Radius of curvature of r^m = a^m cos mθ

Short answer

From the pedal equation p = r^(m+1)/a^m we get dp/dr = (m+1)r^m/a^m. Hence radius of curvature ρ = r dr/dp = a^m / ((m+1) r^(m-1)).

Show that the radius of curvature at any point (r,θ) of the curve r^m = a^m cos mθ is a^m/((m+1) r^(m−1)).

Use the pedal form ρ = r dr/dp.

Pedal equation. From r^m = a^m cos mθ: (1/r)(dr/dθ) = −tan mθ, so tan φ = r dθ/dr = −cot mθ = tan(π/2 + mθ), giving φ = π/2 + mθ.

p = r sin φ = r cos mθ = r · r^m / a^m = r^(m+1)/a^m.

Differentiate with respect to r:

dp/dr = (m+1) r^m / a^m

Radius of curvature:

ρ = r (dr/dp) = r · a^m / ((m+1) r^m)

ρ = a^m / ((m+1) r^(m−1)). Hence shown.

(Special cases: m = 1 gives the circle r = a cos θ with ρ = a/2; m = 2 gives the lemniscate with ρ = a²/(3r).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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