---
title: "Chord of curvature through pole of cardioid"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:curvature-chord-through-pole-cardioid"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-chord-through-pole-cardioid
site: BCT Engineering
---

# Chord of curvature through pole of cardioid

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2073 Chaitra
- Video: [B.Sc1:CURVATURE: CHORD OF CURVATURE THROUGH THE POLE OF CARDIOD r=a(1+cosθ ),Differential Calculas](https://www.youtube.com/watch?v=acO4_jFaE-g) — vikas saxena maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-chord-through-pole-cardioid
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Find the chord of curvature through the pole for the curve r = a(1+cosθ).

## Short answer

φ = π/2 + θ/2, sin φ = cos(θ/2); ρ = (4a/3) cos(θ/2). Chord through pole = 2ρ sin φ = (8a/3) cos²(θ/2) = 4r/3.

## Answer

The chord of curvature through the pole is 2ρ sin φ, where φ is the angle between the radius vector and the tangent.

Cardioid: r = a(1 + cos θ) = 2a cos²(θ/2).

**Angle φ:** dr/dθ = −a sin θ, so

tan φ = r dθ/dr = a(1 + cos θ)/(−a sin θ) = −cot(θ/2) = tan(π/2 + θ/2)

so φ = π/2 + θ/2 and sin φ = cos(θ/2).

**Radius of curvature:** ρ = (r² + r₁²)^(3/2) / (r² + 2r₁² − r r₂), with r₁ = −a sin θ, r₂ = −a cos θ:

```text
r² + r₁²          = a²(2 + 2cos θ) = 2a r
r² + 2r₁² − r r₂  = a²[(1+cosθ)² + 2sin²θ + cosθ(1+cosθ)]
                  = 3a²(1 + cos θ) = 3a r
ρ = (2ar)^(3/2) / (3ar) = (2/3)√(2ar)
```
Since √(2ar) = 2a cos(θ/2), ρ = (4a/3) cos(θ/2).

**Chord of curvature through the pole:**

2ρ sin φ = 2 · (4a/3) cos(θ/2) · cos(θ/2) = (8a/3) cos²(θ/2)

Since r = 2a cos²(θ/2), cos²(θ/2) = r/2a, so the chord = (8a/3)(r/2a).

**Answer: chord of curvature through the pole = 4r/3 = (4a/3)(1 + cos θ).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The chord of curvature through the pole is
   A. 2 rho sin(phi)
   B. 2 rho cos(phi)
   C. rho sin(phi)
   D. rho / (2 sin(phi))
   Answer: A. The chord through the pole is 2 rho sin(phi), phi being the angle between radius vector and tangent. (Key point: Chord formula)

2. For r = a(1+cos θ), the angle phi satisfies
   A. phi = θ/2, so sin(phi) = sin(θ/2)
   B. phi = π/2 + θ/2, so sin(phi) = cos(θ/2)
   C. phi = θ, so sin(phi) = sin θ
   D. phi = π/2 - θ/2, so sin(phi) = sin(θ/2)
   Answer: B. tan(phi) = r dθ/dr = -cot(θ/2) = tan(π/2 + θ/2). (Key point: Angle phi)

3. The radius of curvature of r = a(1+cos θ) is
   A. (4a/3) sin(θ/2)
   B. (2a/3) cos(θ/2)
   C. (2/3) sqrt(2ar) = (4a/3) cos(θ/2)
   D. (8a/3) cos(θ/2)
   Answer: C. With r^2 + r1^2 = 2ar and denominator 3ar, rho = (2ar)^(3/2)/(3ar). (Key point: Radius of curvature)

4. The chord of curvature through the pole is
   A. 2r/3
   B. 8r/3
   C. r/3
   D. 4r/3
   Answer: D. 2 rho sin(phi) = (8a/3) cos^2(θ/2) = (8a/3)(r/2a) = 4r/3. (Key point: Final result)

## Related in Mathematics I

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