc length of a curve that is the straight line between (3.0) and (0,3). In that case a relevant equation is
##L = \int_a^b \sqrt{ r^2 + (\frac{dr}{d\theta})^2 } d\theta ## and you must express ##r## as function of ##\theta## so that the path from ##\theta = a## to ##\theta = b## goes along the line from (3,0) to (0,3).

Stephen Tashi said:
##L = \int_a^b \sqrt{ r^2 + (\frac{dr}{d\theta})^2 } d\theta ## and you must express ##r## as function of ##\theta## so that the path from ##\theta = a## to ##\theta = b## goes along the line from (3,0) to (0,3).

I wrote the equation of the line in Cartesian coordinates: y=-x+3. Plugging in ##x=rcos\theta## and ##y=rsin\theta## and rearranging I got ##r=3/(sin\theta+cos\theta)##. Then ##\frac {dr} {d\theta}=-\frac {3(cos\theta-sin\theta)} {(sin\theta+cos\theta)^2}##. The integral should go from 0 to pi/2. Is this correct so far?

SamRoss said:
I wrote the equation of the line in Cartesian coordinates: y=-x+3. Plugging in ##x=rcos\theta## and ##y=rsin\theta## and rearranging I got ##r=3/(sin\theta+cos\theta)##. Then ##\frac {dr} {d\theta}=-\frac {3(cos\theta-sin\theta)} {(sin\theta+cos\theta)^2}##. The integral should go from 0 to pi/2. Is this correct so far?

You can simpilfy that to $$r = \frac{3}{\sqrt{2}} \sec(\theta - \tfrac{\pi}4)$$ and use $$\frac{d}{d\theta} \sec \theta = \sec \theta \tan \theta, \qquad \frac{d}{d\theta} \tan \theta = \sec^2 \theta$$

pasmith said:
You can simpilfy that to $$r = \frac{3}{\sqrt{2}} \sec(\theta - \tfrac{\pi}4)$$ and use $$\frac{d}{d\theta} \sec \theta = \sec \theta \tan \theta, \qquad \frac{d}{d\theta} \tan \theta = \sec^2 \theta$$

I'm trying that in Wolfram (this is just for fun and not actual homework anyway) but I seem to be doing something wrong. Where is my mistake?

Orodruin said:
In your case you (inadvertently) computed the length of a quarter circle

That makes sense. Thank you.

SamRoss said:
I'm trying that in Wolfram (this is just for fun and not actual homework anyway) but I seem to be doing something wrong. Where is my mistake?

View attachment 245597

The derivative of $A\sec\theta$ is $A\sec\theta \tan\theta$.

## What are polar coordinates?

Polar coordinates are a system of representing points in a two-dimensional plane using a distance from the origin and an angle from a reference axis.

## How do you convert from Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the formulas r = √(x^2 + y^2) and θ = tan^-1(y/x).

## Can you use polar coordinates to find the distance between two points?

Yes, you can use the distance formula in polar coordinates, which is given by d = √(r1^2 + r2^2 - 2r1r2cos(θ1-θ2)).

## What is the reference axis in polar coordinates?

The reference axis in polar coordinates is the horizontal axis, also known as the polar axis, with an angle of 0 degrees.

## Can you use polar coordinates to find the distance between two points in three-dimensional space?

No, polar coordinates are only used to represent points in a two-dimensional plane. To find the distance between two points in three-dimensional space, you would need to use spherical coordinates.

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