# Deriving the formula for arc length of a polar function

## Homework Statement

Derive ∫(dr/dθ)^2 + R^2 )^0.5 dθ

x = Rcosθ
y = Rsinθ

## The Attempt at a Solution

Arc length is the change in rise over run, which can be found using Pythagorean's Theorem. Rise is dy/dθ while run is dx/dθ. The arc length is [(dy/dθ)^2 + (dx/dθ)^2 ]^1/2

dx/dθ = (cosθ -Rsinθ)
dy/dθ = (sinθ + Rcosθ)
dx/dθ ^2 + dy/dθ ^2 = (cos - Rsinθ)^2 + cos^2θ - 2Rsinθcosθ + R^2sin^2θ + sin^2θ + 2Rsinθcosθ + R^2cos^θ

This simplifies to [R^2(cos^2θ + sin^2θ) + sin^2 θ+cos^2θ]^2/4
Which leaves ∫√R^2 + 1 dθ
But that is not the right formula!

Mark44
Mentor

## Homework Statement

Derive ∫(dr/dθ)^2 + R^2 )^0.5 dθ

x = Rcosθ
y = Rsinθ

## The Attempt at a Solution

Arc length is the change in rise over run, which can be found using Pythagorean's Theorem.
No, not at all. The arc length is the length along the curve between two points (r1, θ1) and (r2, θ2). You can approximate this length using the chord between these two points.
Calpalned said:
Rise is dy/dθ while run is dx/dθ. The arc length is [(dy/dθ)^2 + (dx/dθ)^2 ]^1/2

dx/dθ = (cosθ -Rsinθ)
dy/dθ = (sinθ + Rcosθ)
dx/dθ ^2 + dy/dθ ^2 = (cos - Rsinθ)^2 + cos^2θ - 2Rsinθcosθ + R^2sin^2θ + sin^2θ + 2Rsinθcosθ + R^2cos^θ

This simplifies to [R^2(cos^2θ + sin^2θ) + sin^2 θ+cos^2θ]^2/4
Which leaves ∫√R^2 + 1 dθ
But that is not the right formula!

dx/dθ = (cosθ -Rsinθ)
dy/dθ = (sinθ + Rcosθ)
This is certainly not correct. Could you check your differentiation?

x = Rcosθ
y = Rsinθ

R is a constant, so by the multiplication rule of derivatives,
dx/dθ = (1)cosθ + R(-sinθ) = cosθ - Rsinθ
dy/dθ = (1)sinθ) + R(cosθ) = sinθ + Rcosθ

I still get the same differentiation.

Dick
Science Advisor
Homework Helper
x = Rcosθ
y = Rsinθ

R is a constant, so by the multiplication rule of derivatives,
dx/dθ = (1)cosθ + R(-sinθ) = cosθ - Rsinθ
dy/dθ = (1)sinθ) + R(cosθ) = sinθ + Rcosθ

I still get the same differentiation.

R is not a constant, it's a function of theta. And even if it were, the derivative wouldn't be 1.